C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetTNG HP 63 THI TUYN SINH LP 10 TRONG TON QUCNM HC 2012 2013MN TONS GIO DC V O TO K THI TUYN SINH LP 10 THPTH NINm hoc: 2012 2013Mn thi: TonNgy thi: 21 thng 6 nm 2012Thi gian lm bi: 120 phtBi I (2,5 im)1) Cho biu thc x 4Ax 2++. Tnh gi tr ca A khi x = 362) Rt gn biu thc x 4 x 16B :x 4 x 4 x 2 _+ + + + ,(vi x 0; x 16 )3) Vi cc ca biu thc A v B ni trn, hy tm cc gi tr ca x nguyn gi tr ca biu thc B(A 1) l s nguynBi II (2,0 im). Gii bi ton sau bng cch lp phng trnh hoc h phng trnh:Hai ngi cng lm chung mt cng vic trong 125 gi th xong. Nu mi ngi lm mt mnh th ngi th nht hon thnh cng vic trong t hn ngi th hai l 2 gi. Hi nu lm mt mnh th mi ngi phi lm trong bao nhiu thi gian xong cng vic?Bi III (1,5 im)1) Gii h phng trnh: 2 12x y6 21x y+ ' 2) Cho phng trnh: x2 (4m 1)x + 3m2 2m = 0 (n x). Tm m phng trnh c hai nghim phn bit x1, x2 tha mn iu kin : 2 21 2x x 7 + Bi IV (3,5 im)Cho ng trn (O; R) c ng knh AB. Bn knh CO vung gc vi AB, M l mt im bt k trn cung nh AC (M khc A, C); BM ct AC ti H. Gi K l hnh chiu ca H trn AB.1) Chng minh CBKH l t gic ni tip.2) Chng minh ACM ACK 3) Trn an thng BM ly im E sao cho BE = AM. Chng minh tam gic ECM l tam gic vung cn ti C4) Gi d l tip tuyn ca (O) ti im A; cho P l im nm trn d sao cho hai im P, C nm trong cng mt na mt phng b AB v AP.MBRMA . Chng minh ng thng PB i qua trung im ca on thng HKBi V (0,5 im). Vi x, y l cc s dng tha mn iu kin x 2y , tm gi tr nh nht ca biu thc: 2 2x yMxy+1 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetGI P NBai I: (2,5 im)1) Vi x = 36, ta c :A = 36 4 10 58 4 36 2+ +2) Vi x, x 16 ta c :B = x( x 4) 4( x 4) x 2x 16 x 16 x 16 _ + ++ + , = (x 16)( x 2) x 2(x 16)(x 16) x 16+ + + + 3)Ta c:2 4 2 2 2( 1) . 1 .16 16 162 2x x xB Ax x xx x _+ + + + + ,. ( 1) B A nguyn, x nguynth16 xl c ca 2, m (2) = } {1; 2 t tTa c bng gi tr tng ng:16 x 1 1 2 2 x 17 15 18 14Kt hp K 0, 16 x x , ( 1) B A nguyn th} {14;15;17;18 xBai II: (2,0 im)Gi thi gian ngi th nht hon thnh mt mnh xong cng vic l x (gi), K 125x>Th thi gian ngi th hai lm mt mnh xong cng vic l x + 2 (gi)Mi gi ngi th nht lm c1x(cv), ngi thhai lm c12 x+(cv)V c hai ngi cng lm xong cng vic trong 125gi nn mi gi c hai i lm c121:5=512(cv)Do ta c phng trnh1 1 5x x 2 12+ +2 5( 2) 12x xx x+ + + 5x2 14x 24 = 0 = 49 + 120 = 169, ,13 => 7 13 65 5x (loi) v + 7 13 2045 5x (TMK)Vy ngi th nht lm xong cng vic trong 4 gi,ngi th hai lm xong cng vic trong 4+2 = 6 gi.Bai III: (1,5 im) 1)Gii h: 2 126 21x yx y+ ' , (K: , 0 x y).H 4 24 6 104 2 4 1 522 12 1 2 1 2 6 2 12 21 2xx x yx x xyyx y x yx y + + + ' ' ' ' '+ + + .(TMK)2C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetVy h c nghim (x;y)=(2;1).2) + Phng trnh cho c = (4m 1)2 12m2 + 8m = 4m2 + 1 > 0,mVy phng trnh c 2 nghim phn bit m+ Theo L Vi t, ta c: 1 221 24 13 2x x mxx m m+ ' .Khi : 2 2 21 2 1 2 1 27 ( ) 2 7 x x x x xx + + (4m 1)2 2(3m2 2m) = 7 10m2 4m 6 = 0 5m2 2m 3 = 0Ta thy tng cc h s: a + b + c = 0=>m = 1 hay m = 35.Tr li: Vy....Bai IV: (3,5 im)1) Ta c 090 HCB ( do chn na ng trn k AB)090 HKB (do K l hnh chiu ca H trn AB)=> 0180 HCB HKB + nn t gic CBKH ni tip trong ng trn ng knh HB.2) Ta c ACM ABM (do cng chn AM ca (O))v ACK HCK HBK (v cng chnHK.ca trn k HB)Vy ACM ACK 3) V OC AB nn C l im chnh gia ca cung AB AC = BC v 090 sd AC sd BC Xt 2 tam gic MAC v EBC cMA= EB(gt), AC = CB(cmt) v MAC = MBC v cng chn cung MC ca (O)MAC v EBC (cgc) CM = CE tam gic MCE cn ti C (1)Ta li c045 CMB (v chn cung 090 CB ). 045 CEM CMB (tnh cht tam gic MCE cn ti C)M 0180 CME CEM MCE + + (Tnh cht tng ba gc trong tam gic)090 MCE (2)T (1), (2) tam gic MCE l tam gic vung cn ti C (pcm).3A B C M

H K O E C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net4) Gi S l giao im ca BM v ng thng (d), N l giao im ca BP vi HK.Xt PAM v OBM :Theo gi thit ta c. AP MB AP OBRMA MA MB (v c R = OB).Mt khc ta c PAM ABM (v cng chn cung AMca (O)) PAM OBM 1AP OBPA PMPM OM.(do OB = OM = R)(3)V 090 AMB(do chn na trn(O)) 090 AMStam gic AMS vung ti M. + 090 PAMPSMv + 090 PMAPMS PMS PSM PS PM(4)M PM = PA(cmt) nn PAM PMAT (3) v (4) PA = PS hay P l trung im ca AS.V HK//AS (cng vung gc AB) nn theo L Ta-lt, ta c: NK BN HNPA BP PShayNK HNPA PSm PA = PS(cmt) NK NHhay BP i qua trung im N ca HK. (pcm)Bai V: (0,5 im)Cch 1(khng s dng BT C Si)Ta c M = 2 2 2 2 2 2 2( 4 4 ) 4 3 ( 2 ) 4 3 x y x xy y xy y x y xy yxy xy xy+ + + + = 2( 2 ) 34x y yxy x+ V (x 2y)2 0, du = xy ra x = 2yx 2y 1 3 32 2y yx x , du = xy ra x = 2yT ta c M 0 + 4 -32=52, du = xy ra x = 2yVy GTNN ca M l 52, t c khi x = 2y4A B C M

