S GIO DC O TO BC NINH THI GII TON TRN MY TNH CASIO 2004 Thi gian 150 pht ------------------------------------------------------------- ( kt qu tnh ton gn nu khng c quy nh c th c ngm hiu l chnh xc ti 9 ch s thp phn ) Bi 1 : Cho hm sf(x) = a, Tnh gn ng n 5 ch s thp phn gi tr hm s ti x = 1 +b, Tnh gn ng n 5 ch s thp phn gi tr cc s a , b sao cho ng thng y =ax +b l tip tuyn ca th hm s ti im c honh x = 1 + Bi 2 : Tnh gn ng n 5 ch s thp phn gi tr ln nht ca hm s f(x)= trn tpcc s thc S={x: } Bi 3 : Cho ;vi0 n 998 , Tnh gn ng gi tr nh nht [ ] Bi 4 : Tnh gn ng n 5 ch s thp phn gi tr ca im ti hn ca hm sf(x) = trn on[0;2 ] Bi 5 : Trong mt phng to Oxy , cho hnh ch nht c cc nh (0;0) ; (0;3) ; (2;3) ; (2;0) c di n v tr mi bng vic thc hin lin tip 4 php quay gc theo chiu kimng h vi tm quay ln lt l cc im (2;0) ; (5;0) ; (7;0) ; (10;0) . Hy tnh gnng n 5 ch s thp phn gi tr din tch hnh phng gii hn bi ng cong do im(1;1) vch ln khi thc hin cc php quay k trn v bi cc ng thng : trc Ox ; x=1; x=11 Bi 6 : Mt bn c vung gm 1999x1999 mi c xp 1 hoc khng xp qun c no .Tm s b nht cc qun c sao chokhi chn mt trng bt k , tng s qun c tronghng v trong ct cha t nht l 199 Bi 7 : Tam gic ABC c BC=1 , gc . Tnh gn ng n 5 ch s thp phn gi trkhong cch gia tm ng trn ni tip v trng tm ca tam gic ABC. Bi 8 : Tnh gn ng n 5 ch s thp phn gi tr cc h s a, b ca ng thng y=ax+b ltip tuyn ti M(1;2) ca Elp =1 bit Elp i qua im N(-2; ) Bi 9 : Xt cc hnh ch nht c lt kht bi cc cp gch lt hnh vung c tng din tch l1 ,vic c thc hin nh sau : hai hnh vung c xp nm hon ton trong hnh ch nhtm phn trong ca chng khng ln nhau cc cnh ca 2 hnh vung th nm trn hocsong song vi cc cnh ca hnh ch nht . Tnh gn ng khng qu 5 ch s thp phngi tr nh nht din tch hnh ch nht k trn Bi 10 : Cho ng congy =, m l tham s thc. a, Tnh gn ng n 5 ch s thp phn gi tr ca m tim cn xin ca th hm sTo vi cc trc to tam gic c din tch l 2b, Tnh gn ng n 5 ch s thp phn gi tr m ng thng y=m ct th ti haiim A, B sao cho OA vung gc vi OB HT UBND TNH BC NINH THI HC SINH GII THPT S GIO DC O TO Gii ton trn MTT CASIO nm 2004 2005 Thi gian : 150pht ----------------------------------------------------------------- Bi 1 ( 5 im ) Trong cc s sau 2; ; ;6 3 4 3 s no l nghim dng nh nht ca phng trnh : 2sin sin 2 cos 2cos x x x + = + xBi 2 ( 5 im ) Gii h :22log 4.3 67. log 5.3 1xxxx + =+ = Bi 3 ( 5 im ) Cho a thc : ( )3 22 5 1 f x x x x = +a, Tnh ( gn ng n 5 ch s thp phn ) s d ca php chia f(x) cho 12x + b, Tnh ( gn ng n 5 ch s thp phn ) nghim ln nht ca phng trnh : f(x) = 0 Bi 4 ( 5 im )