H K O S P E NC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCch 2:Ta c M = 2 2 2 23( )4 4x y x y x y x y xxy xy xy y x y x y+ + + + +V x, y > 0 , p dng bdt C si cho 2 s dng ;4x yy x ta c2 . 14 4x y x yy x y x+ ,du = xy ra x = 2yVx 2y 3 6 32 .4 4 2x xy y , du = xy ra x = 2yT ta c M 1 +32=52,du = xy ra x = 2yVy GTNN ca M l 52, t c khi x = 2yCch 3:Ta c M = 2 2 2 24 3( )x y x y x y x y yxy xy xy y x y x x+ + + + V x, y > 0 , p dng bdt C si cho 2 s dng 4;x yy x ta c 4 42 . 4x y x yy x y x+ ,du = xy ra x = 2yVx 2y 1 3 32 2y yx x , du = xy ra x = 2yT ta c M 4-32=52,du = xy ra x = 2yVy GTNN ca M l 52, t c khi x = 2yCch 4:Ta c M = 2 2 2 2 22 2 2 22 2 24 33 34 4 4 4 44 4x x x x xy y y yx y x xxy xy xy xy xy xy y+ + + + ++ + +V x, y > 0 , p dng bdt Co si cho 2 s dng 22;4xyta c 2 22 22 .4 4x xy y xy + ,du = xy ra x = 2yVx 2y 3 6 32 .4 4 2x xy y , du = xy ra x = 2yT ta c M xyxy +32= 1+32=52,du = xy ra x = 2yVy GTNN ca M l 52, t c khi x = 2y5C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TO K THI TUYN SINH LP 10 THPT TP.HCMNm hoc: 2012 2013MN: TONThi gian lm bi: 120 phtBai 1: (2 im)Gii cc phng trnh v h phng trnh sau:a)22 3 0 x xb) 2 3 73 2 4 '+ x yx yc) 4 212 0 + x xd) 22 2 7 0 x xBai 2: (1,5 im)a) V th (P) ca hm s 214 y xv ng thng (D): 122 + y xtrn cng mt h trc to .b) Tm to cc giao im ca (P) v (D) cu trn bng php tnh.Bai 3: (1,5 im)Thu gn cc biu thc sau:1 2 11 + + xAx x x x xvi x > 0;1 x(2 3) 26 15 3 (2 3) 26 15 3 + + BBai 4: (1,5 im)Cho phng trnh 22 2 0 + x mx m(x l n s)a) Chng minh rng phng trnh lun lun c 2 nghim phn bit vi mi m.b) Gi x1, x2 l cc nghim ca phng trnh.Tm m biu thc M = 2 21 2 1 2246+ x x x x t gi tr nh nhtBai 5: (3,5 im)Cho ng trn (O) c tm O v im M nm ngoi ng trn (O). ng thng MO ct (O) ti E v F (ME 0;1 x(2 3) 26 15 3 (2 3) 26 15 3 + + B7M E F K SA B T P Q C H O V C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net1 1(2 3) 52 30 3 (2 3) 52 30 32 2 + + 2 21 1(2 3) (3 3 5) (2 3) (3 3 5)2 2 + + 1 1(2 3)(3 3 5) (2 3)(3 3 5) 22 2 + + Cu 4:a/ Phng trnh (1) c = m2 - 4m +8 = (m - 2)2 +4 > 0 vi mi m nn phng trnh (1) c 2 nghim phn bit vi mi m.b/ Do , theo Viet, vi mi m, ta c: S =2bma ; P =2 cmaM = 21 2 1 224( ) 8+ x x x x= 2 224 64 8 16 2 4 + + m m m m26( 1) 3 + m. Khi m = 1 ta c 2( 1) 3 + m nh nht26( 1) 3 +Mm ln nht khi m = 126( 1) 3 +Mm nh nht khi m = 1Vy M t gi tr nh nht l- 2 khi m = 1Cu 5a) V ta c do hai tam gic ng dng MAE v MBFNn MA MFME MB MA.MB = ME.MF(Phng tch ca M i vi ng trn tm O)b) Do h thc lng trong ng trn ta cMA.MB = MC2, mt khc h thc lngtrong tam gic vung MCO ta cMH.MO = MC2 MA.MB = MH.MOnn t gic AHOB ni tip trong ng trn.c) Xt t gic MKSC ni tip trong ngtrn ng knh MS (c hai gc K v C vung).Vy ta c :MK2 = ME.MF = MC2nn MK = MC.Do MF chnh l ng trung trc ca KCnn MS vung gc vi KC ti V.d) Do h thc lng trong ng trn ta c MA.MB = MV.MS ca ng trn tm Q.Tng t vi ng trn tm P ta cng c MV.MS = ME.MF nn PQ vung gc vi MS v l ng trung trc ca VS (ng ni hai tm ca hai ng trn). Nn PQ cng i qua trung im ca KS (do nh l trung bnh ca tam gic SKV). Vy 3 im T, Q, P thng hng.8C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TO K THI TUYN SINH LP 10 THPT TP. NNGNm hoc: 2012 2013MN: TONThi gian lm bi: 120 phtBi 1: (2,0 im)1) Gii phng trnh:(x + 1)(x + 2) = 02) Gii h phng trnh: 2 12 7+ ' x yx yBi 2: (1,0 im)Rt gn biu thc ( 10 2) 3 5 + ABi 3: (1,5 im)Bit rng ng cong trong hnh v bn l mt parabol y = ax2.1) Tm h s a.2) Gi M v N l cc giao im ca ng thngy = x + 4 vi parabol. Tm ta ca cc im M v N.Bi 4: (2,0 im)Cho phng trnh x2 2x 3m2 = 0, vi m l tham s.1) Gii phng trnh khi m = 1.2) Tm tt c cc gi tr ca m phng trnh c hai nghim x1, x2 khc 0 v tha iu kin 1 22 183 x xx x.Bi 5: (3,5 im)Cho hai ng trn (O) v (O) tip xc ngoi ti A. K tip tuyn chung ngoi BC,B (O),C(O). ng thng BO ct (O) ti im th hai l D.1) Ch`ng minh rng t gic COOB l mt hnh thang vung.2) Chng minh rng ba im A, C, D thng hng.3) T D k tip tuyn DE vi ng trn (O) (E l tip im). Chng minh rng DB = DE.BI GIIBi 1:1)(x + 1)(x + 2) = 0 x + 1 = 0 hay x + 2 = 0 x = -1 hay x = -22) 2 1 (1)2 7 (2)+ ' x yx y 5y 15 ((1) 2(2))x 7 2y ' + y 3x 1 ' Bi 2: ( 10 2) 3 5 + A = ( 5 1) 6 2 5 + =2( 5 1) ( 5 1) + =( 5 1)( 5 1) += 4Bi 3:1)Theo th ta c y(2) = 2 2 = a.22 a = 2) Phng trnh honh giao im ca y = 212xv ng thng y = x + 4 l :x + 4 = 212x x2 2x 8 = 0 x = -2 hay x = 4y(-2) = 2 ; y(4) = 8. Vy ta cc im M v N l (-2 ; 2) v (4 ; 8).Bi 4:1) Khi m = 1, phng trnh thnh : x2 2x 3 = 0 x = -1 hay x = 3 (c dng ab + c = 0)901 22y=ax2yx CHNH THCBCEDAOOC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net2) Vi x1, x2 0, ta c : 1 22 183 x xx x 2 21 2 1 23( ) 8 x x x x 3(x1 + x2)(x1 x2) = 8x1x2Ta c : a.c = -3m2 0 nn 0, mKhi 0 ta c : x1 + x2 =2 ba v x1.x2 = 23 cma 0iu kin phng trnh c 2 nghim 0 m m 0 > 0 v x1.x2 < 0 x1 < x2Vi a = 1 x1 = ' ' b v x2 = ' ' + b x1 x2 = 22 ' 2 1 3 + mDo , ycbt 2 23(2)( 2 1 3 ) 8( 3 ) + m m v m 0 2 21 3 2 + m m(hin nhin m = 0 khng l nghim) 4m4 3m2 1 = 0 m2 = 1 hay m2 = -1/4 (loi) m = t1Bi 5:1) Theo tnh cht ca tip tuyn ta c OB, OC vung gc vi BC t gic COOB l hnh thang vung.2) Ta c gc ABC = gc BDC gc ABC + gc BCA = 900 gc BAC = 900Mt khc, ta c gc BAD = 900 (ni tip na ng trn)Vy ta c gc DAC = 1800 nn 3 im D, A, C thng hng.3) Theo h thc lng trong tam gic vung DBC ta c DB2 = DA.DCMt khc, theo h thc lng trong ng trn (chng minh bng tam gic ng dng) ta c DE2 = DA.DC DB = DE.10C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GD&TVNH PHCK THI TUYN SINH LP 10 THPT NM HC 2012-2013 THI MN : TONThi gian lm bi 120 pht (khng k thi gian giao )Ngy thi: 21 thng 6 nm 2012Cu 1 (2,0 im). Cho biu thc :P=23 6 41 1 1x xx x x+ + 1. Tm iu kin xc nh ca biu thc P.2. Rt gn PCu 2 (2,0 im). Cho h phng trnh :2 4ax 3 5x ayy+ ' 1. Gii h phng trnh vi a=12. Tm a h phng trnh c nghim duy nht.Cu 3 (2,0 im). Mt hnh ch nht c chiu rng bng mt na chiu di. Bit rng nu gim mi chiu i 2m th din tch hnh ch nht cho gim i mt na. Tnh chiu di hnh ch nht cho.Cu 4 (3,0 im). Cho ng trn (O;R) (im O c nh, gi tr R khng i) v im M nm bn ngoi (O). K hai tip tuyn MB, MC (B,C l cc tip im ) ca (O) v tia Mxnm gia hai tia MO v MC. Qua B k ng thng song song vi Mx, ng thng ny ct (O) ti imth hai l A. V ng knh BB ca (O). Qua O k ng thng vung gc vi BB,ng thng ny ct MC v BC ln lt ti K v E. Chng minh rng:1. 4 im M,B,O,C cng nm trn mt ng trn.2. on thng ME = R.3. Khi im M di ng m OM = 2R th im K di ng trn mt ng trn c nh, ch r tm v bn knh ca ng trn .Cu 5 (1,0 im). Cho a,b,c l cc s dng tha mn a+ b + c =4. Chng minh rng :3 3 3 4 4 42 2 a b c + + >S GD&T VNH PHC K THI TUYN SINH LP 10 THPT NM HC 2012-2013P N THI MN : TONNgy thi: 21 thng 6 nm 2012Cu p n, gi imC1.1(0,75 im)Biu thc P xc nh ' + 0 10 10 12xxx0,50,2511 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net' 11xxC1.2 (1,25 im)P=) 1 )( 1 () 4 6 ( ) 1 ( 3 ) 1 () 1 )( 1 (4 6131 + + + +++ x xx x x xx xxx xx) 1 (11) 1 )( 1 () 1 () 1 )( 1 (1 2) 1 )( 1 (4 6 3 322 2t + + ++ ++ + +x voixxx xxx xx xx xx x x x0,250,50,5C2.1 (1,0 im)Via = 1, h phng trnh c dng: ' +5 34 2y xy x' ' ' ' +215 3 115 37 75 312 3 6yxyxy xxy xy xVy vi a = 1, h phng trnh c nghim duy nht l: ' 21yx0,250,250,250,25C2.2 (1,0 im)-Nu a = 0, h c dng: ' ' 3525 34 2yxyx=> c nghim duy nht-Nu a0 , h c nghim duy nht khi v ch khi: 32aa62 a(lun ng, v02 avi mi a)Do , vi a0 , h lun c nghim duy nht.Vy h phng trnh cho c nghim duy nht vi mi a.0,250,250,250,2512C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetC3 (2,0 im)Gi chiu di ca hnh ch nht cho l x (m), vix > 4.V chiu rng bng na chiu di nn chiu rng l: 2x (m)=> din tch hnh ch nht cho l: 2 2.2x xx (m2)Nu gim mi chiu i 2 m th chiu di, chiu rng ca hnh ch nht ln lt l: 222 xva x(m)khi , din tch hnh ch nht gim i mt na nn ta c phng trnh: 2 21) 22)( 2 (2x xx 0 16 1244 2222 2 + + x xxx xx.=>5 2 61+ x(tho mn x>4);5 2 62 x (loi v khng tho mn x>4)Vy chiu di ca hnh ch nht cho l 5 2 6 +(m).0,250,250,250,250,250,50,25C4.1 (1,0 im)1) Chng minh M, B, O, C cng thuc 1 ng trnTa c: 090 MOB (v MB l tip tuyn)090 MCO (v MC l tip tuyn)=>MBO +MCO == 900 + 900 = 1800=> T gic MBOC ni tip(v c tng 2 gc i =1800)=>4 im M, B, O, C cng thuc 1 ng trn0,250,250,250,25C4.2 (1,0 im)2) Chng minh ME = R:Ta c MB//EO (v cng vung gc vi BB)=>O1 =M1(so le trong)MM1 =M2 (tnh cht 2 tip tuyn ct nhau) =>M2 =O1 (1)C/m c MO//EB (v cng vung gc vi BC)=>O1 =E1 (so le trong)(2)T (1), (2) =>M2 =E1 => MOCE ni tip=>MEO =MCO = 900=>MEO =MBO =BOE = 900 => MBOE l hnh ch nht=> ME = OB = R (iu phi chng minh)0,250,250,250,25C4.3 (1,0 im)3) Chng minh khi OM=2R th K di ng trn 1 ng trn c nh:Chng minh c Tam gic MBC u =>BMC = 600=>BOC = 1200=>KOC = 600 -O1 = 600 -M1 = 600 300 = 300Trong tam gic KOC vung ti C, ta c: 33 223:300RRCosOCOKOKOCCosKOC M O c nh, R khng i => K di ng trn ng trn tm O, bn knh = 33 2 R (iu phi chng minh)0,250,250,250,2513MOBCKEB12 11C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetC5 (1,0 im)( ) ( ) ( )3 3 3 4 4 43 3 34 4 44 4 4 4 4 44 4 44a b ca b c a a b c b a b c ca b ca b c+ + + + + + + + + +> + + + +Do , 3 3 3 4 4 444 42 24 2a b c + + > 0,250,250,250,25Ch : -Cu 4, tha gi thit tia Mx v im A gy ri.-Mi cu u c cc cch lm khccu 5Cach 2: t x = 4 4 4 a;y b;z c=> x, y , z > 0 v x4 + y4 + z4 = 4.BT cn CM tng ng: x3 + y3 + z3 > 2 2hay 2(x3 + y3 + z3 ) > 4 = x4 + y4 + z4 x3(2-x) + y3(2-y)+ z3(2-z) > 0 (*).Ta xt 2 trng hp:- Nu trong 3 s x, y, z tn ti it nht mt s 2 , gi s x2 th x3 2 2 .Khi o: x3 + y3 + z3 > 2 2 ( do y, z > 0).- Nu c 3 s x, y, z u nh2 < th BT(*) lun ung.Vy x3 + y3 + z3 > 2 2c CM.Cach 3: C th dng BT thc Csi kt hp phng php lm tri v nh gi cng cho kt qu nhng hi di, phc tp).14C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GD V O TO K THI TUYN SINH VO 10 THPT NM HC 2012-2013KLK MN THI : TONThi gian lm bi: 120 pht,(khng k giao )Ngy thi: 22/06/2012Cu 1. (2,5)1) Gii phng trnh:a) 2x2 7x + 3 = 0. b) 9x4 + 5x2 4 = 0.2) Tm hm s y = ax + b, bit th hm s ca n i qua 2 im A(2;5) ; B(-2;-3).Cu 2. (1,5)1) Hai t i t A n B di 200km. Bit vn tc xe th nht nhanh hn vn tc xe th hai l 10km/h nn xe th nht n B sm hn xe th hai 1 gi. Tnh vn tc mi xe.2) Rt gn biu thc: ( )1A= 1 x x ;x 1 _ + + , vi x 0.Cu 3. (1,5 )Cho phng trnh: x2 2(m+2)x + m2 + 4m +3 = 0.1) Chng minh rng : Phng trnh trn lun c hai nghim phn bit x1, x2 vi mi gi tr ca m.2) Tm gi tr ca m biu thc A = 2 21 2x x +t gi tr nh nht.Cu 4. (3,5)Cho tam gic ABC c ba gc nhn ni tip ng trn tm O (AB < AC). Hai tip tuyn ti B v C ct nhau ti M. AM ct ng trn (O) ti im th hai D. E l trung im on AD. EC ct ng trn (O)ti im th hai F. Chng minh rng:1) T gic OEBM ni tip.2) MB2 = MA.MD.3) BFC MOC .4) BF // AMCu 5. (1)Cho hai s dng x, y tha mn: x + 2y = 3. Chng minh rng: 1 23x y+ Bi gii s lc:Cu 1. (2,5)1) Gii phng trnh:a) 2x2 7x + 3 = 0. = (-7)2 4.2.3 = 25 > 0= 5. Phng trnh c hai nghim phn bit: 127 5x 3.47 5 1x4 2+ b) 9x4 + 5x2 4 = 0. t x2 = t , k : t 0.Ta c pt: 9t2 + 5t 4 = 0.a b + c = 0 t1 = - 1 (khng TMK, loi)t2 = 49 (TMK)t2 = 49 x2 = 49 x =4 29 3 t .15 CHNH THCEFDAMOCBC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetVy phng trnh cho c hai nghim: x1,2 = 23t2) th hm s y = ax + b i qua hai im A(2;5) v B(-2;-3) 2a b 5 a 22a b 3 b 1 + ' ' + Vy hm s cn tm l : y = 2x + 1Cu 2.1) Gi vn tc xe th hai l x (km/h). k: x > 0Vn tc xe th nht l x + 10 (km/h)Thi gian xe th nht i qung ng t A n B l : 200x 10 +(gi)Thi gian xe th hai i qung ng t A n B l : 200x (gi)Xe th nht n B sm 1 gi so vi xe th hai nn ta c phng trnh: 200 2001x x 10 +Gii phng trnh ta c x1 = 40 , x2 = -50 ( loi)x1 = 40 (TMK). Vy vn tc xe th nht l 50km/h, vn tc xe th hai l 40km/h.2) Rt gn biu thc: ( ) ( )1 x 1 1A 1 x x x xx 1 x 1 _ _ + + + + + , ,= ( )xx x 1x 1 _+ + ,= x, vi x 0.Cu 3. (1,5 )Cho phng trnh: x2 2(m+2)x + m2 + 4m +3 = 0.1) Chng minh rng : Phng trnh trn lun c hai nghim phn bit x1, x2 vi mi gi tr ca m.Ta c 22(m 2) m 4m 3 1 1 + ]> 0 vi mi m.Vy phng trnh cho lun c hai nghim phn bit x1, x2 vi mi gi tr ca m.2) phng trnh cho lun c hai nghim phn bit x1, x2 vi mi gi tr ca m. Theo h thc Vi-t ta c : 1 221 2x x 2(m 2)x .x m4m3 + +' + +A = 2 21 2x x += (x1 + x2)2 2 x1x2= 4(m + 2)2 2(m2 + 4m +3) = 2m2 + 8m+ 10= 2(m2 + 4m) + 10= 2(m + 2)2 + 2 2 vi mi m.Suy raminA= 2 m + 2 = 0 m = - 2Vy vi m = - 2 th A t min = 2Cu 4.1) Ta c EA = ED (gt) OEAD ( Quan h gia ng knh v dy)OEM = 900; OBM = 900 (Tnh cht tip tuyn)E v B cng nhn OM di mt gc vung T gic OEBM ni tip.2) Ta c 1MBD2 s BD( gc ni tip chn cung BD)1MAB2 s BD ( gc to bi tia tip tuyn v dy cung chn cung BD) MBD MAB . Xt tam gic MBD v tam gic MAB c:16C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetGc M chung, MBD MAB MBD ng dng viMAB MB MDMA MBMB2 = MA.MD3) Ta c: 1MOC2BOC= 12s BC ( Tnh cht hai tip tuyn ct nhau); 1BFC2 s BC(gc ni tip) BFC MOC .4) T gic MFOC ni tip ( $ F C + = 1800) MFC MOC ( hai gc ni tip cng chn cung MC), mt khc MOC BFC (theo cu 3) BFC MFC BF // AM.Cu 5. ( )22 2a ba bx y x y++ +Ta c x + 2y = 3 x = 3 2y , v x dng nn 3 2y > 0Xt hiu 1 23x y+ = 21 2 y 6 4y 3y(3 2y) 6(y 1)33 2y y y(3 2y) y(3 2y)+ + 0 ( v y > 0 v 3 2y > 0)1 13x 2y+ du = xy ra x 0,y 0 x 0,y 0x 1x 3 2y x 1y 1y 1 0 y 1 > > > > ' ' ' 17C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC VO O TO HI DNG-----------------K THI TUYN SINH LP 10 THPTNM HC 2012-2013MN THI: TONThi gian lm bi 120 pht (khng k thi gian giao )Ngy thi: Ngy 12 thng 7 nm 2012( thi gm: 01 trang)Cu 1 (2,0 im):Gii cc phng trnh sau:a) x(x-2)=12-x.b)228 1 116 4 4xx x x + + Cu 2 (2,0 im):a) Cho h phng trnh 3 2 95x y mx y+ + '+ c nghim (x;y). Tm m biu thc (xy+x-1) t gii tr ln nht.b) Tm m ng thng y = (2m-3)x-3 ct trc honh ti im c honh bng 23.Cu 3 (2,0 im):a) Rt gn biu thc( )3 1. 22 1P xx x x _ + + , vi0 x v4 x .b) Nm ngoi, hai n v sn xut nng nghip thu hoch c 600 tn thc. Nm nay, n v th nht lm vt mc 10%, n v th hai lm vt mc 20% so vi nm ngoi. Do c hai n v thu hoch c 685 tn thc. Hi nm ngoi, mi n v thu hoch c bao nhiu tn thc?Cu 4 (3,0 im):Cho tam gic ABC c ba gc nhn, ni tip ng trn (O). V cc ng cao BE, CF ca tam gic y. Gi H l giao im ca BE v CF. K ng knh BK ca (O) .a) Chng minh t gic BCEF l t gic ni tip.b) Chng minh t gic AHCK l mnh bnh hnh.c) ng trn ng knh AC ct BE M, ng trn ng knh AB ct CF N. Chng minh AM = AN.Cu 5 (1,0 im):Cho a, b, c, d l cc s thc tha mn: b + d 0 v 2acb d +. Chng minh rng phng trnh (x2 + ax +b)(x2 + cx + d)=0 (x l n) lun c nghim.---------------------Ht--------------------18 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetHNG DN - P NCu 1:a ) x = - 3 v x = 4. b) x = - 2; loi x = 4.Cu 2:a) H => x = m + 2 v y = 3 - m => A = (xy+x-1) = = 8 - ( m -1)2Amax = 8 khi m = 1.b) Thayx = 2/3 v y = 0 vo pt ng thng => m = 15/4Cu 3:a) A = 1b) x + y = 600 v0,1x + 0,2y = 85 hay x + 2y = 850.T tnh cy = 250 tn, x = 350 tnCu 4 (3,0 im):a) 090 C E B C F Bb) AH//KC ( cng vung gc vi BC)CH // KA ( cng vung gc vi AB)c) C AN2 = AF.AB; AM2 =AE.AC( H thc lng trong tam gic vung)AF. AF.ABACAEAEF ABC AE ACAB :AM = ANNMKHFEOC BACu 5 (1,0 im) Xt 2 phng trnh:x2 + ax + b = 0 (1) vx2 + cx + d = 0(2)[ ] [ ] ) ( 2 2 ) ( ) ( 2 2 2 ) 4 ( ) 4 (2 2 2 2 22 1d b ac c a d b ac c ac a d c b a + + + + + + + + Vi b+d 0 hoc2>0 pt cho c nghim+ Vi0 + d b. T 2acb d + ac > 2(b + d) =>02 1 + => t nht mt trong hai biu gi tr2 1, 0 => t nht mt trong hai pt (1) v (2) c nghim.Vy vi a, b, c, d l cc s thc tha mn: b + d 0 v 2acb d +,phng trnh (x2 + ax +b)(x2 + cx + d)=0 (x l n) lun c nghim.19C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC VO O TO HI DNG-----------------K THI TUYN SINH LP 10 THPTNM HC 2012-2013MN THI: TONThi gian lm bi 120 pht (khng k thi gian giao )Ngy thi: Ngy 14 thng 7 nm 2012( thi gm: 01 trang)Cu 1 (2,0 im): Gii cc phng trnh sau:a) 2 45 3 03 5x x _ _ + , ,b) | 2x 3 | = 1.Cu 2 (2,0 im): Cho biu thc:A = :2a a a ab a a b a b a b ab _ _+ + + + + + , ,vi a v b l cc s dng khc nhau.a) Rt gn biu thc A 2 a b abb a+ +.b) Tnh gi tr ca A khi a =7 4 3 v b =7 4 3 + .Cu 3 (2,0 im):a) Tm m cc ng thng y = 2x + m v y = x 2m + 3 ct nhau ti mt im nm trn trc tung.b) Cho qung ng t a im A ti a im B di 90 km. Lc 6 gi mt xe my i t A ti B Lc 6 gi 30 pht cng ngy, mt t cng i t A ti B vi vn tc ln hn vn tc xe my 15 km/h (Hai xe chy trn cng mt con ng cho). Hai xe ni trn u n B cng lc. Tnh vn tc mi xe.Cu 4 (3,0 im): Cho na ng trn tm O ng knh AB = 2R (R l mt di cho trc). Gi C, D l hai im trn na ng trn sao cho C thuc cung AD v COD = 1200 . Gi giao im ca hai dy AD v BCl E, giao im ca cc ng thng AC v BD l F.a) Chng minh rng bn im C, D, E, F cng nm trn mt ng trn.b)Tnh bn knh ca ng trn i qua C, E, D, F ni trn theo R.c)Tm gi tr ln nht ca in tch tam gic FAB theo R khi C, D thay i nhung vn tha mn gi thit bi tonCu 5 (1,0 im): Khng dng my tnh cm tay , tm s nguyn ln nht khng vt qu S, trong S =( )62 3 +-------------------- Ht --------------------HNG DN GII .Cu 1.20 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net2 4) 5 3 03 52 155 02 153 24 4 15 153 05 4a x xx xxxx x _ _ + , , >