Bi5 ( 5 im ) 1. Tm tt c cc cp s t nhin (x,y) sao cho x l c ca v y l c ca2. Chng minh rng phng trnh c nghim t nhin khi v ch khi a=3 Tm tt c cc cp s t nhin (x,y) l nghim ca phng trnh3. Tm tt c cc b s t nhin (x,y,z) l nghim ca phng trnh :Bi 6 ( 5 im ) : T mt phi hnh nn chiu cao12 3 h =v bn knh yR=5 2 c th tin c mt hnh tr cao nhng y hp hoc hnh tr thp nhng y rng . Hy tnh ( gn ng 5 ch s thpphn ) th tch ca hnh tr trong trng hp tin b t vt liu nht .Bi 7 ( 5 im ) : Cho hm sy=c th (C) , ngi ta v hai tip tuyn ca th ti im chonh v ti im cc i ca th hm s . Hy tnh ( gn ng 5 ch s thp phn )din tch tam gic tao bi trc tung v hai tip tuyn cho. Bi 8 ( 5 im ) Hy tnh ( gn ng 4 ch s thp phn ) l nghim ca phng trnh:

Bi 9 ( 5 im ) Hy tnh ( gn ng 4 ch s thp phn )Bi 10 ( 5 im ) Tm ch s hng n v ca s HT CHN I TUYN TRUNG HC C S (S GIO DC BC NINH NM 2005) Bi 1 : 1.1: Tm tt c cc s c 10 ch s c ch s tn cng bng 4 v l lu tha bc 5 ca mt s t nhin. S : 1073741824, 2219006624, 4182119424 , 7330402241.2 : Tm tt c cc s c 10 ch s c ch s u tin bng 9 v llu tha bc nm ca mt s t nhin. S : 9039207968 , 9509900499 Bi 2 :2.1. Tm s c 3 ch s l lu tha bc 3 ca tng ba ch s ca n. S : 512 2.2. Tm s c 4 ch s l lu tha bc 4 ca tng bn ch s c n. S : 2401 2.3. Tn ti hay khng mt s c nm ch s l lu tha bc 5 ca tng nm ch s ca n ? S : khng c s no c 5 ch s tho mn iu kiu bi Bi 3 : 3.1. Cho a thc bc 4f(x) = x4+bx3+cx2+dx+43 c f(0) = f(-1);f(1) = f(-2) ; f(2) = f(-3) . Tm b, c, d S : b = 2 ; c = 2 ; d = 1 3.2. Vi b, c, d va tm c, hy tm tt c cc s nguyn n sao cho f(n) = n4+bn3+cn2+n+43 l s chnh phng. S : n = -7 ; - 2 ; 1 ; 6 Bi 4 : T th trn A n Bc Ninh c hai con ng tovi nhau gc 600 . Nu i theo ng lin tnh bn tri n th trn B th mt 32 km ( k t th trn A), sau r phi theo ng vung gc v i mt on na th s n Bc Ninh.Cn nu t A i theo ng bn phi cho n khi ct ng cao tc th c ng na qung ng, sau r sang ng cao tc v i nt na qung ng cn li th cng s n Bc Ninh .Bit hai con ng di nh nhau. 4.1. Hi i theo hng c on ng cao tc n Bc Ninh t th trn A thi nhanh hn i theo nglin tnh bao nhiu thi gian( chnh xc n pht), bit vn tc xe my l 50 km/h trn ng lin tnh v 80 km/ h trn ng cao tc. S :10 pht 4.2. Khong cch t th trn A n Bc Ninh l bao nhiu mt theo ng chim bay. S :34,235 km Bi 5 : Vi n l s t nhin, k hiu an l s t nhin gn nht can.Tnh 2005 2 1 2005... a a a S + + + =. S :598652005 = SBi 6 : 6.1. Gii phng trnh : 22333 1 5 33 5 355 5 9x xx xxx x ++ + = + + +S : ( )22 5 32 , 1 = x ; ( )5 22 5 36 , 5 , 4 , 3 = x 6.2. Tnh chnh xc nghim n 10 ch s thp phn. S : ; ; 618033989 , 11 x 381966011 , 12 x ;850650808 , 04 , 3 x 7861511377 , 06 , 5 xBi 7 : 7.1. Trc cn thc mu s : 3 39 3 2 2 12 += MS :1 2 9 723 6+ + + = M7.2 Tnh gi tr ca biu thc M ( chnh xc n 10 ch s) S : 533946288 , 6 = MBi 8 : 8.1 Cho dy s,11 0= = a a1211++=nnnaaa Chng minh rngvi mi 0 1 31221= + ++ + n nnna a a a0 n8.2. Chng minh rng vi mi1 13 + =n n na a a 1 n8.3.Lp mt quy trnh tnh aiv tnh ai vi i = 2 , 3 ,,25 Bi 9 : 9.1. Tm tt c cc cp s t nhin (x,y)sao cho x l c ca y2+1 v y l c ca x2+1 9.2. Chng minh rng phng trnh x2 + y2 axy + 1 = 0 c nghim t nhin khi v ch khi a = 3. Tm tt c cc cp s t nhin ( x, y, z ) l nghim ca phng trnh x2 + y2 3xy + 1 = 0 9.3 .Tm tt c cc cp s t nhin ( x, y, z ) l nghim ca phng trnh x2(y2 - 4) = z2 + 4 S : , y = 3 , na x =12 3 =n na a zBi 10 : Cho mt s t nhin c bin i nh mt trong cc php bin i sau Php bin i 1) : Thm vo cui s ch s 4 Php bin i 2) : Thm vo cui s ch s 0 Php bin i 3) : Chia cho 2 nu ch s chn Th d: T s 4, sau khi lm ccphp bin i 3) -3)-1) -2) ta c 140 14 1 2 4) 2 ) 1 ) 13 ) 3 10.1. Vit quy trnh nhn c s 2005 t s 4 10.2. Vit quy trnh nhn c s 1249 t s 4 10.3. Chng minh rng, t s 4 ta nhn c bt k s t nhin no nh 3 phpbin s trn. HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN THTHCS, lp 9, 2001-2002 Bi 1: Tnh ( lm trn n 6 ch s thp phn): 4 3 5 6 7 8 9 11 2 3 4 5 6 7 8 9 1 A = + + + + 00Bi 2: Tnh 2 24100, 6 1, 256 1 3 25 3551 5 1 1 5 2 50.61 6 3 225 9 4 17 + + Bi 3: Tnh ( lm trn n 4 ch s thp phn): 9 8 7 6 5 4 39 8 7 6 5 4 3 2 C =Bi 4: Tm phn d ca php chia a thc: 5 4 3 2(2 1, 7 2, 5 4, 8 9 1) ( 2, 2) x x x x x x + Bi 5: Tm cc im c ta nguyn dng trn mt phng tha mn: 2x + 5y = 200 Bi 6: Phn tch a thc thnh nhn t 4 3 2( ) 2 15 26 120 Px x x x x = + +Bi 7: Mt ngi b bi vo hp theo quy tc: ngy u 1 vin, mi ngy sau b vo s bi gp i ngy trc . Cng lc cng ly bi ra khi hp theo quy nguyn tc: ngy u v ngy th hai ly mt vin, ngy th ba tr i mt ngy ly ra s bi bng tng hai ngy trc 1) Tnh s bi c trong hp sau 15 ngy. 2) s bi c trong hp ln hn 2000 cn bao nhiu ngy? Bi 8: Vit quy trnh tm phn d ca php chia 26031913 cho 280202. Bi 9: Tnh ( cho kt qu ng v kt qu gn ng vi 5 ch s thp phn): 11121314151617189++++++++ Bi 10: Tm s nguyn dng nh nht tha: chia 2 d 1, chia 3 d 2, chia 4 d 3, chia 5 d 4, chia 6 d 5, chia 7 d 6, chia 8 d 7, chia 9 d 8, chia 10 d 9. Bi 11: Tm nghim gn ng vi su ch s thp phn ca 22 3 3 1, 5 x x 0 + = Bi 