+ Vy nghim ca phng trnh cho l S = {15 15;2 4} b) 2 3 1 2 4 22 3 12 3 1 2 2 1x x xxx x x >

Vy nghim ca phng trnh cho l S = {1;2}Cu 2 .Ta c :( )( )( )222:2:( )( )( ) ( ):( )( ).( )( )a a a aAb a a b a b a b aba a a aAa b b a b a a ba ba b a a a a b aAb a b aa ba babAb a b a aba bAb a _ _ + + + + + , , 1 _ 1 + 1+ + + , + 1 ] + + + +++ +a) Ta c :22 22( )( ) ( )0a b abAb aa b a bb a b aa b a bb a+ ++ + + + Vy 2 a b abAb a+ + = 0b) Ta c :( )27 4 34 4 3 32 32 3aaaa + > 21C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net( )27 4 34 4 3 32 32 3bbbb + + + +> +Thay2 3; 2 3 a b +vo biu thc a bAb a+ ta c :2 3 2 32 3 2 342 32 33AAA + ++ +Vy vi a = 7 -4 3 ; b = 7 + 4 3th A = 2 33.Cu 3 .a) hai ng thng y = 2x + m v y = x 2m + 3 ct nhau ti mt im trn trc tung th m = -2m + 3 => 3m = 3 => m = 1.Vy vi m = 1 th hai ng thng y = 2x + m v y = x 2m + 3 ct nhau ti mt im trn trc tung.b) Xe my i trc t thi gian l : 6 gi 30 pht - 6 gi = 30 pht = 12h .Gi vn tc ca xe my l x ( km/h ) ( x > 0 )V vn tc t ln hn vn tc xe my 15 km/h nnvn tc ca t l x + 15 (km/h)Thi gian xe my i ht qung ng AB l : 90( ) hxThi gian t i ht qung ng AB l : 90( )15hx +Do xe my i trc t 12 gi v hai xe u ti B cng mt lc nn ta c phng trnh :2290 1 902 1590.2.( 15) ( 15) 90.2180 2700 15 18015 2700 0x xx x x xx x x xx x +> + + + + Ta c :215 4.( 2700) 11025 011025 105 > 115 105602x ( khng tha mn iu kin )215 105452x + ( tha mn iu kin )Vy vn tc ca xe my l 45 ( km/h ) , vn tc ca t l 45 + 15 = 60 ( km/h ).Cu 4.a) Ta c : C, D thuc ng trn nn :22C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net 090 ACB ADB ( gc ni tip chn na ng trn )=> 0 090 ; 90 FCE FDE ( gc k b )Hai im C v D cng nhn on thng FE di mt gc bng nhaubng 900 nn 4 im C,D,E,F cng thuc ng trn ng knh EF.b) Gi I l trung im EF th ID = IC l bn knh ng trn i qua4 im C, D, E, F ni trn.Ta c : IC = ID ; OC = OD ( bn knh ng trn tm O )suy ra IO l trung trc ca CD => OI l phn gic ca COD=> 00120602IOD Do O l trung im AB v tam gic ADB vung ti D nn tam gic ODB cn ti O=> ODB OBD (1)Do ID = IF nn tam gic IFD cn ti I => IFD IDF (2)Tam gic AFB c hai ng cao AD, BC ct nhau ti E nn E l trc tm tam gic => FE l ng cao th ba => FE vung gc AB ti H => 0IF 90 OBD D + (3)T (1) , (2) , (3) suy ra 090 IDF ODB + => 090 IDO .Xt tam gic vung IDO c 060 IOD .Ta c : ID = OD.tanIOD = R.tan600 = R 3 .Vy bn knh ng trn i qua 4 im C,D,E,F l R 3 .c) Theo phn b) : OI = 2 2 2 23 2 ID OD R R R + + .t OH = x th0 x R => IH = 2 24R x .=> FH = R 3+ 2 24R x .2 22 2 21 1. . .2 .( 3 4 )2 23 4FABFABS AB FH R R R xS R R R x + + Ta c : 4R2 - x2 4R2 . Du bng xy ra khi x = 0.Khi : SFAB = R23+ 2R2v H O => O, I, F thng hng => CD // AB => 015 ADO DAO => BD = AC = 2RSin150 .Vy din tch ln nht t c ca tam gic AFB l R23+ 2R2khi AC = BD = 2Rsin150 .Cu 5Xt hai s a = 2 +3v b = 2 -3 .Ta c : a + b = 4 v ab = 1, 0< b < 1.(a+b)3 = 43 = 64 => a3 + b3 = 64 - 3ab(a + b) = 64 - 3.1.4 = 52(a3+b3)(a3 + b3) = 52.52 => a6 + b6 = 2704 - 2(ab)3 = 2704 - 2 = 2702=> a6 = S = 2702 - b6 (*).Do 0 0Tm gi tr nh nht ca biu thcA = 2248bab a++---------------------------------------HT ----------------------------------36 THI CHNH THC A C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net37C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net38C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net39C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net40C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net41C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC O TOK THI VO LP 10 CHUYN LAM SNTHANH HONM HC 2012 - 2013Mn thi : TON( gm c 01 trang)(Mn chung cho tt cc th sinh)Thi gian lm bi :120 pht (Khng k thi gian giao )Ngy thi : 17 thng 6 nm 2012Cu 1: (2.0 im )Cho biu thc :1 1 141 1 2a aP aa a a a _+ + + ,, (Vi a > 0 ,a 1)1. Chng minh rng: 21Pa2. Tm gi tr ca a P = aCu 2 (2,0 im ) : Trong mt phng to Oxy, cho Parabol (P) : y = x2 v ng thng (d) : y = 2x + 31. Chng minh rng (d) v (P) c hai im chung phn bit2. Gi A v B l cc im chung ca (d) v (P) . Tnh din tch tam gic OAB ( O l gc to )Cu 3 (2.0 im) : Cho phng trnh : x2 + 2mx + m2 2m + 4 = 01. Gii phng trnh khi m = 42. Tm m phng trnh c hai nghim phn bitCu 4 (3.0 im) : Cho ng trn (O) c ng knh AB c nh, M l mt im thuc (O) ( M khc A v B ) . Cc tip tuyn ca (O) ti A v M ct nhau C. ng trn (I) i qua M v tip xc vi ng thng AC ti C. CD l ng knh ca (I). Chng minh rng:1. Ba im O, M, D thng hng2. Tam gic COD l tam gic cn3. ng thng i qua D v vung gc vi BC lun i qua mt im c nh khi M di ng trn ng trn (O)Cu 5 (1.0 im) : Cho a,b,c l cc s dng khng m tho mn : 2 2 23 a b c + + Chng minh rng : 2 2 212 3 2 3 2 3 2a b ca b b c c a+ + + + + + + +42 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetHNG DN GII:CU NI DUNG IM11. Chng minh rng : 21Pa1 1 141 1 2a aP aa a a a _+ + + ,( ) ( ) ( ) ( )( ) ( )2 21 1 4 1 11.21 1a a a a aPa aa a+ + + + ( ) ( )2 1 2 1 4 4 1.21 1a a a a a a aPa aa a+ + + + + 4 1 2.1 1 2a aPa a a a (PCM)1.02. Tm gi tr ca a P = a. P = a=> 222 01a a aa > .Ta c 1 + 1 + (-2) = 0, nn phng trnh c 2 nghima1 = -1 < 0 (khng tho mn iu kin) -Loia2 = 221ca (Tho mn iu kin)Vy a = 2 th P = a1.02 1. Chng minh rng (d) v (P) c hai im chung phn bitHonh giao im ng thng (d) v Parabol (P) l nghim ca phng trnhx2 = 2x + 3 => x2 2x 3 = 0 c a b + c = 0Nn phng trnh c hai nghim phn bitx1 = -1 v x2 = 331ca Vi x1 = -1 => y1 = (-1)2 = 1 => A (-1; 1)Vi x2 = 3 => y2 = 32 = 9 => B (3; 9)Vy (d) v (P) c hai im chung phn bit A v B1.02. Gi A v B l cc im chung ca (d) v (P) . Tnh din tch tam gic OAB ( O l gc to )Ta biu din cc im A v B trn mt phng to Oxy nh hnh v1D CBA93 -101.043C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net1 9. .4 202 2ABCDAD BCS DC+ + . 9.313, 52 2BOCBC COS . 1.10, 52 2AODAD DOS Theo cng thc cng din tch ta c:S(ABC) = S(ABCD) - S(BCO) - S(ADO)= 20 13,5 0,5 = 6 (vdt)31. Khi m = 4, ta c phng trnhx2 + 8x + 12 = 0 c = 16 12 = 4 > 0Vy phng trnh c hai nghim phn bitx1 = - 4 + 2 = - 2 v x2 = - 4 - 2 = - 61.02. Tm m phng trnh c hai nghim phn bitx2 + 2mx + m2 2m + 4 = 0C D = m2 (m2 2m + 4) = 2m 4 phng trnh c hai nghim phn bit th D > 0=> 2m 4 > 0 => 2(m 2) > 0 => m 2 > 0 => m > 2Vy vi m > 2 th phng trnh c hai nghim phn bit1.0412NKHDI COA BM1. Ba im O, M, D thng hng:Ta c MC l tip tuyn ca ng trn (O) MC MO (1)Xt ng trn (I) : Ta c 090 CMD MC MD (2)T (1) v (2) => MO // MD MO v MD trng nhau O, M, D thng hng1.02. Tam gic COD l tam gic cnCA l tip tuyn ca ng trn (O) CA AB(3)ng trn (I) tip xc vi AC ti C CA CD(4)T (3) v (4) CD // AB => DCO COA (*)( Hai gc so le trong)CA, CM l hai tip tuyn ct nhau ca (O) COA COD (**)T (*) v (**) DOC DCO Tam gic COD cn ti D1.03. ng thng i qua D v vung gc vi BC lun i qua mt im c nh khi M di1.044C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netng trn ng trn (O)* Gi chn ng vung gc h t D ti BC l H. 090 CHD H (I) (Bi ton qu tch)DH ko di ct AB ti K.Gi N l giao im ca CO v ng trn (I)=>090 can tai DCNDNC NOCOD 'Ta c t gic NHOK ni tipV c 2 1H O DCO ( Cng b vi gc DHN) 0180 NHO NKO + (5)* Ta c : NDH NCH (Cng chn cung NH ca ng trn (I)) ( )CBO HND HCD DHNCOB(g.g)......HN OBHD OCOB OA HN ONOC OC HD CDOA CN ONOC CD CD ) M ONH CDH NHO DHC (c.g.c) 090 NHO M 0180 NHO NKO + (5) 090 NKO , NK AB NK // AC K l trung im ca OA c nh (PCM)5Cu 5 (1.0 im) : Cho a,b,c l cc s dng khng m tho mn : 2 2 23 a b c + + Chng minh rng : 2 2 212 3 2 3 2 3 2a b ca b b c c a+ + + + + + + +* C/M b : ( )22 2a ba bx y x y++ + v ( )22 2 2a b ca b cx y x x y z+ ++ + + + .Tht vy( )( ) ( ) ( ) ( )22 22 22 20a ba ba y b x x y xy a b ay bxx y x y++ + + + +(ng) PCMp dng 2 ln , ta c: ( )22 2 2a b ca b cx y x x y z+ ++ + + +* Ta c: 2 22 3 2 1 2 2 2 2 a b a b a b + + + + + + +, tng t Ta c: 2 2 22 3 2 3 2 3 2 2 2 2 2 2 2 2 2a b c a b cAa b b c c a a b b c c a + + + ++ + + + + + + + + + + + 1 (1)2 1 1 1Ba b cAa b b c c a _ + + + + + + + + ,1 4 4 4 4 4 2 4 4 4 4 4 3Ta chng minh11 1 1a b ca b b c c a+ + + + + + + +1.045C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net( )( ) ( )( )( ) ( )( )( ) ( )2 2 231 1 1 21 1 11 1 121 1 11 1 121 1 11 1 12(2)1 1 1 1 1 1Ba b ca b b c c ab c aa b b c c ab c aa b b c c ab c aa b b b c c c a a + + + + + + + + + + + + + + + ++ + + + + + + + + + ++ + + + + + + + + + + + + +1 4 4 4 4 4 4 4 4 442 4 4 4 4 4 4 4 4 4 43* p dng B trn ta c: ( )( ) ( ) ( ) ( ) ( ) ( )2331 1 1 1 1 1a b cBa b b b c c c a a+ + + + + + + + + + + + + +( )22 2 233 (3)3( ) 3a b cBa b c ab bc ca a b c+ + + + + + + + + + + +* M:( )( )2 2 22 2 22 2 2 2 2 22 2 2222 2 2 2 3( ) 32 2 2 2 2 2 6 6 6 62 2 2 2 2 2 6 6 6 6 ( : 3)2 2 2 6 6 6 9333( )a b c ab bc ca a b ca b c ab bc ca a b ca b c ab bc ca a b c Do a b ca b c ab bc ca a b ca b ca b ca b c ab bc ca a b c 1 + + + + + + + + + ] + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + ++ + ++ + + + + + + +2 (4)3 +T (3) v (4) (2)Kt hp (2) v (1) ta c iu phi chng minh.Du = xy ra khi a = b = c = 146C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TOTHNH PH CN THK THI TUYN SINH LP 10 THPTNM HC 2012-2013Kha ngy:21/6/2012MN: TONThi gian lm bi: 120 pht (khng k thi gian pht )Cu 1: (2,0 im)Gii h phng trnh , cc phng trnh sau y:1. 433 2 19x yx y+ ' 2. 5 2 18 x x + 3. 212 36 0 x x + 4. 2011 4 8044 3 x x + Cu 2: (1,5 im)Cho biu thc: 21 1 12 :1aKa a a a _+ _ , ,(vi 0, 1 a a > )1. Rt gn biu thc K.2. Tm a 2012 K .Cu 3: (1,5 im)Cho phng trnh (n s x):( )2 24 3 0 * x x m + .1. Chng minh phng trnh (*) lun c hai nghim phn bit vi mi m.2. Tm gi tr ca m phng trnh (*) c hai nghim 1 2, x x tha 2 15 x x .Cu 4: (1,5 im)Mt t d nh i t A n B cch nhau 120 km trong mt thi gian quy nh. Sau khi i c 1 gi th t b chn bi xe cu ha 10 pht. Do n B ng hn xe phi tng vn tc thm 6 km/h. Tnh vn tc lc u ca t.Cu 5: (3,5 im)Cho ng trn( ) O, t imA ngoi ng trn v hai tip tuynAB v AC (, B Cl cc tip im).OAct BC ti E.1. Chng minh t gicABOC ni tip.2. Chng minhBCvung gc viOA v. . BA BE AE BO .3. Gi I l trung im caBE , ng thng qua I v vung gcOI ct cc tia , AB ACtheo th t tiDvF . Chng minh IDO BCO vDOF cn tiO.4. Chng minhFl trung im ca AC .GI GII:Cu 1: (2,0 im)Gii h phng trnh , cc phng trnh sau y:1. 43 2 2 86 5 105 213 2 19 3 2 19 43 22x y x y x xx y x y x y y+ + ' ' ' ' + 2.5 2 18 ; : 9 x x K x + 47 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net23( )5 2 18135 2 18 ( )3x TMKx xx x x KTMK +