12: S no trong cc s 33; ; 3;1, 87 l nghim ca phng trnh 4 3 22 5 3 1, 5552 0 x x x + =Bi 13: Cho 20cotA=21. Tnh 2Asin os2Acos sin 23A cBA=+ Bi 14: Cho tam gic ABC c AH l ng cao. Tnh di BH v CH bit .3; 5; 7 AB AC BC = = =Bi 15: Tnh din tch phn hnh nm gia tam gic v cc hnh trn bng nhau c bn knh l 3cm ( phn mu trng ) HTS GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN THTHCS, lp 8, 2001-2002 Bi 1: So snh cc phn s sau: 19 1919 191919 19191919; ; ;27 2727 272727 27272727 Bi 2: Tnh 2 24100, 6 1, 256 1 3 25 3551 5 1 1 5 2 50.61 6 3 225 9 4 17 + + Bi 3: Tm x v lm trn n bn ch s thp phn: 1 1 1 1 1... 140 1, 08 [0, 3 ( -1)] 1121 22 22 23 23 24 28 29 29 30x + + + + + + = Bi 4: Tnh: 1313131313133+++ Bi 5: Tm cc c chung ca cc s sau: 222222;506506;714714;999999 Bi 6: Chia s 19082002 cho 2707 c s d lr . Chia cho 209 c s d l. Tmr . 1 1r2r2Bi 7: Hi c bao nhiu s gm 6 ch s c vit bi cc ch s 2, 3, 5 v chia ht cho 9? Bi 8: Vit quy trnh tm phn d ca php chia 19052002 cho 20969. Bi 9: Tm s nguyn dng nh nht tha: chia 2 d 1, chia 3 d 2, chia 4 d 3, chia 5 d 4, chia 6 d 5, chia 7 d 6, chia 8 d 7, chia 9 d 8, chia 10 d 9. Bi 10: Tam gic ABC c y BC = 10. ng cao AH = 8. Gi I v O ln lt l trung im AH v BC . Tnh din tch ca tam gic IOA v IOC. Bi 11: Phn tch a thc thnh nhn t4 3 2( ) 2 13 14 2 Px x x x x = + + 4Bi 12: Tm mt s gm ba ch s dngxyzbit tng ca ba ch s bng php chia 1000 cho xyzBi 13: Mt ngi b bi vo hp theo quy tc: ngy u 1 vin, mi ngy sau b vo s bi gp i ngy trc . Cng lc cng ly bi ra khi hp theo quy nguyn tc: ngy u v ngy th hai ly mt vin, ngy th ba tr i mt ngy ly ra s bi bng tng hai ngy trc 1) Tnh s bi c trong hp sau 10 ngy. 2) s bi c trong hp ln hn 1000 cn bao nhiu ngy? Bi 14: Cho hnh thang vung ABCD( ) AB CD , F l im nm gia CD, AF ct BC ti E. Bit . Tnh din tch tam gic BEF.1, 482; 2, 7182; 2 AD BC AB = = =Bi 15: Tnh din tch phn hnh ( mu trng ) gii hn bi 4 hnh trn bng nhau c bn knh l 13cm . HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN THTHPT, lp 10, 2001-2002 Bi 1: Tm x ( , pht, giy), bit 18v tanx = 0,7065193280 270ox < > >> 2.2 41128 1028233300 8325E = =A > B C > D 0,5 0,5 1,0 2 F l s l, nn c s ca n khng th l s chn. F l s nguyn t nu n khng c c s no nh hn 106.0047169 F = . gn 1 cho bin m D, thc hin cc thao tc: ALPHAD,ALPHA=,ALPHAD+2,ALPHA:, 11237ALPHA D, bm = lin tip (my 570ES th bmCALCsaumibm=).Nut3chon 105phpchiakhngchn,thktlunFls nguyn t. Qui trnh bm phm Kt qu: F: khng phi l s nguyn t. 11237= 17*661 0,5 0,5 3 (1897, 2981) 271 UCLN = . Kim tra thy 271 l s nguyn t. 271 cn l c ca3523. Suy ra: ( )5 5 5 5271 7 11 13 M = + +Bm my tnh. 5 5 57 11 13 549151 A = + + =gn 1 cho bin m D, thc hin cc thao tc: ALPHAD,ALPHA=,ALPHAD+2,ALPHA:, 549151ALPHAD,bm=lintip,phpchia chn vi D = 17. Suy ra: 17 32303 A = Bngthutgiikimtrasnguyntnhtrn,ta bit 32303 l s nguyn t. Vy cc c nguyn t ca M l: 17; 271;32303 0,5 0,5 2 Ta c: 1 2345103 3(mod10); 103 9(mod10);103 3 9 27 7(mod10);103 21 1(mod10);103 3(mod10); = Nh vy cc lu tha ca 103 c ch s tn cng lin tip l: 3,9,7,1 (chu k 4). 2006 2(mod 4) , nn c ch s hng n v l 9. 20061030,5 0,5 4 1 23 45 629 29( 1000); 29 841(mod1000);29 389(mod1000); 29 281(mod1000);29 149(mod1000); 29 321(mod1000);Mod ( )210 5 220 240 8029 29 149 201(mod1000);29 201 401(mod1000);29 801(mod1000); 29 601(mod1000);= 100 20 8029 29 29 401 601 1(mod1000); = ( )202000 100 202007 2000 6 129 29 1 1(mod1000);29 29 29 29 1 321 29(mod1000)309(mod1000);= = = Ch s hng trm ca P l 3. 1,0 2 Gii thut: 1 STO A, 0 STO D, ALPHA D, ALPHA =, ALPHA D + 1, ALPHA : , ALPHA A, ALPHA =, ALPHA A + (-1)(D-1) x ((D-1)D2. Sau bm = lin tip, theo di s m D ng vi ch s ca uD, ta c: 4 5 6113 3401 967; ;144 3600 1200u u u = = = ; 1,0 5 200, 8474920248; u u25 0,8895124152; u30 0.8548281618 1,0 2 u10 = 28595 ;u15 = 8725987 ;u21 = 98848794231,0 6 S10 = 40149 ;S15 = 13088980 ;S20 = 4942439711 Qui trnh bm phm: 1STOA,2STOB,3STOM,2STOD,ALPHAD, ALPHA=, ALPHA D+1, ALPHA : , ALPHA C, ALPHA =, 3 ALPHA A, +, 2 ALPHA B, ALPHA : , ALPHA M, ALPHA=,ALPHAM+ALPHAC,ALPHA:ALPHA A, ALPHA =, ALPHA B, ALPHA : , ALPHA B, ALPHA =, ALPHA C, ALPHA : ,ALPHA D, ALPHA=, ALPHA D+1, ALPHA : , ALPHA C, ALPHA =, ALPHA 2 ALPHA A, +, 3 ALPHA B, ALPHA : , ALPHA M, ALPHA =, ALPHA M + ALPHA C, ALPHA : ALPHA A, ALPHA =, ALPHA B, ALPHA : , ALPHA B, ALPHA =, ALPHA C, sau bm = lin tip, D l ch s, C l uD , M l SD 1,0 2 7 7.1Qui trnh0,5 100000 STO A, 100000 STO B, 1 STO D, ALPHA D, ALPHA =, ALPHA D + 1, ALPHA : , ALPHA B, ALPHA=, ALPHA B+20000, ALPHA : , ALPHA A, ALPHA =, ALPHA A1.006 + B, bm = lin tip cho n khi A vt qu 5000000 th D l s thng phi gi tit kim. D l bin m, B l s tin gp hng thng, A l s tin gp c thng th D. D = 18 thng 0,5 7.2Thng th nht, sau khi gp cn n: A = 5000000 -100000 = 4900000 (ng). 4900000 STO A, 100000 STO B, th: Thng sau gp: B = B + 200000 (gi tr trong nh B cng thm 20000), cn n: A= A1,007 -B. Thc hin qui trnh bm phm sau: 4900000 STO A, 100000 STO B, 1 STO D, ALPHA D, ALPHA =, ALPHA D+1, ALPHA : , ALPHA B, ALPHA =, ALPHA B + 20000, ALPHA : , ALPHA A, ALPHA =, ALPHA A1,007 - ALPHA B, sau bm = lin tip cho n khi D = 19 (ng vi thng 19 phi tr gp xong cn n: 84798, bm tip =, D = 20, A m. Nh vy ch cn gp trong 20 thng th ht n, thng cui ch cn gp : 847981,007 = 85392 ng. Cch gii Kt qu cui cng ng 0,5 0,5 8.1Gii h phng trnh: 4 3 5450 62xa x b xc x x + + = (h s ng vi x ln lt thay bng 2, 3, 5; n s l a, b, c). Dng chc nng gii h 3 phng trnh, cc h s ai, bi, ci, di c th nhp vo trc tip mt biu thc, v d cho h s di ng vi x = 2.6 2 ^ 5 2 ^ 2 450 S lc cch gii Kt qu a = -59 b = 161 c = -495 0.5 0.5 8 8.2P(x) = (x-2)(x-3)(3x+5)(x-5)(2x-3) 1 2 3 4 53 52; 3; 5; ;2 3x x x x x= = = = =0.5 0,5 2 9 5 253 19(72 ) 240677 (*)3 2406777219x x yxx y = = Xt 53 2406777219xy x= (iu kin:)9 x >9STOX,ALPHAX,ALPHA=,ALPHAX+1, ALPHA:,72ALPHAX-((3ALPHAX^5-240677)19),bm=lintip.KhiX=32thc kt qu ca biu thc nguyn y = 5. Thay x = 32 vo phng trnh (*), gii pt bc 2 theo y, ta c thm nghim nguyn dng y2 =4603. ( )( )32; 5;32; 4603x yx y= == = Li gii Kt qu x = 32 0,5 0,5 1,0 2 10Bng ca to nh BC c xem l vung gc vi BC0,5 nn tam gic CBH vung ti B. Do cc tia sng c xem nh song song vi nhau, nn nn1 07.32tan 6 28'64.58BCH AOB = = = Khong cch gia hai thnh ph A v B: 2 2 6485.068731.9461924( )360 360Rkm = 0,5 1,0 HT S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 9 THCS - Nm hc 2006-2007 Thi gian: 120 pht- Ngy thi: 02/12/2006. Ch :- thi gm4 trang - Th sinh lm bi trc tip vo bn thi ny. - Nu khng ni g thm, hy tnh chnh xc n 10 ch s. im ton bi thi Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng thi ghi) GK1 Bng sBng ch GK2 Bi 1: Tnh gi tr ca cc biu thc: 5 0 4 033 7 0 3 0235, 68 cot 23 35' os 69 43'62, 06 69 55' sin 77 27'g cAtg= . Lm trn n 5 ch s l thp phn. 4 42 2 2 2 2 23 2 16 164 9 6 4 4x y x y x yBx y x xy y x y + = + + + + khi: A B =a/( . 5; 16) x y = = B b/( 1, 245; 3, 456). x y = = Bi 2: a = ; b = c = ; d = e = ; f = g = a/Bit 20062007 1120081111abcdefg= ++++++, , , , , , a b c d e fg. Tm cc s t nhin. b/Cho dy s 1 1 1 11 1 1 12 4 8 2n nu = . Tnh(chnh xc) v (gn ng) 5u10 15 20, , u u u Bi 3: a/ Phn tch thnh tha s nguyn t cc s sau: 252633033 v 8863701824. b/ Tm cc ch s sao cho s567abcdal s chnh phng. a/ 252633033 = 8863701824 = b/ Cc s cn tm l: Bi 4:Khai trin biu thc ta c a thc gi tr chnh xc ca biu thc: (1521 2 3 x x + +)0Tnh vi2 30 1 2 30... . a a x a x a x + + + +. 0 1 2 3 29 302 4 8 ... 