+ + 3. 2 212 36 0 ( 6) 0 6 x x x x + 4. 2011 4 8044 3; : 20113 2011 3 2012( )x x K xx x TMK + Cu 2: (1,5 im)Cho biu thc: 21 1 12 :1aKa a a a _+ _ , ,(vi 0, 1 a a > )( )21 1 1 1 12 : 2 :( 1) 1 ( 1)1 1 12 : 2 : ( 1) 2( 1) ( 1) ( 1)a a a aKa a a a a a a aa a aa a a a a a _ _ _+ + + _ , , , , _ _ _ , , ,2012 K 2 a = 2012

a = 503 (TMK)Cu 3: (1,5 im)Cho phng trnh (n s x):.1. ( )2 22 24 3 0 *16 4 12 4 4 4 0;x x mm m m + + + > Vy (*) lun co hai nghim phn bit vi mi m.2. Tm gi tr ca m phng trnh (*) c hai nghim 1 2, x x tha 2 15 x x .Theo h thc VI-ET co :x1.x2 = - m2 + 3;x1+ x2 = 4; ma 2 15 x x => x1 = - 1 ; x2 = 5Thayx1 = - 1 ; x2 = 5 vao x1.x2 = - m2 + 3=> m =2 2 tCu 4: (1,5 im)Goi x (km/h)la vt d inh; x > 0 => Thi gian d inh : 120( ) hxSau 1 h t i c x km => quang ng con lai 120 x ( km)Vt luc sau: x + 6 ( km/h)Pt 1 120 12016 6xx x+ + +=> x = 48 (TMK) => KLHD C3Tam giac BOC cn tai O => goc OBC = goc OCBT giac OIBD co goc OID = goc OBD = 900 nn OIBD ni tip =>goc ODI = goc OBIDo o IDO BCO Lai co FIOC ni tip ; nn goc IFO = goc ICOSuy ra goc OPF = goc OFP ; vyDOF cn tiO.HD C4Xet t giac BPFE co IB = IE ; IP = IF ( Tam giac OPF cn co OI la ng cao=> )Nn BPEF la Hinh binh hanh => BP // FETam giac ABC co EB = EC ; BA // FE; nn EF la TB cua tam giac ABC => FA = FC48C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GD T NGHAN thi vo THPT nm hc 2012 - 2013Mn thi: TonThi gian 120 phtNgy thi 24/ 06/ 2012Cu 1: 2,5 im:Cho biu thc A = 1 1 2.2 2xx x x _+ + ,a) Tm iu kin xc nh v t gn A.b) Tm tt c cc gi tr ca x 12A >c) Tm tt c cc gi tr ca x 73B A t gi tr nguyn.Cu 2: 1,5 im:Qung ng AB di 156 km. Mt ngi i xe my t A, mt ngi i xe p t B. Hai xe xut pht cng mt lc v sau 3 gi gp nhau. Bit rng vn tc ca ngi I xe my nhanh hn vn tc ca ngi I xe p l 28 km/h. Tnh vn tc ca mi xe?Cu 3: 2 im:Chjo phng trnh: x2 2(m-1)x + m2 6 =0 ( m l tham s).a) GiI phng trnh khi m = 3b) Tm m phng trnh c hai nghim x1, x2 tha mn 2 21 216 x x + Cu 4: 4 imCho im M nm ngoi ng trn tm O. V tip tuyn MA, MB vi ng trn (A, B l cc tip im). V ct tuyn MCD khng I qua tm O ( C nm gia M v D), OM ct AB v (O) ln lt ti H v I. Chng minh.a) T gic MAOB ni tip.b) MC.MD = MA2c) OH.OM + MC.MD = MO2d) CI l tia phn gic gc MCH.49 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetHNG DN GIICu 1: (2,5 im)a, Vi x > 0 v x4, ta c:A = 1 1 2.2 2xx x x _+ + , = 2 2 2.( 2)( 2)x x xx x x + + + = ... = 22 x +b, A = 22 x +

22 x + > 12 ... x > 4.c, B = 73. 22 x + = 143( 2) x + l mt s nguyn ... 2 x +l c ca 14 hay2 x +=t1, 2 x +=t7,2 x +=t14.(Gii cc pt trn v tm x)Cu 2: (1,5 im)Gi vn tc caxe p l x (km/h), iu kin x > 0Th vn tc ca xe my l x + 28 (km/h)Trong 3 gi:+ Xe p i c qung ng 3x (km),+ Xe my i c qung ng 3(x + 28) (km), theo bi ra ta c phng trnh:3x + 3(x + 28) = 156Gii tm x = 12 (TMK)Tr li: Vn tc ca xe p l 12 km/h v vn tc ca xe my l 12 + 28 = 40 (km/h)Cu 3: (2,0 im)a, Thay x = 3 vo phng trnh x2 - 2(m - 1)x + m2 - 6 = 0 v gii phng trnh:x2 - 4x + 3 = 0 bng nhiu cch v tm c nghim x1 = 1, x2 = 3.b, Theo h thc Vit, gi x1, x2 l hai nghim ca phng trnhx2 - 2(m - 1)x + m2 - 6 = 0 , ta c:1 221 22( 1). 6x x mx x m+ ' v x12 + x22 = (x1 + x2)2 - 2x1.x2 = 1650C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetThay vo gii v tm c m = 0, m = -4Cu 4: (4,0 im).T vit GT-KLADCMIHBa, V MA, MB l cc tip tuyn ca ng trn (O) ti A v B nn cc gc ca t gic MAOB vung ti A v B, nn ni tip c ng trn.b,MAC vMDA c chung M v MAC = MDA (cng chn AC), nn ng dng. T suy ra 2.MA MDMC MD MAMC MA (fcm)c,MAO vAHO ng dng v c chung gc O v AMO HAO (cng chn hai cung bng nhau ca ngtrn ni tip t gic MAOB). Suy ra OH.OM = OA2p dng nh l Pitago vo tam gic vung MAO v cc h thc OH.OM = OA2 MC.MD = MA2 suy ra iu phi chng minh.d, T MH.OM = MA2, MC.MD = MA2 suy ra MH.OM = MC.MD MH MCMD MO(*)TrongMHC vMDO c (*) v DMO chung nn ng dng. M OMC MO MOHC D A hay OMC MOCH A(1)Ta li c MAI IAH (cng chn hai cung bng nhau) AI l phn gic ca MAH.Theo t/c ng phn gic ca tam gic, ta c: AMI MAIH H(2)MHA vMAO c OMA chung v 090 MHA MAO do ng dng (g.g) O AMO MAA H(3)T (1), (2), (3) suy ra MC MICH IHsuy ra CI l tia phn gic ca gc MCH51HOC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TOH NAMK THI TUYN SINH LP 10 THPTNM HC 2012 2013Mn: TonThi gian lm bi: 120 phtNgy thi : 22/06/2012Cu 1 (1,5 im) Rt gn cc biu thc sau:a) A 2 5 3 45 5008 2 12b) B 83 1 + Cu 2: (2 im)a) Gii phng trnh: x2 5x + 4 = 0b) Gii h phng trnh: 3x y 1x 2y 5 '+ Cu 3: (2 im)Trong mt phng to Oxy cho Parabol (P) c phng trnh: y = x2 v ng thng (d) c phng trnh: y = 2mx 2m + 3 (m l tham s)a) Tm to cc im thuc (P) bit tung ca chng bng 2b) Chng minh rng (P) v (d) ct nhau ti hai im phn bit vi mi m.Gi 1 2y , yl cc tung giao im ca (P) v (d), tm m 1 2y y 9 + 5Cu 2 (2)a) Giai h phng trinh ' +7 23 3y xy xb) Chng minh rng 762 312 31++Cu 3 (2)Cho phng trinh x2 2(m 3)x 1 = 0a) Giai phng trinh khi m = 1b) Tim m phng trinh co nghim x1 ; x2 ma biu thcA = x12 x1x2 + x22at gia tri nho nht? Tim gia tri nho nht o.Cu 4 (3)Cho tam giac ABC vung tai A. Ly B lam tm ve ng tron tm B ban kinh AB.Ly C lam tm ve ng tron tm C ban kinh AC, hai ng tron nay ct nhau tai im th 2 la D.Ve AM, AN ln lt la cac dy cung cua ng tron (B) va (C) sao cho AM vung goc vi AN va D nm gia M; N.a) CMR:ABC=DBCb) CMR: ABDC la t giac ni tip.c) CMR: ba im M, D, N thng hangd) Xac inh vi tri cua cac dy AM; AN cua ng tron (B) va (C) sao cho oan MN co dai ln nht.Cu 5 (1)Giai H PT '+ + y x y x y x y xy y x2 ) 3 2 4 ( 1 2 ) 1 4 2 (3 8 52 2---------------------------Ht--------------------------GI GIICu 1 (2) a) Giai phng trinh 2x 5 = 1b) Giai bt phng trinh 3x 1 > 5ap an a) x = 3;b) x > 2Cu 2 (2) a) Giai h phng trinh ' +7 23 3y xy x68S GIAO DUC VA AO TAOPHU THO CHNH THC KY THI TUYN SINHVAO LP 10 TRUNG HOC PH THNGNM HOC 2012-2013Mn toanThi gian lam bai: 120 phut, khng k thi gian giao thi co 01 trang-------------------------------------------C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netb) Chng minh rng 762 312 31++ap an a) x = 2 ; y = 3b) VT =762 92 3 2 3+ + =VP (pcm)Cu 3 (2) Cho phng trinh x2 2(m 3)x 1 = 0c) Giai phng trinh khi m = 1d) Tim m phng trinh co nghim x1 ; x2 ma biu thcA = x12 x1x2 + x22at gia tri nho nht? Tim gia tri nho nht o.ap an a) x1 =5 2 ; x2 =5 2 + e) Thy h s cua pt : a = 1 ; c = A 1 pt lun co 2 nghimTheo vi- et ta cox1 + x2 =2(m 3); x1x2 = 1Ma A=x12 x1x2 + x22= (x1 + x2 )2 3x1x2= 4(m 3)2 + 3 3 GTNN cua A = 3 m = 3Cu 4 (3)Hng dna) Co AB = DB; AC = DC; BC chung ABC = DBC (c-c-c)b) ABC = DBC goc BAC =BDC = 900

ABDC la t giac ni tipc) Co gocA1 = gocM1 ( ABM cn tai B)gocA4 = gocN2 ( ACN cn tai C)gocA1 = gocA4 ( cung phu A2;3) gocA1 = gocM1 =gocA4= gocN2gocA2 = gocN1 ( cung chn cung AD cua (C) )Lai co A1+A2 + A3 = 900 => M1 + N1 + A3 = 900Ma AMN vung tai A => M1 + N1 + M2 = 900=> A3 = M2=> A3 = D1CDN cn tai C => N1;2 = D4 D2;3 + D1 + D4 =D2;3 + D1 + N1;2 = D2;3 + M2 + N1 + N2= 900 + M2 + N1 + M1 ( M1 = N2)= 900 + 900 = 1800 M; D; N thng hang.d) AMN ng dang ABC (g-g)Ta co NM2 = AN2 +AM2 NM ln nht thi AN ; AM ln nhtMa AM; AN ln nht khi AM; AN ln lt la ng kinh cua (B) va (C)Vy khi AM; AN ln lt la ng kinh cua (B) va (C) thi NM ln nht.Cu 5 (1):Giai H PT '+ + y x y x y x y xy y x2 ) 3 2 4 ( 1 2 ) 1 4 2 (3 8 52 2Hng dn'+ + y x y x y x y xy y x2 ) 3 2 4 ( 1 2 ) 1 4 2 (3 8 52 2692143212 1432121MDNCBAC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net '+ > < > + < ) 2 ( 2 ) 1 1 2 2 ( 1 2 ) 1 2 2 () 1 ( 3 8 52 2y x y x y x y xy y xT (2) t x +2y = a ;2xy 1 = b(a:b0)Ta dc(2a-1) b =(2b 1) a ( b a )(2) 1 + ab= 0 a = b x = 3y + 1 thay vao (1) ta dc2y2 y 1= 0 => y1 = 1 ; y2 = 1/2=> x1 = 4 ; x2 = 1/2Thy x2 + 2y2 = 1 < 0 (loai)Vy h co nghim (x; y) = (4 ; 1)S gio dc v o toHng yn( thi c 01 trang)k thi tuyn sinh vo lp 10 thpt chuynNm hc 2012 - 2013Mn thi: Ton(Dnh cho th sinh d thi cc lp chuyn: Ton, Tin)Thi gian lm bi: 150 phtBi 1: (2 im)a) Cho A =2 2 2 22012 2012 .2013 2013 + +. Chng minh A l mt s t nhin.b) Gii h phng trnh 221 xx 3y y1 xx 3y y+ + '+ + Bi 2: (2 im)a) Cho Parbol (P): y = x2 v ng thng (d): y = (m +2)x m + 6. Tm m ng thng (d) ct Parabol (P) ti hai im phn bit c honh dng.b) Gii phng trnh: 5 + x +2 (4 x)(2x 2) 4( 4 x 2x 2) + Bi 3: (2 im)a) Tm tt c cc s hu t x sao cho A = x2 + x+ 6 l mt s chnh phng.b) Cho x > 1 v y > 1. Chng minh rng : 3 3 2 2(x y ) (x y )8(x 1)(y 1)+ + Bi 4 (3 im)70 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCho tam gic ABC nhn ni tip ng trn tm O, ng cao BE v CF. Tip tuyn ti B v C ct nhau ti S, gi BC v OS ct nhau ti Ma) Chng minh AB. MB = AE.BSb) Hai tam gic AEM v ABS ng dngc) Gi AM ct EF ti N, AS ct BC ti P. CMR NP vung gc vi BCBi 5: (1 im)Trong mt gii bng c 12 i tham d, thi u vng trn mt lt (hai i bt k thi u vi nhau ng mt trn).a) Chng minh rng sau 4 vng u (mi i thi u ng 4 trn) lun tm c ba i bng i mt cha thi u vi nhau.b) Khng nh trn cn ng khng nu cc i thi u 5 trn?HNG DN GIIBi 1: (2 im)a) Cho A =2 2 2 22012 2012 .2013 2013 + +t 2012 = a, ta c 2 2 2 22012 2012 .2013 2013 + + 2 2 2 2a a (a 1) (a 1) + + + +2 2 2(a a 1) a a 1 + + + +b) t xay1x by'+ Ta c 221 xx 3y y1 xx 3y y+ + '+ + 21 xx 3y y1 xx 3y y _+ , '+ + nn 2 2b a 3 b b 6 0b a 3 b a 3 + ' '+ + a 6 a 1vb 3 b 2 ' ' Bi 2:a) ycbt tng ng vi PT x2 = (m +2)x m + 6 hay x2 - (m +2)x + m 6 = 0 c hai nghim dng phn bit.b) t t = 4 x 2x 2 + Bi 3:a) x = 0, x = 1, x= -1 khng tha mn. Vi x khc cc gi tr ny, trc ht ta chng minh x phi l s nguyn.+) x2 + x+ 6 l mt s chnh phng nn x2 + x phi l s nguyn.+)Gi s mxnvi m v n c c nguyn ln nht l 1.Ta c x2 + x =2 22 2m m m mnn n n++ l s nguyn khi 2m mn + chia ht cho n271C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netnn 2m mn + chia ht cho n, v mn chia ht cho n nn m2 chia ht cho n v do m v n c c nguyn ln nht l 1, suy ra m chia ht cho n( mu thun vi m v n c c nguyn ln nht l 1). Do x phi l s nguyn.t x2 + x+ 6 = k2Ta c 4x2 + 4x+ 24 = 4 k2 hay (2x+1)2 + 23 = 4 k2 tng ng vi 4 k2 - (2x+1)2 = 233 3 2 2 2 2(x y ) (x y ) x (x 1) y (y 1)(x 1)(y 1) (x 1)(y 1)+ + + = 2 2x yy 1 x 1+ 2 2(x 1) 2(x 1) 1 (y 1) 2(y 1) 1y 1 x 1 + + + + + 2 2(x 1) (y 1) 2(y 1) 2(x 1) 1 1y 1 x 1 x 1 y 1 y 1 x 1 111 + + + + + 111 ] ] ].Theo BT Csi2 2 2 2(x 1) (y 1) (x 1) (y 1)2 . 2 (x 1)(y 1)y 1 x 1 y 1 x 1 + 2(y 1) 2(x 1) 2(y 1) 2(x 1). 4x 1 y 1 x 1 y 1 + 1 1 1 12 .y 1 x 1 y 1 x 1+ 1 1 1 12 . (x 1)(y 1) 2.2 . . (x 1)(y 1) 4y 1 x 1 y 1 x 1 1+ 1 ]Bi 4a) Suy ra t hai tam gic ng dng l ABE v BSMb) T cu a) ta c AE MBAB BS(1)PNFEMSOABCQ72C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetM MB = EM( do tam gic BEC vung ti E c M l trung im ca BCNn AE EMAB BSC 0 0MOB BAE, EBA BAE 90 , MBO MOB 90 + + Nn MBO EBA do MEB OBA( MBE) Suy ra MEA SBA (2)T (1) v (2) suy ra hai tam gic AEM v ABS ng dng(pcm.)c) D thy SM vung gc vi BC nn chng minh bi ton ta chng minh NP //SM.+ Xt hai tam gic ANE v APB:T cu b) ta c hai tam gic AEM v ABS ng dng nn NAE PAB ,M AEN ABP ( do t gic BCEF ni tip)Do hai tam gic ANE v APB ng dng nn AN AEAP ABLi c AM AEAS AB ( hai tam gic AEM v ABS ng dng)Suy ra AM ANAS APnn trong tam gic AMS c NP//SM( nh l Talet o)Do bi ton c chng minh.Bi 5a. Gia s kt lun cua bai toan la sai, tc la trong ba i bt ky thi co hai i a u vi nhau ri. Gia s i a gp cac i 2, 3, 4, 5. Xet cac b (1; 6; i) vi i {7; 8; 9;;12}, trong cac b nay phai co it nht mt cp a u vi nhau, tuy nhin 1 khng gp 6 hay i nn 6 gp i vi mi i {7; 8; 9;;12} , v ly vi i 6 nh th a u hn 4 trn. Vy co pcm.b. Kt lun khng ung. Chia 12 i thanh 2 nhom, mi nhom 6 i. Trong mi nhom nay, cho tt ca cac i i mt a thi u vi nhau. Luc nay ro rang mi i a u 5 trn. Khi xet 3 i bt ky, phai co 2 i thuc cung mt nhom, do o 2 i nay a u vi nhau. Ta co phan vi du.Co th giai quyt n gian hn cho cu a. nh sau:Do mi i a u 4 trn nn tn tai hai i A, B cha u vi nhau. Trong cac i con lai, vi A vaB chi u 3 trn vi ho nn tng s trn cua A, B vi cac i nay nhiu nht la 6 va do o, tn tai i C trong s cac i con lai cha u vi ca A va B. Ta co A, B, C la b ba i i mt cha u vi nhau.73C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TOHNG YNK THI TUYN SINH VO LP 10 THPTNM HC 2012 - 2013Thi gian lm bi: 120 pht (khng k thi gian giao )PHN A: TRC NGHIM KHCH QUAN (2 im)T cu 1 n cu 8, hy chn phng n ng v vit ch ci ng trc phng n vo bi lmCu 1: gi tr ca biu thc2 8 +bng:A. 10B.3 2 C. 6D.2 4 +Cu 2: Biu thc1 2 x x + c ngha khi:A. x < 2 B. 2 x C. 1 x D. 1 x Cu 3: ng thng y = (2m 1)x + 3 song song vi ng thng y = 3x 2 khi:A. m = 2 B. m = - 2 C. 2 m D. 2 m Cu 4: H phng trnh 2 33x yx y '+ c nghim (x;y) l:A. (-2;5) B. (0;-3) C. (1;2) D. (2;1)Cu 5: Phng trnh x2 6x 5 = 0 c tng hai nghim l S v tch hai nghim l P th:A. S = 6; P = -5 B. S = -6; P = 5 C. S = -5; P = 6 D. S = 6; P = 5Cu 6: th hm s y = -x2 i qua im:A. (1;1) B. (-2;4) C. (2;-4)D. (2;-1)Cu 7: Tam gic ABC vungti A c AB = 4cm; AC = 3cm th di ng cao AH l:74 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetA.34cm B.125cm C.512cm D.43cmCu 8: Hnh tr c bn knh y v chiu cao cng bng R th th tch lA.32 R B.2R C.3R D.22 R PHN B: T LUN ( 8,0 im)Bi 1: (1 im)a) Tm x bit( )3 2 2 2 x x + +b) Rt gn biu thc: ( )21 3 3 A Bi 2: (1,5 im)Cho ng thng (d): y = 2x + m 1a) Khi m = 3, tm a im A(a; -4) thuc ng thng (d).b) Tm m ng thng (d) ct cc trc ta Ox, Oy ln lt ti M v N sao cho tam gic OMN c din tch bng 1.Bi 3: (1,5 im) Cho phng trnh x2 2(m + 1)x + 4m = 0 (1)a) Gii phng trnh (1) vi m = 2.b) Tm m phng trnh (1) c nghim x1, x2 tha mn (x1 + m)(x2 + m) = 3m2 + 12Bi 4: (3 im) T im A bn ngoi ng trn (O), k cc tip tuyn Am, AN vi ng trn (M, N l cc tip im). ng thng d i qua A ct ng trn (O) ti hai im phn bit B,C (O khng thuc(d), B nm gia A v C). Gi H l trung im ca BC.a) Chng minh cc im O, H, M, A, N cng nm trn mt ng trn,b) Chng minh HA l tia phn gic ca MHN.c) Ly im E trn MN sao cho BE song song vi AM. Chng minh HE//CM.Bi 5 (1,0 im) Cho cc s thc dng x, y , z tha mn x + y + z = 4.Chng minh rng 1 11xy xz+ HNG DN GII:Phn trc nghim:CuCu Cu Cu Cu Cu Cu CuB D A D A B B CPhn t lun:Bi 1:a) Tm x bit( )3 2 2 2 x x + +3 2 2 2 2 x x + + 2 x . Vy 2 x 75C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netb) Rt gn biu thc: ( )21 3 3 1 3 3 3 1 3 1 A . Vy 1 A Bi 2:a) Thay m = 3 vo phng trnh ng thng ta c: y = 2x + 2. im A(a; -4) thuc ng thng (d) khi v ch khi: -4 = 2a + 2 suy ra a = -3.b) Cho x = 0 suy ra y = m 1suy ra: 1 ON m , cho y = 0 suy ra 12mxsuy ra 1 12 2m mOM hayOM din tch tam gic OMN = 1 khi v ch khi: OM.ON = 2 khi v ch khi 1 m.122mKhi v ch khi (m 1)2 = 4 khi v ch khi: m 1 = 2 hoc m 1 = -2 suy ra m = 3 hoc m = -1Vy din tch tam gic OMN = 1 khi v ch khi m = 3 hoc m = -1.Bi 3: Cho phng trnh x2 2(m + 1)x + 4m = 0 (1)a) Gii phng trnh (1) vi m = 2.b) Tm m phng trnh (1) c nghim x1, x2 tha mn (x1 + m)(x2 + m) = 3m2 + 12HD:a) Thay m = 2 vo phng trnh (1) ta c phng trnh:x2 6x + 8 = 0 Khi v ch khi (x 2)(x 4) = 0 khi v ch khi x = 2 hoc x = 4Vy vi m = 2 th phng trnh c 2 nghim x1 = 2 , x2 = 4.b) Ta c( ) ( )2 2' 1 4 1 0 m m m + vy phng trnh lun c nghim vi mi m.p dng nh l Vi-et ta c: ( ) 2 14S mP m + ' (x1 + m)(x2 + m) = 3m2 + 12 khi v ch khi x1x2 + (x1 + x2) m - 2 m2 12 = 0. S khi v ch khi : 4m + m.2(m + 1) 2m2 12 = 0 khi v ch khi 6m = 12 khi v ch khi m= 2Bi 5 :a) Theo tnh cht tip tuyn ct nhau ta c : 090 AMO ANO Do H l trung im ca BC nn ta c:090 AHO Do 3 im A, M, H, N, O thuc ng trn ng knh AOb) Theo tnh cht hai tip tuyn ct nhau ta c: AM = ANDo 5 im A, M, H, O, N cng thuc mt ng trn nn: AHM AHN (gc ni tip chn hai cung bng nhau)Do HA l tia phn gic ca MHN76EBHNO AMCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netc) Theo gi thit AM//BE nn MAC EBH ( ng v) (1)Do 5 im A, M, H, O, N cng thuc mt ng trn nn: MAH MNH (gc ni tip chn cung MH) (2)T (1) v (2) suy ra ENH EBH Suy ra t gic EBNH ni tipSuy ra EHB ENB M ENB MCB (gc ni tip chn cung MB)Suy ra: EHB MCB Suy ra EH//MC.Bi 5 (1,0 im) Cho cc s thc dng x, y , z tha mn x + y + z = 4.Chng minh rng 1 11xy xz+ Hng dn:V x + y + z = 4 nn suy ra x = 4 (y + z)Mt khc: 1 1 1 1 1 1 11 1 xxy xz x y z y z _+ + + , do x dng.(*)Thay x = 4 (y + z) vo (*) ta c : ( )221 1 1 1 1 14 2 2 0 0 y z y z y zy z y z y z _ _+ + + + + + , ,Lun ng vi mi x, y, z dng, du bng xy ra khi v ch khi : y = z = 1, x = 2.77EBHNO AMCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TO K THI TUYN SINH VO LP 10 THPT NM HC 2012NG NAIKha ngy : 29 , 30 / 6 / 2012Mn thi : TON HCThi gian lm bi : 120 pht( ny c 1 trang , 5 cu )Cu 1 : ( 1,5 im )1 / Gii phng trnh : 7x2 8x 9 = 0 .2 / Gii h phng trnh : 3x+2y =14x +5y = 6'Cu 2 : ( 2,0 im )1 / Rt gn cc biu thc : 12 +3 3 2 2M ; N3 2 1 2 / Cho x1 ; x2 l hai nghim ca phng trnh : x2 x 1 = 0 .Tnh : 1 21 1+x x .Cu 3 : ( 1,5 im )Trong mt phng vi h trc ta Oxy cho cc hm s :y = 3x2 c th ( P ) ;y = 2x 3 c th l ( d ) ;y = kx+ n c th l ( d1 ) vi k v n l nhng s thc .1 / V th ( P ) .2 / Tm k v n bit ( d1 ) i qua imT( 1 ; 2 ) v ( d1 ) // ( d ) .Cu 4 : ( 1,5 im )Mt tha t hnh ch nht c chu vi bng 198 m , din tch bng 2430 m2 . Tnh chiu di v chiu rng ca tha t hnh ch nht cho .Cu 5 : ( 3,5 im )78 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCho hnh vung ABCD . Ly im E thuc cnh BC , vi E khng trng B v E khng trng C . V EF vung gc vi AE , vi F thuc CD . ng thng AF ct ng thng BC ti G . V ng thng a i qua im A v vung gc vi AE , ng thng a ct ng thng DE ti im H .1 / Chng minh AE CDAF DE.2 / Chng minh rng t gic AEGH l t gic ni tip c ng trn .3 / Gi b l tip tuyn ca ng trn ngoi tip tam gic AHE ti E , bit b ct ng trung trc ca on thng EG ti im K . Chng minh rng KG l tip tuyn ca ng trn ngoi tip tam gic AHE .HNG DN GII:Cu 1 : ( 1,5 im )1 / Gii phng trnh : 7x2 8x 9 = 0 ( x1,2 = 4 797t )2 / Gii h phng trnh : 3x +2y =14x +5y = 6' ( x ; y ) = (1 ; 2 )Cu 2 : ( 2,0 im )1 / Rt gn cc biu thc : 12 +3 2 3 3M 2 33 3+ +( )22 13 2 2N 2 12 1 2 1 2 / Cho x1 ; x2 l hai nghim ca phng trnh : x2 x 1 = 0 .S = b1a ;P = c1a Nn : 1 21 2 1 2111x x 1 1+x x x x+ Cu 3 : ( 1,5 im )1 / V th ( P ) .2 / ( d1 ) // ( d ) nn k = 2 ; n 3 v i qua imT( 1 ; 2 ) nn x = 1 ; y = 2 . Ta c phng trnh : 2 = 1.2 + nn = 0Cu 4 : ( 1,5 im )Gi x ( m ) l chiu di tha t hnh chnht ( 49,5 < x < 99 )Chiu rng ca tha t hnh ch nht l : 99 x ( m )Theo bi ta c phng trnh : x ( x 99 ) = 2430Gii c : x1 = 54 ( nhn ); x2 = 45 ( loi )Vy chiu di tha t hnh chnht l 54 ( m )Chiu rng ca tha t hnh ch nht l : 99 54 = 45 ( m )Cu 5 : ( 3,5 im )1 / Chng minh t gic AEFD ni tip 1 1A D AEFDCE( g g )791211K IbaGHFEDCBAC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetAE AF=DC DEAE DC=AF DE2 / Ta c 2A ph vi1ATa c1Eph vi 1DM 1 1A D 2 1A E Suy ra t gic AEFD ni tip ng trn ng knh HEGi I trung im ca HEI l tm ng trn ngoi tip t gic AEFD cng l ng trn ngoi tipAHEI nm trn ng trung trc EG IE = IGV K nm trn ng trung trc EG KE = KGSuy raIEK =IGK ( c-c-c ) 0IGK IEK 90 KG IG ti G ca ng trn ngoi tipAHEKG l tip tuyn ca ng trn ngoi tipAHETHI TUYN SINH VO LP 10 CHUYN TNH NG NAINM HC 2012 - 2013Mn thi: Ton chungThi gian lm bi: 120 pht ( khng k thi gian giao )( thi ny gm mt trang, c bn cu)Cu 1: ( 2,5 im) .1/ Gii cc phng trnh :a/ 4 220 0 x x b/1 1 x x + 2/ Gii h phng trnh : 3 13x yy x + ' Cu 2 : ( 2,0 im) .Cho parabol y = x2 (P) v ng thng y = mx (d), vi m l tham s.1/ Tm cc gi tr ca m (P) v (d) ct nhau ti im c tung bng 9.2/ Tm cc gi tr ca m (P) v (d) ct nhau ti 2 im, m khong cch gia hai im ny bng6Cu 3 : ( 2,0 im)1/ Tnh : 1 1 3 1( ).2 3 2 3 3 3P + 2/ Chng minh : 5 5 3 2 2 3a b a b a b + + , bit rng0 a b + .Cu 4 : (3,5 im)Cho tam gic ABC vung A, ng cao AH. V ng trn tm O, ng knh AH, ng trn ny ct cc cnh AB, AC theo th t ti D v E .80 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net1/ Chng minh t gic BDEC l t gic ni tip c ng trn.2/ Chng minh 3 im D, O, E thng hng.3/ Cho bit AB = 3 cm, BC = 5 cm. Tnh din tch t gic BDEC.--------HT------THI TUYN SINH VO LP 10 CHUYN TNH NG NAINM HC 2012 - 2013Mn thi: Ton ( mn chuyn)Thi gian lm bi: 150 pht ( khng k thi gian giao )( thi ny gm mt trang, c nm cu)Cu 1. (1,5 im)Cho phng trnh 4 216 32 0 x x + ( vix R )Chng minh rng6 3 2 3 2 2 3 x + + + l mt nghim ca phng trnh cho.Cu 2. (2,5 im)Gii h phng trnh 2 ( 1)( 1) 62 ( 1)( 1) yx 6x x y xyy y x+ + + '+ + + ( vi , x R y R ).Cu 3.(1,5 im)Cho tam gic u MNP c cnh bng 2 cm. Ly n im thuc cc cnh hoc pha trong tam gic u MNP sao cho khong cch gia hai im tu ln hn 1 cm ( vi n l s nguyn dng). Tm n ln nht tho mn iu kin cho.Cu 4. (1 im)Chng minh rng trong 10 s nguyn dng lin tip khng tn ti hai s c c chung ln hn 9.Cu 5. (3,5 im)Cho tam gic ABC khng l tam gic cn, bit tam gic ABC ngoi tip ng trn (I). Gi D,E,F ln lt l cc tip im ca BC, CA, AB vi ng trn (I). Gi M l giao im ca ng thng EF v 81 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netng thng BC, bit AD ct ng trn (I) ti im N (N khng trng vi D), gii K l giao im ca AI v EF.1) Chng minh rng cc im I, D, N, K cng thuc mt ng trn.2) Chng minh MN l tip tuyn ca ng trn (I).----------HT-----------GII THI VO LP 10CHUYN LNG TH VINH NG NAINM 2012 2013Mn: Ton chung-----------------Cu 1: ( 2,5 im) .1/ Gii cc phng trnh :a/ 4 220 0 x x (*)t 2; ( 0) x t t (*) t2 t 20 = 0 (t1 = 5 (nhn) v t2 = - 4 ( loi)); Vi t = 5 => x2 = 5 x =5 tVy phng trnh c hai nghim x = 5v x = -5b/1 1 x x + ( iu kin1 x )2 2 2 2( 1) ( 1) 1 2 1 3 0 x x x x x x x + + + x(x-3) = 0 x = 0 ( loi) v x = 3 ( nhn).Vy phng trnh c mt nghim x = 3.2/ Gii h phng trnh : 3 13x yy x + ' T 3 3 3 0 3 3 y x y x y y y 13 1 3 1 4 2 123 3 3 3 72xx y x y x y xy x y x y x y xy t + + + ' ' ' ' ' + (nhn)82C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetVy h phng trnh c 2 nghim (x; y): 1 7 1 7( ; ), ( ; )2 2 2 2Cu 2 : ( 2,0 im) .1/ P.trnh honh giao im(P) v (d) :1 2200 ( ) 0xx mx x x mx m V giao im 2 2( ) : P y x y m . Vi y = 9 => m2 = 9 (m = 3 v m = -3)Vy vi3 m tth (P) v (d) ct nhau ti im c tung bng 9.2/ T cu 1 => (P) v (d) lun ct nhau ti hai im phn bit khi0 m .Khi giao im th nht l gc to O ( x = 0; y = 0), giao im th 2 l im A c ( x = m; y = m2).Khong cch gia hai giao im : AO = 2 4 4 26 6 0 m m m m + + (1)t 2; ( 0) t m t (1) 26 0 t t + (t1 = 3 ( nhn ) v t2 = - 2 ( loi))Vi t1 = 3 m2 = 3 , 3 m t( nhn)Vy vi3 m tth (P) ct (d) ti hai im c khong cch bng6 .Cu 3 : ( 2,0 im)1/ Tnh:1 1 3 1 2 3 2 3 3 1( ). . 24 3 2 3 2 3 3 3 3( 3 1)P + + + 2/ Ta c:5 5 3 2 2 3 5 5 3 2 2 3 3 2 2 3 2 2 3 3 2 22 2 20 ( ) ( ) 0 ( )( ) 0( ) ( )( ) 0a b a b a b a b a b a b a a b b a b a b a ba b a b a b ab+ + + + + + V : 2( ) 0 a b (vi mi a, bR ).0 a b + ( theo gi thit)2 20 a b ab + + ( vi mi a, b R )Nn bt ng thc cui ng.Vy5 5 3 2 2 3a b a b a b + +vi0 a b + (pcm)Cu 4 : (3,5 im)EDOHC BA1/ Ni H vi E .+090 HEA ( v AH l ng knh), 090 AHC ( AH l ng cao)=>AHE ACB (cng ph viEHC )(1)+ADE AHE ( gc ni tip cng chn cung AE) (2)T (1) v (2) =>ADE =ACB=>T gic BDEC ni tip ng trn ( c gc i bng gc k b gc i)2/ V 090 DAE => DE l ng knh => D, O, E thng hng (pcm).3/ Ta c BDEC ABC ADES S S 83C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net+ ABC vungc AH l ng cao:2 24 AC BC AB cm =>.62ABCAB ACs (cm2). 125AB ACDE AHBC (cm)( cng l ng knh t O).+ADEv ABCc : A chung ,ADE =ACB ( cu 1)=> ADE~ABC(g.g) => t s din tch bng bnh phng t ng dng : 222.ABC AEDAEDABCS DE S DESS BC BC _ ,+ 2 22 2 212(1 ) 6(1 )5 .5BDEC ABC ADE ABCDES S S SBC = 4,6176 (cm2)---------HT---------GII THI VO LP 10CHUYN LNG TH VINH NG NAINM 2012 2013Mn: Ton chuyn-----------------Cu 1: Phng trnh cho : 4 216 32 0 x x + ( vix R ) 2 2( 8) 32 0 x (1)Vi 6 3 2 3 2 2 3 x + + + 3 2 2 3 2 2 3 x + + + => 28 2 2 3 2 3 2 3 x + Th x vo v phi ca (1) ta c: 2 2 2( 8) 32 (8 2 2 3 2 3 2 3 8) 32 4(2 3) 4 3 12(2 3) 32 x + + + + =8 4 3 8 3 24 12 3 32 0 + + + ( v phi bng v tri)Vy 6 3 2 3 2 2 3 x + + + l mt nghim ca phng trnh cho ( pcm)Cu 2: H pt cho 2 ( 1)( 1) 62 ( 1)( 1) yx 6x x y xyy y x+ + + '+ + + (1)(2) ' ) 2 ( 1)( 1) 62 ( 1)( 1) 6x x y xyy y x xy+ + '+ + Thayx = 0,y = 0 th hkhng tho . Thay x = -1 v y = -1 vo, h khng tho =>( ; ) (0; 0); 0; 1 0; 1 0 6 0 x y xy x y xy + + (*)- Chia tng v ca hai phng trnh cho nhau : => 6( ) 6( )6x xyxy x y x yy xy +84C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetThay x = y, h pt c v phi bng nhau, v tri khc nhau(khng tho) =>0 x y ) (**)=> 6( ) x yxyx y+ (3)- Cng tng v (1) v (2)ca h ta c pt: 2(x+y)(x+1)(y+1) + 2xy = 0(4) (x + y) ( x + y + xy + 1) + xy = 06( ) 6( )( )( 1 ) 0x y x yx y x yx y x y+ ++ + + + + 6( 1)( )( 1 ) 0x yx y x yx y+ ++ + + + 6( )( 1)(1 ) 0 x y x yx y+ + + + 01 061 0x yx yx y+