536870912 1073741824 E a a a a a a = + + + E = Bi 5: Tm ch s l thp phn th11k t du phy ca s thp phn v hn tun hon ca s hu t 20071000029. Ch s l thp phn th 11ca 2007 1000029 l: Bi6:Tmccstnhin(2000 60000) n n < < saochovimisth 354756 15na = + ncng l s t nhin. Nu qui trnh bm phm c kt qu. n = Qui trnh bm phm: Bi 7: Cho dy s: 1 2 3 41 1 1 12 ; 2 ; 2 ; 21 122 2 21 122 21222u u u u = + = + = + = ++ + ++ ++1; ... 1212...122nu = ++ (biu thc c cha tng phn s).nTnh gi tr chnh xc cauu v gi tr gn ng ca. 5 9 1, , u0 15 20, u uu5 = ----------------------u9 = -----------------------u10 = ------------------------ Bi8:Choathcbit 3 2( ) P x ax bx cx d = + + + (1) 27; (2) 125; (3) 343 P P P = = = v .(4) 735 P =a/ TnhP P (Ly kt qu chnh xc).( 1); (6); (15); (2006). P P b/ Tm s d ca php chia( ) 3 5 P x cho x . u15 = ----------------------u20 = ----------------------- S d ca php chia( ) 3 5 P x cho x l:r =( 1) ; (6))(15) ; (2006)P PP P = == = Bi 9: Li sut ca tin gi tit kim ca mt s ngn hng hin nay l 8,4% nm i vi tin gi c k hn mt nm. khuyn mi, mt ngn hng thng mi A a ra dch v mi: Nu khch hng gi tit kim nm u th vi li sut 8,4% nm, sau li sut nm sau tng thm so vi li sut nm trc l 1%. Hi nu gi 1.000.000 ng theo dchvthstinsnhnclbaonhiusau:10nm?;15nm?Nuslc cch gii. S tin nhn c sau 10 nm l: S tin nhn c sau 15 nm l: S lc cch gii: Bi10:Cho3ngthng 1 2 3( ) : 3 2 6 ; ( ) :2 3 15; ( ) : 3 6 d x y d x y d x y = + = + =1( ) d3) d).Hai ng thng v(ct nhau ti A; hai ng thng v(ct nhau ti B; hai ng thng(v ct nhau ti C. 1( ) d2d) (d)2d3a) Tm ta ca cc im A, B, C (vit di dng phn s). Tam gic ABC l tam gic g? Gii thch. b) Tnh din tch tam gic ABC (vit di dng phn s) theo on thng n v trn mi trc ta l 1 cm. d) Tnh s o ca mi gc ca tam gic ABC theo n v o (chnh xc n pht). V th v in kt qu tnh c vo bng sau: Ht S Gio dc v o to k thi chn hoc sinh gii tnh Tha Thin Hulp 9 thCS nm hc 2005 - 2006 Mn :MY TNH B TIp n v thang im: BiCch giiim TP im ton bi 3, 01541 A 0,75 Rt gn biu thc ta c: ( )3 3 2 22 24 7 18 49 6 4x y xy x yBx xy y +=+ +. 0,5 1 286892( 5; 16)769x y B = = = ( 1, 245; 3, 456) -33.03283776 x B = = 0,50 0,25 2 a/ 9991; 25; 2; 1; 6. a b c d e f g = = = = = = = 1,0 2 b/0 SHIFT STO X; 1 SHIFT STO A; ALPHA X ALPHA = ALPHA X+1: ALPHA A ALPHA = ALPHA A ( 1 12X ). Bm phm = lin tip (570MS) hoc CALC v bm = lin tip (570ES). Kt qu: 5 1015 209765; 0.2890702984;327680.2887969084; u 0.2887883705 u= u u 1,0 2 a) 3 26 2252633033=3 53 3331; 8863701824=2 101 1171 0,5 0,5 3 b) Ta c: 56700000 567 56799999 7529 567 7537 