+ +

+

- Vi x + y = 0 x = - y. Th vo h=> -2y2 = 0 (y = 0v x = 0) khng tho (*)- Vi x + y +1 =0 x = -y - 1 th vo phng trnh (1) ca h ta c :3 2 22 3 6 0 ( 2)(2 3) 0 y y y y y y + + + + + 22 0 22 3 0( )y yy y vn+

+ Vi y = - 2 => x = 1.Th voh tho, vy c nghim 1: (x; y) = (1; - 2)- Vi 61 0 6 0 6 x y x yx y+ + Th x = y -6 vo pt (2)ca h :(2) 3 22 7 16 6 0 y y y 222 1 0(2 1)( 4 6) 04 6 0yy y yy y+ + y2 - 4y - 6 = 0 122 102 10yy

+

2y +1 = 0 y3 = 12T ba gi tr ca y trn ta tm c ba gi tr x tng ng: 1234 104 10132xxx

+

Th cc gi tr(x; y) tm c vo h(tho).Vy h phng trnh choc 4 nghim ( x;y):(1; -2), (13 14 10; 2 10), ( 4 10; 2 10), ( ; ).2 2 + + Cu 3. (Cch 1)Tam gic u c cnh bng 2 cm th din tch bng 3 cm2 , tam gic u c cnh bng 1 cmth din tch bng 34 cm2 . Nu tam gic u c cnh > 1cm th din tch > 34 cm2Gi t ls tam gic u c cnh bng> 1cmcha c trong tam gic u c cnh 2 cm:1 4 t p ( vi t l s nguyn dng) => tmax =3.Theo nguyn l Drichen s c 1 trong t tam gic u c cnh > 1cm cha ti a 2 im tho mn khong cch gia hai im bt k lun > 1 cm.Vy s im tho yu cu bi ton l :2 4 n Vy nmax = 485C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net(Cch 2): Gii theo kin thc hnh hcNu ta chn 3 im 3 nh ca tam gic u cnh bng 2 cmv 3 ng trn ng knh 1 cm, cc ng trn ny tip xc vi nhau trung im mi cnh tam gic. => Cc imkhc trong tam gic cch 3 nh > 1cmch c th nm trongphn din tch cn li ca tam gic (ngoi phn din tch b ba hinh trn che ph), c gii hn bi 3 cung trn bn kinh 1 cm.V 3 dy cung l 3 ng trung bnh ca tam gic c di 1 cm => khong cch gia hai im bt k nm trong phn din tch cn li ca tam gic lun 1 cm.=> trong phn din tch ch ly c 1 im m khong cch n 3 nh ca tam gic lun > 1 cm.Vy s imln nht tho mn khong cch gia hai im bt k > 1cml :nmax = 3 + 1 = 4 im.Cu 4. Gi a v b l hai s bt k trong 10 s nguyn dng lin tip vi a > b ( a; b nguyn dng) 1 9 a b .Gi n l c chung ca a v b, khi : a = n.x v b = n.y( n, x, yl s nguyn dng).Va > b => x > y => 1 x y

1 91 . . 9 n x n y x yn n 91 9 nn Vy trong 10 s nguyn dng lin tip khng tn ti hai s c c chung ln hn 9.Cu 5.DKFNEMICBA1)Ni N v F, D v F.-XtANF v AFD c: AFN =ADF ( v AF l tt) vFAD chung =>ANFAFD (g.g) =>2AFAF .AFANAN ADAD (1)- XtAFI c: AF IF( v AF tip tuyn, FI l bn knh) vFK AI ( v AF v AE tt chungv AI ni tm) =>AFIvung ti F c FK l ng cao) => AK.AI = AF2(2)86C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net- XtANK vAID c:+IAD chung.+ T (1) v (2) => AN.AD = AK.AI => AN AIAK AD=>ANKAID (c.g.c) =>NKA =IDN (3)- T (3) => t gic DIKNni tip t (v c gc i bng gc k b gc i)=> cc im I,D,N,K cng thuc mt ng trn. (pcm).2)Ta cID DM( DM l tip tuyn, DI l bn knh) v IK KM( cu 1) => t gic DIKM ni tip ng trn ng knh MI.V 4 im D, I, K, N cng thuc mt ng trn ( cu 1)=> hai ng trn ny cng ngoi tip DIK => hai ng trn trng nhau => N cng nm trn ng trn ng knh MI=>INM = 900 .VIN l bn knh ng trn (I),MN IN => MN l tip tuyn ca ng trn (I) ti tip im N.(pcm).-----------HT----------87 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetGI GII:Cu 1c C = 1Cu 2a ( 2;1); Cu 2bb = - 1Cu 3aa = 1Cu 3bA ( -1 ; 1 ) ; B (2 ; 4 )Cu 4a1 12 0 > ; nn pt lun co 2 nghim phn bitvi moi xCu 4 a2 => x1 + x2 = - 5 ; x1x2 = 3Cu 4bGoi x ( km/h) la vt xe II => vt xe I la x + 10 ( km/h ) ; x> 0Th gian xe I i ht qg : 100x(h)Th gian xe II i ht qg : 10010 x +(h)PT100x - 10010 x + = 12 => x = 40KLCu 5: a1. MH = 20 ( cm ) ; ME = 12 ( cm)2. NPFE la h thang cnb )b1b288C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetTam giac ABC vung tai A co AH la g cao => AB2 = BH.BC (1)Tam giac BHE g dang vi tam giac BDC =>. .BH BEBH BC BD BEBD BC > (2)T (1) va (2) => AB2 = BD . BECu 1:(2,0 im)1. Cho biu thc P = x + 5. Tnh gi tr biu thc P ti x = 1.2. Hm s bc nht y = 2x + 1 ng bin hay nghch bin trn R? V sao?3. Gii phng trnh x2 + 5x + 4 = 0Cu 2: (2,5 im)1. Gii h phng trnh: 2 13 2 5x yx y+ ' 2. Cho biu thc Q = 1 1 1 2:1 1 1 x x x x x _ _+ + + , , vi x > 0 v x1.a) Rt gn Q.b) Tnh gi tr ca Q vi x = 7 43.Cu 3: (1,5 im)Khong cch gia hai bn sng A v b l 30 km. Mt ca n i xui dng t bn A n bn B ri li ngc dng t bn B v bn A. Tng thi gian ca n i xui dng v ngc dng l 4 gi . Tm vn tc ca ca n khi nc yn lng, bit vn tc ca dng nc l 4 km/h.Cu 4: (3,0 im)S GIO DC - O TOTNH NINH BNHK THI TUYN SINH VO LP 10 THPTNM HC 2012 2013Mn thi: TONThi gian lm bi: 120 pht (khng k thi gian giao ) thi gm 05 cu trong 01 trang89 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCho ng trn tm O bn knh R. Mt ng thng d khng i qua O v ct ng trn ti hai im phn bit A v B. Trn d ly im M sao cho A nm gia M v B. T M k hai tip tuyn MC v MD vi ng trn (C, D l cc tip im).1. Chng minh rng MCOD l t gic ni tip.2. Gi I l trung im ca AB. ng thng IO ct tia MD ti K. Chng minh rng KD. KM = KO. KI3. Mt ng thng i qua O v song song vi CD ct cc tia MC v MD ln lt ti E v F. Xc nh v tr ca M trn d sao cho din tch tam gic MEF t gi tr nh nht.Cu 5: (1,0 im)Cho a, b, c l cc s thc dng. Chng minh rng:4b c c a a b a b ca b c b c c a a b _ + + ++ + + + + + + ,------------------- Ht ----------------------HNG DN GII:Cu 1:1) Thay x = 1 vo biu thc P c: P = x + 5 = 1 + 5 = 6.2) Hm s ng bin trn R v a = 2 > 03)Ta thy a b + c = 1 5 + 4 = 0 nn pt c 2 nghim: x1 = 1;x2 = 4Cu 2:1. Vy h ptc nghim : x = 1 v y = 1.2. Vi x > 0 v x1, ta c:a) Qb)ViSuy ra :Cu 3:Gi vn tc ca ca n khi nc yn lng l x(km/h) (k:)90C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetVn tc ca ca n khi xui dng: x + 4 (km/h)Vn tc ca ca n khi ngc dng: x 4 (km/h)Thi gian ca n i xui dng:304 x + (h)Thi gian ca n i ngc dng: 304 x (h)Tng thi gian ca n i xui dng v ngc dng l 4h nn ta c phng trnh:304 x + + 304 x = 4 x2 15x 16 = 0Gii phng trnh trn ta c:

121( )16( )x khongthoa Kx thoa KVy vn tc ca ca n khi nc yn lng l 16km/h91C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCu 5: (cch 2)p dng bt ng thc Csi ta c:: du bng xy ra a = bTng t ta c :du = xy ra a = b = c92C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCu 1 (2 im). Cho phng trnh bc hai n x, tham s m: x2 + 2mx 2m 3 = 0(1)a) Gii phng trnh (1) vi m = -1.b) Xc nh gi tr ca m phng trnh (1) c hai nghim x1, x2 sao cho2221x x +nh nht. Tm nghim ca phng trnh (1) ng vi m va tm c.Cu 2 (2,5 im).1. Cho biu thcA=

,_

++

,_

+ ++xxxx xxxx33 13 3 14 3 2 338 3 34 633a) Rt gn biu thc A.b) Tm cc gi tr nguyn ca x biu thc A nhn gi tr nguyn.2. Gii phng trnh: ( ) 1 1 1 + + x x x xCu 3 (1,5 im). Mt ngi i xe p t A ti B, qung ng AB di 24 km. Khi i t B tr v A ngi tng vn tc thm 4 km/h so vi lc i, v vy thi gian v t hn thi gian i l 30 pht. Tnh vn tc ca xe p khi i t A ti B.S GIO DC - O TOTNH NINH BNHK THI TUYN SINH VO LP 10 THPT CHUYNMn thi: TONNgy thi: 26 / 6 / 2012Thi gian lm bi: 120 pht93 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCu 4 (3 im). Cho ABC nhn ni tip (O). Gi s M l im thuc on thng AB (M A, B); N l im thuc tia i ca tia CA sao cho khi MN ct BC ti I th I l trung im ca MN. ng trn ngoi tip AMN ct (O) ti im P khc A.1. C MR cc t gic BMIP v CNPI ni tip c.2. Gi s PB = PC. Chng minh rng ABC cn.Cu 5 (1 im). Cho x; y R , tha mn x2 + y2 = 1. Tm GTLN ca : 2 +yxPHNG DN GII:2) Gii pt :1 ) 1 ( 1 + + x x x x K :1 0 xt0 1 ; 0 b x a xTa c ' + + +(**) 1(*) 12 2b aab b aT tm c nghim ca pt l x = 0Cu 5 :T2 1 2 1 2 1 , 1 12 2+ + + y y x y xV ) 2 (2+ + y P xyxPthay vo12 2 + y xa v pt:0 1 2 2 2 ) 1 (2 2 2 2 + + + P y P y PDng iu kin c nghim ca pt bc hai1 P 22122MaxxPy ' S GIO DC V O TO K THI TUYN SINH VO 10 - THPTTNH LO CAI NM HC: 2012 2013MN: TONThi gian: 120 pht (khng k thi gian giao )Cu I: (2,5 im)1. Thc hin php tnh: ( ) ( )2 333a) 2 10 36 64 b) 2 3 2 5 . + + 2. Cho biu thc: P = 232a 4 1 11 a 1 a 1 a+ + a) Tm iu kin ca a P xc nh b) Rt gn biu thc P.Cu II: (1,5 im)1. Cho hai hm s bc nht y = -x + 2 v y = (m+3)x + 4. Tm cc gi tr ca m th ca hm s cho l:a) Hai ng thng ct nhaub) Hai ng thng song song.2. Tm cc gi tr ca a th hm s y = ax2 (a0) i qua im M(-1; 2).Cu III: (1,5 im)1. Gii phng trnh x 2 7x 8 = 094 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net2. Cho phng trnh x2 2x + m 3 = 0 vi m l tham s. Tm cc gi tr ca m phng trnh c hai nghim x1; x2 tha mn iu kin 3 31 2 1 2x x x x 6 + Cu IV: (1,5 im)1. Gii h phng trnh 3x 2y 1.x 3y 2 ' + 2. Tm m h phng trnh 2x y m 13x y 4m 1 '+ + c nghim (x; y) tha mn iu kin x + y > 1.Cu V: (3,0 im) Cho na ng trn tm O ng knh AB = 2R v tip tuyn Ax cng pha vi na ng trn i vi AB. T im M trn Ax k tip tuyn th hai MC vi na ng trn (C l tip im). AC ct OM ti E; MB ct na ng trn (O) ti D (D khc B).a) Chng minh AMOC l t gic ni tip ng trn.b) Chng minh AMDE l t gic ni tip ng trn.c) Chng mnh ADE ACO -------- Ht ---------HNG DN GII:Cu I: (2,5 im)1. Thc hin php tnh:3 3a) 2 10 36 64 8 100 2 10 12 + ( ) ( )2 33b) 2 3 2 5 2 3 2 5 3 2 2 5 2 + + + 2. Cho biu thc: P = 232a 4 1 11 a 1 a 1 a+ + a) Tm iu kin ca a P xc nh:P xc nh khi a 0 v a 1 b) Rt gn biu thc P.P =232a 4 1 11 a 1 a 1 a+ + =( ) ( )( ) ( )( ) ( )2 2 222a 4 1 a a a 1 1 a a a 11 a a a 1+ + + + + + + +=( ) ( )2 2 2 222a 4 a a 1 a a a a a a 1 a a a a a1 a a a 1+ + + + + +=( ) ( )22 2a1 a a a 1 + +=22a a 1 + +Vy vi a 0 v a 1 th P = 22a a 1 + +Cu II: (1,5 im)1. Cho hai hm s bc nht y = -x + 2 v y = (m+3)x + 4. Tm cc gi tr ca m th ca hm s cho l:a) hm sy = (m+3)x + 4 l hm s bc nht th m + 30 suy ra m-3. th ca hai hm s cho l hai ng thng ct nhauaa-1 m+3m-4Vy vi m-3 v m-4 th th ca hai hm s cho l hai ng thng ct nhau.b) th ca hm s cho l Hai ng thng song song95C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Neta a ' 1 m 3m 4b b' 2 4 + ' ' tha mn iu kin m-3Vy vi m = -4 th th ca hai hm s cho l hai ng thng song song.2. Tm cc gi tr ca a th hm s y = ax2 (a0) i qua im M(-1; 2).V th hm s y = ax2 (a0) i qua im M(-1; 2) nn ta thay x = -1 v y = 2 vo hm s ta c phng trnh 2 = a.(-1)2 suy ra a = 2 (tha mn iu kin a0)Vy vi a = 2 th th hm s y = ax2 (a0) i qua im M(-1; 2).Cu III: (1,5 im)1. Gii phng trnh x 2 7x 8 = 0 c a b + c = 1 + 7 8 = 0 suy ra x1= -1 v x2= 82. Cho phng trnh x2 2x + m 3 = 0 vi m l tham s. Tm cc gi tr ca m phng trnh c hai nghim x1; x2 tha mn iu kin 3 31 2 1 2x x x x 6 + . phng trnh c hai nghim x1; x2 th 0 1 m + 3 0 m 4Theo viet ta c: x1+ x2 =2 (1) v x1. x2 = m 3 (2)Theo u bi: 3 31 2 1 2x x x x 6 + ( )21 2 1 2 1 2x x x x 2x x + = 6(3)Th (1) v (2) vo (3) ta c: (m - 3)(2)2 2(m-3)=6 2m =12 m = 6 Khng tha mn iu kin m 4 vy khng c gi tr no ca m phng trnh c hai nghim x1; x2 tha mn iu kin 3 31 2 1 2x x x x 6 + .Cu IV: (1,5 im)1. Gii h phng trnh 3x 2y 1.x 3y 2 ' + ( ) 3 3y 2 2y 1 7y 7 y 1x 3y 2 x 1x 3y 2 ' ' ' 2. Tm m h phng trnh 2x y m 13x y 4m 1 '+ + c nghim (x; y) tha mn iu kin x + y > 1.2x y m 1 5x 5m x m x m3x y 4m 1 2x y m 1 2m y m 1 y m 1 ' ' ' '+ + + M x + y > 1 suy ra m + m + 1 > 12m > 0m > 0.Vy vi m > 0 th h phng trnh c nghim (x; y) tha mn iu kin x + y > 1.Cu V: (3,0 im) Cho na ng trn tm O ng knh AB = 2R v tip tuyn Ax cng pha vi na ng trn i vi AB. T im M trn Ax k tip tuyn th hai MC vi na ng trn (C l tip im). AC ct OM ti E; MB ct na ng trn (O) ti D (D khc B).a) Chng minh AMCO l t gic ni tip ng trn.b) Chng minh AMDE l t gic ni tip ng trn.c) Chng mnh ADE ACO Gii.a) 0MAO MCO 90 nn t gic AMCO ni tipb) 0MEA MDA 90 . T gic AMDE cD, E cng nhn AM di cng mt gc 900Nn AMDE ni tipc) V AMDE ni tip nn ADE AMEcngchan cung AE V AMCO ni tip nn ACO AMEcngchan cung AO Suy ra ADE ACO 96DOEMCBAC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TOGIA LAI chnh thcNgy thi: 26/6/2012K THI TUYN SINH VO LP 10 CHUYNNm hc 2012 2013Mn thi: Ton (khng chuyn)Thi gian lm bi: 120 phtCu 1. (2,0 im)Cho biu thc ( )x 2 x 2Q x xx 1x 2 x 1 _+ + + + ,, vi x 0, x 1 > a. Rt gn biu thc Qb. Tm cc gi tr nguyn ca x Q nhn gi tr nguyn.Cu 2. (1,5 im)Cho phng trnh 2x 2(m 1)x m 2 0 + + , vi x l n s,m R a. Gii phng trnh cho khi m 297 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netb. Gi s phng trnh cho c hai nghim phn bit 1x v 2x. Tm h thc lin h gia 1x v 2x m khng ph thuc vo m.Cu 3. (2,0 im)Cho h phng trnh (m 1)x (m 1)y 4mx (m 2)y 2+ + '+ , vim R a. Gii h cho khi m 3b. Tm iu kin ca m phng trnh c nghim duy nht. Tm nghim duy nht .Cu 4. (2,0 im)Cho hm s 2y x c th (P). Gi d l ng thng i qua im M(0;1) v c h s gc k.a. Vit phng trnh ca ng thng db. Tm iu kin ca k t d ct th (P) ti hai im phn bit.Cu 5. (2,5 im)Cho tam gic nhn ABC (AB < AC < BC) ni tip trong ng trn (O). Gi H l giao im ca hai ng cao BD v CE ca tam gic ABC (D AC, E AB) a. Chng minh t gic BCDE ni tip trong mt ng trnb. Gi I l im i xng vi A qua O v J l trung im ca BC. Chng minh rng ba im H, J, I thng hngc. Gi K, M ln lt l giao im ca AI vi ED v BD. Chng minh rng 2 2 21 1 1DK DA DM +HNG DN GII:Cu 1.a. ( )x 2 x 2Q x xx 1x 2 x 1 _+ + + + , ( )( ) ( )( ) _+ + + + ,2x 2 x 2x x 1x 1 x 1x 1 _+ + ,x 2 x 2xx 1 x 1 _+ + + ,x 1 1 x 1 1xx 1 x 1 _ + + + ,1 11 1 xx 1 x 1 _ + + ,1 1xx 1 x 1 + +x 1 x 1. xx 12 x. xx 12xx 1Vy2xQx 1b.Q nhn gi tr nguyn + + 2x 2x 2 2 2Q 2x 1 x 1 x 1 Q khi2x 1 khi 2 chia ht cho x 1 t tx 1 1x 1 2

x 0x 2x 1x 3 i chiu iu kin th x 2x 3

Cu 2. Cho pt 2x 2(m 1)x m 2 0 + + , vi x l n s,m R a.Gii phng trnh cho khi m 298C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetTa c phng trnh 2x 2x 4 0 + 2 2x 2x 4 0 x 2x 1 5 + + + ( )( )22x 1 5 5 + x 1 5 + x 1 5 x 1 5x 1 5 x 1 5 + + + Vy phng trinh c hai nghimx 1 5 vx 1 5 +b.Theo Vi-et, ta c 1 21 2x x 2m 2 (1)x x m 2 (2)+ + ' 1 21 2x x 2m 2m x x 2+ + ' +( )1 2 1 21 2x x 2 x x 2 2m x x 2 + + + ' +Suy ra( )1 2 1 2x x 2 x x 2 2 + + +1 2 1 2x x 2x x 6 0 + Cu 3.Cho h phng trnh (m 1)x (m 1)y 4mx (m 2)y 2+ + '+ , vim R a. Gii h cho khi m 3Ta c h phng trnh 2x 2y 12x 5y 2 + '

x y 6x 5y 2 + '

x 7y 1 'Vy h phng trnh c nghim( ) x; y vi( ) 7;1b. iu kin c nghim ca phng trnh( ) m 1m 11 m 2 ++( ) ( ) ( ) m 1 m 2 m 1 + +( ) ( ) ( ) m 1 m 2 m 1 0 + + + ( ) ( ) m 1 m 1 0 + m 1 0m 1 0+ ' m 1m 1 'Vy phng trnh c nghim khim 1 vm 1 Gii h phng trnh (m 1)x (m 1)y 4mx (m 2)y 2+ + '+ khi m 1m 1 '(m 1)x (m 1)y 4mx (m 2)y 2+ + '+ + '+ 4mx ym 1x (m 2)y 2 + + '+ 4mx ym 12ym 1 + '+ 4m 2xm 12ym 1.Vy h c nghim (x; y) vi _ + + ,4m 2 2;m 1 m 1Cu 4.a. Vit phng trnh ca ng thng dng thng d vi h s gc k c dng y kx b +ng thng d i qua im M(0; 1) nn 1 k.0 b + b 1 Vy d : y kx 1 +b.Phng trnh honh giao im ca (P) v d2x kx 1 +2x kx 1 0 + + , c 2k 4 d ct (P) ti hai im phn bit khi0 >2k 4 0 >2k 4 >2 2k 2 >k 2 >k 2k 2< >99C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetCu 5.a.BCDE ni tip 0BEC BDC 90 Suy ra BCDE ni tip ng trn ng knh BCb.H, J, I thng hngIB AB; CE AB (CH AB)Suy ra IB // CHIC AC; BD AC (BH AC)Suy ra BH // ICNh vy t gic BHCI l hnh bnh hnhJ trung im BC J trung im IHVy H, J, I thng hngc. 1ACB AIB AB2 ACB DEA cng b vi gc DEB ca t gic ni tip BCDE 0BAI AIB 90 + v ABI vung ti BSuy ra 0BAI AED 90 + , hay 0EAK AEK 90 + Suy ra AEK vung ti KXt ADM vung ti M (suy t gi thit)DK AM (suy t chng minh trn)www.VNMATH.Nh vy 2 2 21 1 1DK DA DM +S GIO DC V O TO K THI TUYN SINH LP 10 THPTQUNG NINHNM HC 2012 2013MN: TON(Dng cho mi th sinh d thi)Ngy thi: 28/6/2012Thi gian lm bi: 120 pht (Kh ngkt hig i a ng i a o )( thi ny c 01 trang)C uI. (2,0 im)1) Rt gn cc biu thc sau:a) A = 12 182 + b) B = 1 1 21 1 1 x x x+ + vi x 0, x 1100 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net2. Gii h phng trnh: 2x 52 4yx y+ '+ Cu II. (2,0 im)Cho phng trnh (n x): x2 ax 2 = 0 (*)1. Gii phng trnh (*) vi a = 1.2. Chng minh rng phng trnh (*) c hai nghim phn bit vi mi gi tr ca a.3. Gi x1, x2 l hai nghim ca phng trnh (*). Tm gi tr ca a biu thc:N= 2 21 1 2 2( 2)( 2) x x x x + + + +c gi tr nh nht.C u I II. (2,0 im)Gii bi ton bng cch lp phng trnh hoc h phng trnh.Qung ng sng AB di 78 km. Mt chic thuyn my i t A v pha B. Sau 1 gi, mt chic ca n i t B v pha A. Thuyn v ca n gp nhau ti C cch B 36 km. Tnh thi gian ca thuyn, thi gian ca ca n i t lc khi hnh n khi gp nhau, bit vntc ca ca n ln hn vn tc ca thuyn l 4 km/h.C uIV . (3,5 im)Cho tam gic ABC vung ti A, trn cnh AC ly im D (D A, D C). ng trn (O) ng knh DC ct BC ti E (E C).1. Chng minh t gic ABED ni tip.2. ng thng BD ct ng trn (O) ti im th hai I. Chng minh ED l tia phn gic ca gc AEI.3. Gi s tg ABC 2Tm v tr ca D trn AC EA l tip tuyn ca ng trnng knh DC.C u V . (0.5 im) Gii phng trnh:7 2 (2 ) 7 x x x x + + HNG DN GII:Cu IV :c. EA l tip tuyn ca .Trn, . knh CD th gc E1 = gc C1(1)M t gic ABED ni tip nn gc E1 = gc B1(2)T (1) v (2)gc C1 = gc B1ta li c gc BAD chung nn ABD ACB ABADACAB AB2 = AC.AD AD = ACAB2 ( I )Theo bi ra ta c : tan (ABC) = ABAC= 2nn 21ACAB( II )T (I) v (II) AD = 2AB.Vy AD = 2AB th EA l tip tuyn ca T, knh CDCu V:Gii phng trnh:7 2 (2 ) 7 x x x x + + tt x 7 ;v x K v, t 0t v v t ). 2 ( 22+ + ... 0 ) 2 )( ( t v t v t hoc t=2Nut= 2 th 2 7 x x = 3 (TM)Nu t = v thx x 7 x = 3,5101C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TO K THI TUYN SINH VO LP 10 THPTKHNH HA NM HC 2011 - 2012Mn thi: TONNgy thi : 21/06/2011102 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetThi gian lm bi: 120 phtBi 1( 2 im)1) n gin biu thc:A2 3 6 8 42 3 4+ + + ++ +2) Cho biu thc:1 1( ); ( 1)1 1P a aa a a a + Rt gn P v chng tP0Bi 2( 2 im)1) Cho phng trnh bc hai x2 + 5x + 3 = 0 c hai nghim x1; x2. Hy lp mt phng trnh bc hai c hai nghim (x12 + 1 ) v ( x22 + 1).2) Gii h phng trnh 2 3424 112x yx y+ ' Bi 3( 2 im)Qung ng t A n B di 50km.Mt ngi d nh i xe p t A n B vi vn tc khng i.Khi i c 2 gi,ngi y dng li 30 pht ngh.Mun n B ng thi gian nh,ngi phi tng vn tc thm 2 km/h trn qung ng cn li.Tnh vn tc ban u ca ngi i xe p.Bi 4( 4 im)Cho tam gic ABC c ba gc nhn v H l trc tm.V hnh bnh hnh BHCD.ng thng i qua D v song song BC ct ng thng AH ti E.1) Chng minh A,B,C,D,E cng thuc mt ng trn2) Chng minhBAE DAC 3) Gi O l tm ng trn ngoi tip tam gic ABC v M l trung im ca BC,ng thng AM ct OH ti G.Chng minh G l trng tm ca tam gicABC.4) Gi s OD = a.Hy tnh di ng trn ngoi tip tam gic BHC theo a103C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetHNG DN GII:Bi 13) A2 3 2 6 8 2 ( 2 3 4)(1 2)1 22 3 4 2 3 4+ + + + + + + + ++ + + +4)21 1( ); 112 1 1 2 1 1; : 1( 1 1) 0; 1a a a aP a aa aa a a a vi aP a a+ + + + Bi 2 x2 + 5x + 3 = 01) C25 12 13 0 >Nn pt lun c 2 nghim phn bit x1+ x2 = - 5 ; x1x2 = 3Do S = x12 + 1 + x22 + 1 = (x1+ x2)2 - 2 x1x2 + 2 = 25 6 + 2= 21V P = (x12 + 1) (x22 + 1) = (x1x2)2 + (x1+ x2)2 - 2 x1x2 + 1 = 9 + 20 = 29Vy phng trnh cn lp lx2 21x + 29 = 02) K 0; 2 x y 2 3144 2 72 232 3 1 4 12 3 343 222xx x yxyyx yx y+ ' ' ' '+ + Vy HPT c nghim duy nht ( x ;y) = ( 2 ;3)Bi 3 :Gi x(km/h) l vtc d nh; x > 0 ; c 30 pht = (h) Th gian d nh : 50( ) hxQung ng i c sau 2h : 2x (km) Qung ng cn li : 50 2x (km)Vn tc i trn qung ng cn li : x + 2 ( km/h)Th gian i qung ng cn li : 50 2( )2xhx+Theo bi ta c PT:1 50 2 5022 2xx x+ + +Gii ra ta c : x = 10 (tha K bi ton)Vy Vn tc d nh : 10 km/hBi 4 :Gii cuc)V BHCD l HBH nn H,M,D thng hngTam gic AHD c OM l TBnh => AH = 2 OMV AH // OM2 tam gic AHG v MOG c ( ) HAG OMG slt AGH MGO ( )( )2AHG MOG G GAH AGMO MG 104AB CEDHOMGC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetHay AG = 2MGTam gic ABC c AM l trung tuyn; G AMDo G l trng tm ca tam gic ABCd)BHC BDC ( v BHCD l HBH)c B ;D ;C ni tip (O) bn knh l aNn tam gic BHC cng ni tip (K) c bn knh aDo C (K) =2 a ( VD)S GIO DC-O TO K THI TUYN SINH VO 10 THPT NM2012BNH NH Kha ngy 29 thng 6 nm 2012Mn thi: TONNgy thi: 30/6/2012Thi gian lm bi: 120 pht (khng k thi gian giao )Bi 1: (3, 0 im)Hc sinh khng s dng my tnh b tia) Gii phng trnh: 2x 5 = 0b) Gii h phng trnh: y x 25x 3y 10 ' c) Rt gn biu thc25 a 3 3 a 1 a 2 a 8Aa 4a 2 a 2 + + + + + vi a 0, a 4 d) Tnh gi tr ca biu thcB 4 2 3 7 4 3 + + Bi 2: (2, 0 im)Cho parabol (P) v ng thng (d) c phng trnh ln lt l2y mx v( ) 2 1 y m x m + (m l tham s, m 0).a) Vi m =1 , tm ta giao im ca (d) v (P).b) Chng minh rng vi mi m 0 ng thng (d) lun ct parabol (P) ti hai im phn bit.Bi 3: (2, 0 im)Qung ng t Quy Nhn n Bng Sn di 100 km. Cng mt lc, mt xe my khi hnh t Quy Nhn i Bng Sn v mt xe t khi hnh t Bng Sn i Quy Nhn. Sau khi hai xe gp nhau, xe my i 1 gi 30 pht na mi n Bng Sn. Bit vn tc hai xe khng thay i trn sut qung ng i v vn tc ca xe my km vn tc xe t l 20 km/h. Tnh vn tc mi xe.Bi 4: (3, 0 im)Cho ng trn tm O ng knh AB = 2R. Gi C l trung im ca OA, qua C k dy MN vung gc vi OA ti C. Gi K l im ty trn cung nh BM, H l giao im ca AK v MN.a) Chng minh t gic BCHK l t gic ni tip.b) Chng minh AK.AH = R2c) Trn KN ly im I sao cho KI = KM, chng minh NI = KB.HNG DN GII:Bi 1:a) 2x 5 = 0 52 5 0 2 52x x x 105 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Netb) y x 2 5x 5y 10 2y 20 y 105x 3y 10 5x 3y 10 y x 2 x 8 + ' ' ' ' c)( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( )( )( ) ( )2222 25 a 3 a 2 3 a 1 a 2 a 2 a 85 a 3 3 a 1 a 2 a 8Aa 4 a 2 a 2a 2 a 2a 8a 165a 10 a 3 a 6 3a 6 a a 2 a 2 a 8 a 8a 16a 2 a 2 a 2 a 2 a 2 a 2 + + + + + + + + + + + ++ + + + + + +( )( )2a 4a 4 4 aa 4 d) ( ) ( )2 2B 4 2 3 7 4 3 3 1 2 3 3 1 2 3 3 1 2 3 3 + + + + + + + + Bi 2:a) Vi1 m ( ) P v( ) d ln lt tr thnh 2; 2 y x y x .Lc phng trnh honh giao im ca( ) P v( ) d l: 2 22 2 0 x x x x + c 1 1 2 0 a b c + + + nn c hai nghim l 1 21; 2 x x .Vi 1 11 1 x y Vi 2 22 4 x y Vy ta giao im ca( ) P v( ) d l( ) 1; 1 v( ) 2; 4 .b) Phng trnh honh giao im ca( ) P v( ) d l: ( ) ( ) ( )2 22 1 2 1 0 * mx m x m mx m x m + + .Vi0 m th( ) * l phng trnh bc hai n x c ( ) ( )22 2 22 4 1 4 4 4 4 5 4 0 m m m m m m m m + + + + >vi mi m. Suy ra( ) * lun c hai nghim phn bit vi mi m. Hay vi mi m 0 ng thng (d) lun ct parabol (P) ti hai im phn bit.Bi 3:i '1 30 1, 5 h h t a im :- Quy Nhn l A- Hai xe gp nhau l C- Bng Sn l BGi vn tc ca xe my l( ) / x km h. K :0 x > .Suy ra :Vn tc ca t l( ) 20 / x km h +.Qung ng BC l :( ) 1, 5x kmQung ng AC l :( ) 100 1, 5x km Thi gian xe my i t A n C l :( )100 1, 5xhxThi gian t my i t B n C l :( )1, 520xhx +V hai xe khi hnh cng lc, nn ta c phng trnh : 100 1, 5 1, 520x xx x+106100-1,5x1,5xAB CC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetGii pt :( ) ( )2 2 22100 1, 5 1, 5100 1, 5 20 1, 5 100 2000 1, 5 30 1, 5203 70 2000 0x xx x x x x x xx xx x + + + 2' 35 3.2000 1225 6000 7225 0 ' 7225 85 + + > Phng trnh c hai nghim phn bit : 135 85403x+ (tha mn K)235 85 503 3x (khng tha mn K)Vy vn tc ca xe my l 40 / km h.Vn tc ca t l( ) 40 20 60 / km h + .Bi 4:a) T gic BCHK l t gic ni tip .Ta c : 090 AKB (gc ni tip chn na ng trn)hay ( )0 090 ; 90 HKB HCB gt T gic BCHK c 0 0 090 90 180 HKB HCB + + t gic BCHK l t gic ni tip.b)2. AK AH R D thy( )2ACH AKB . . . 22AC AH Rg g AK AH AC AB R RAK AB c) NI KB OAM c( ) OA OM R gt OAM cn ti( ) 1 OOAM c MC l ng cao ng thi l ng trung tuyn (gt)OAM cn ti ( ) 2 M( ) ( ) 1 & 2 OAM l tam gic u 0 0 060 120 60 MOA MON MKI KMI l tam gic cn (KI = KM) c 060 MKI nn l tam gic u( ) 3 MI MK .D thyBMK cn ti B c 0 01 1120 602 2MBN MON nn l tam gic u( ) 4 MN MB Gi E l giao im ca AK v MI.D thy 006060NKB NMBNKB MIKMIK ) KB // MI (v c cp gc v tr so le trong bng nhau) mt khc ( ) AK KB cmt nnAK MI ti E 090 HME MHE .Ta c : ( ) ( ) 009090ddHAC AHCHME MHE cmt HAC HMEAHC MHE ) mt khc HAC KMB (cng chn KB) HME KMB hay ( ) 5 NMI KMB T( ) ( ) ( ) ( ) 3 , 4 & 5 . . IMN KMB c g c NI KB (pcm)107EIHNMCAOBKC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TO THI TUYN SINH LP 10 THPTBC GIANG NM HC 2012 2013Mn thi: TonThi gian : 120 pht khng k thi gian giao Ngy thi30 thng 6 nm 2012Cu 1. (2 im) 1.Tnh 122 1--2 .Xc nh gi tr ca a,bit th hm s y = ax - 1 i qua imM(1;5)Cu 2: (3 im)1.Rt gn biu thc: 1 2 3 2( ).( 1)2 2 2a aAa a a a- += - +- - - vi a>0,a 4 2.Gii h pt: 2 5 93 5x yx y- = + =3. Chng minh rng pt:21 0 x mx m + + - =lun c nghim vi mi gi tr ca m.Gi s x1,x2 l 2 nghim ca pt cho,tm gi tr nh nht ca biu thc2 21 2 1 24.( ) B x x x x = + - +Cu 3: (1,5 im)Mt t ti i t A n B vi vn tc 40km/h. Sau 2 gi 30 pht th mt t taxi cng xut pht i t A n B vi vn tc 60 km/h v n B cng lc vi xe t ti.Tnh di qung ng AB.Cu 4: (3 im)Cho ng trn (O)v mt im A sao cho OA=3R. Qua A k 2 tip tuyn AP v AQ ca ng trn(O),vi P v Q l 2 tip im.Ly M thuc ng trn (O) sao cho PM song song vi AQ.Gi N l giao im th 2 ca ng thng AM v ng trn (O).Tia PN ct ng thng AQ ti K.1.Chng minh APOQ l t gic ni tip.2.Chng minh KA2=KN.KP3.K ng knh QS ca ng trn (O).Chng minhtiaNS l tia phn gic ca gcPNM.4. Gi G l giao im ca 2 ngthng AO v PK .Tnh di on thng AG theo bn knh R.Cu 5: (0,5im)Cho a,b,c l 3 s thc khc khng v tho mn:2 2 22013 2013 2013( ) ( ) ( ) 2 01a b c b c a c a b abca b c + + + + + + =+ + =Hy tnh gi tr ca biu thc 2013 2013 20131 1 1Qa b c= + +108 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetHNG DN CHM(tham kho)Cu Ni dung im1 121 2 1 2 12 2 2 2 1 2 12 1 ( 2 1).( 2 1) ( 2) 1)+ +- = - = - = + - =- - + -KL:12 Do th hm s y = ax-1 i qua M(1;5) nn ta c a.1-1=5a=6KL:12 12 ( 1).( 2)( ).( 1)( 2) ( 2) 22 1( ).( 1 1) . 1( 2)a a aAa a a a aaa aa a a- -= - + =- - --= - + = =-KL:0,50,522 5 9 2 5 9 2 5 9 13 5 15 5 25 17 34 2x y x y x y yx y x y x x - = - = - = =- + = + = = = KL:13Xt Pt: 21 0 x mx m + + - =2 2 2 4( 1) 4 4 ( 2) 0 m m m m m = - - = - + = - Vy pt lun c nghim vi mi mTheo h thc Viet ta c1 21 21x x mx x m+ =- = -Theo bi 2 2 21 2 1 2 1 2 1 2 1 22 2 224.( ) ( ) 2 4.( )2( 1) 4( ) 2 2 4 2 1 1( 1) 1 1B x x x x x x x x x xm m m m m m m mm= + - + = + - - += - - - - = - + + = + + += + + VyminB=1 khi v ch khi m = -1KL:0,250,250,53 Gi di qumg ng AB l x (km) x>0Thi gian xe ti i t A n B l 40xhThi gian xe Taxi i t A n B l :60xhDo xe ti xut pht trc2h30pht = 52nn ta c pt540 60 23 2 300300x xx xx- = - = =Gi tr x = 300 c tho mn K0,250,250,250,250,250,25109C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetVy di qung ng AB l 300 km.4 1Xt t gic APOQ c090 APO=(Do AP l tip tuyn ca (O) P)090 AQO= (Do AQ l tip tuyn ca (O) Q) 0180 APO AQO + = ,m hai gc ny l 2 gc i nn t gic APOQ l t gic ni tip0,752Xt AKN v PAK c AKP l gc chung APN AMP = ( Gc ntcng chn cung NP)M NAK AMP =(so le trong ca PM //AQAKN ~ PKA (gg) 2.AK NKAK NK KPPK AK = = (pcm)0,753 K ng knh QS ca ng trn (O)Ta c AQ^QS (AQ l tt ca (O) Q)M PM//AQ (gt) nn PM^QSng knh QS^PM nn QS i qua im chnh gia ca cung PM nh sd PS sdSM = PNS SNM =(hai gc nt chn 2 cung bng nhau)Hay NS l tia phn gic ca gc PNM0,754Chng minh c AQO vung Q, cQG^AO(theo Tnh cht 2 tip tuyn ct nhau)Theo h thc lng trong tam gic vung ta c2 221.3 31 833 3OQ ROQ OI OA OI ROA RAI OA OI R R R= = = = = - = - =Do KNQ ~ KQP (gg)2. KQ KN KP =m 2. AK NK KP = nn AK=KQVy APQ c cc trung tuyn AI v PK ct nhau G nn G l trng tm2 2 8 16.3 3 3 9AG AI R R = = =0,755 Ta c:0,25110GKNSMIQPAOC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net2 2 22 2 2 2 2 22 2 2 2 2 22 22( ) ( ) ( ) 2 02 0( ) ( ) (2 ) 0( ) ( ) ( ) 0( )( ) 0( ).( ).( ) 0a b c b c a c a b abca b a c b c b a c a c b abca b b a c a c b abc b c a cab a b c a b c a ba b ab c ac bca b a c b c+ + + + + + = + + + + + + = + + + + + + = + + + + + = + + + + = + + + =*TH1: nu a+ b=0Ta c 2013 2013 20131 1a b a bc a b c =- =- = + + = ta c 2013 2013 20131 1 11 Qa b c= + + =Cc trnghp cn li xt tng tVy 2013 2013 20131 1 11 Qa b c= + + =0,25S GIO DC V O TO THI TUYN SINH LP 10 THPTYN BI NM HC 2012 2013Mn thi: TONThi gian : 120 pht (khng k thi gian giao )Kha ngy23 thng 6 nm 2012( thi c 01 trang, gm 05 cu)Cu 1: (2,0 im)1. Cho hm s y = x + 3 (1)a. Tnh gi tr ca y khi x = 1b. V th ca hm s (1)2. Gii phng trnh: 4x 7x + 3 = 0Cu 2: (2,0 im)Cho biu thc M =+ 1. Tm iu kin ca x biu thc M c ngha. Rt gn biu thc M.2. Tm cc gi tr ca x M > 1Cu 3: (2,0 im)Mt i th m phi khai thc 260 tn than trong mt thi hn nht nh. Trn thc t, mi ngy i u khai thc vt nh mc 3 tn, do h khai thc c 261 tn than v xong trc thi hn mt ngy.Hi theo k hoch mi ngy i th phi khai thc bao nhiu tn than?Cu 4: (3,0 im)Cho na ng trn tm O, ng knh AB = 12 cm. Trn na mt phng b AB cha na ng trn (O) v cc tia tip tuyn Ax, By. M l mt im thuc na ng trn (O), M khng trng vi A v B. AM ct By ti D, BM ct Ax ti C. E l trung im ca on thng BD.1. Chng minh: AC . BD = AB.2. Chng minh: EM l tip tuyn ca na ng trn tm O.3. Ko di EM ct Ax ti F. Xc nh v tr ca im M trn na ng trn tm O sao cho din tch t gic AFEB t gi tr nh nht? Tm gi tr nh nht .Cu 5: (1,0 im)Tnh gi tr ca biu thc T = x + y + z 7 bit:x + y + z = 2 + 4 + 6 + 45111 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.NetS GIO DC V O TOK THI TUYN SINH L 10 THPTLM NG MN THI : TONKha ngy : 26 thng 6 nm 2012( thi gm 01 trang)Thi gian lm bi : 120 phtCu 1: (0,75)Tnh : 18 2 2 32 + Cu 2: (0,75)Gii h phng trnh : 2 3 14 3 11x yx y '+ Cu 3: (0,75) Cho tam gic ABC vung ti A, ng cao AH. Bit BH = 9cm, Ch = 16cm.Tnh di cc on thng AH, BH, AC.Cu 4: (0,75)Cho hai ng thng (d) : y = (m-3)x + 16 (m 3) v (d): y = x + m2.Tm m (d) v (d) ct nhau ti mt im trn trc tungCu 5: (0,75)Cho AB l dy cung ca ng trn tm O bn knh 12cm. Bit AB = 12cm . Tnhdin tch hnh qut to bi hai bn knh OA, OB v cung nh AB.Cu 6: (1) Cho hm s y = ax2 (a 0) c th (P).a)Tm a bit (P) i qua im A(2;4)b)Tm k ng thng (d) : y = 2x + k lun ct (P) ti 2 im phn bit.Cu 7: (0,75)Hnh nn c th th tch l 320cm3, bn knh ng trn l 8cm. Tnh din tch tonphn ca hnh nn .Cu 8: (1)Cho ng trn (O) ng knh AB, M l trung im ca OA. Qua M v dy cung CDvung gc vi OA.a) Chng minh t gic ACOD l hnh thoi .b) Tia CO ct BD ti I. Chng minh t gic DIOM ni tip.Cu 9: (1) Hai i cng nhn cng o mt con mng . Nu h cng lm th trong 8 gi xongvic. Nu h lm ring th i A hon thnh cng vic nhanh hn i B 12 gi. Hi nulm ring th mi i phi lm trong bao nhiu gi mi xong vic.Cu 10: (0,75)Rt gn :37 20 3 37 20 3 + +Cu 11: (1) Cho phng trnh : x2 2(m-2)x - 3m2 +2 = 0(x l n, m l tham s )Tm m phng trnh c 2 nghim x1; x2tha : x1(2-x2) +x2(2-x1) = -2Cu 12: (0,75) Cho na ng trn (O) ng knh AB, v cc tip tuyn Ax v By cng pha vina ng trn , M l im chnh gia cung AB, N l mt im thuc on OA( ) , N O N A . ng thng vung gc vi MN ti M ct Ax v By ln lt ti C v D.Chng minh :AC = BN112 CHNH THCC:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net113C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Net114C:\Users\ANH BINH\AppData\Local\Temp\Rar$DI00.762\WWW.ToanTrungHocCoSo.ToanCapBa.Ne