abcda abcda < < < 9STOX,ALPHAX,ALPHA=,ALPHAX+1, ALPHA:,72ALPHAX-(3ALPHAX^5-240677),bm=lintip.KhiX=32thckt qu ca biu thc nguyn y = 5. Thay x = 32 vo phng trnh (*), gii pt bc 2 theo y, ta c thm nghim nguyn dng y2 =4603. ( )( )32; 5;32; 4603x yx y= == = Li gii Kt qu x = 32 0,5 0,5 2 4.1 Sau 4 nm, bn Chu n ngn hng: A= 4 3 22000000(1.03 1.03 1.03 1.03) 8618271.62 + + + NmthnhtbnChuphigp12m(ng).Gi 1 0.03 1.03 q = + =Sau nm th nht, Chu cn n: 112 x Aq m = Saunmthhai,Chucnn: ( )2212 12 12 ( 1) x Aq mq m Aq mq = = +...Saunmthnm,Chucnn . 5 4 3 2512 ( 1) x Bq m q q q q = + + + +Gii phng trnh: ,tac 5 4 3 2512 ( 1) 0 x Bq m q q q q = + + + + =156819 m = Cch gii Kt qu cui cng ng 0,5 0,5 4 4.2 Thng th nht, sau khi gp cn n: A = 5000000 -100000 = 4900000 (ng). 4900000 STO A, 100000 STO B, th: Thng sau gp: B = B + 200000 (gi tr trong nh B cng thm 20000), cn n: A= A1,007 -B. Thc hin qui trnh bm phm sau: 4900000 STO A, 100000 STO B, 1 STO D, ALPHA D, ALPHA =, ALPHA D+1, ALPHA : , ALPHA B, ALPHA =, ALPHA B + 20000, ALPHA : , ALPHA A, ALPHA =, ALPHA A1,007 - ALPHA B, sau bm = lin tip cho n khi D = 19 (ng vi thng 19 phi tr gp xong cn n: 84798, bm tip =, D = 20, A m. Nh vy ch cn gp trong 20 thng th ht n, thng cui ch cn gp : 847981,007 = 85392 ng. Cch gii Kt qu cui cng ng 0,5 0,5 2 5 32013' 18"cbaaaABCD a = 3,84 ; c = 10 (cm) 2 22 cos 7.055029796 b a c ac D = + 2 222cos 0, 68773889942a bBa= n0133 27' 5" ABC 15.58971171ABCDS 0,5 0,5 2 .27.29018628; 4.992806526SH MHSH IHMH MS= = =+=R(bnknhmtcu ni tip). Thtchhnhcu(S1): 3343521.342129( )V Rcm =. 28, 00119939 SM 6, 27; MH IK I = = H0,5 0,5 6 Khong cch t tm I n mt phng i qua cc tip im ca (S1) vi cc mt bn ca hnh chp: 24.866027997IHd EISH IH= = = Bn knh ng trn giao tuyn:2 21,117984141 r EK R d = = Din tch hnh trn giao tuyn: 274, 38733486( ) S c m0,5 0,5 2 F l s l, nn c s ca n khng th l s chn. F l s nguyn t nu n khng c c s no nh hn 106.0047169 F = . gn 1 cho bin m D, thc hin cc thao tc: ALPHAD,ALPHA=,ALPHAD+2,ALPHA:, 11237ALPHA D, bm = lin tip (my 570ES th bmCALCsaumibm=).Nut3chon 105phpchiakhngchn,thktlunFls nguyn t. Qui trnh bm phm Kt qu: F: khng nguyn t 0,5 0,5 (1897, 2981) 271 UCLN = . Kim tra thy 271 l s nguyn t. 271 cn l c ca3523. Suy ra: ( )5 5 5 5271 7 11 13 M = + +Bm my tnh. 5 5 57 11 13 549151 A = + + =gn 1 cho bin m D, thc hin cc thao tc: ALPHAD,ALPHA=,ALPHAD+2,ALPHA:, 549151ALPHAD,bm=lintip,phpchia chn vi D = 17. Suy ra: 17 32303 A = Bngthutgiikimtrasnguyntnhtrn,ta bit 32303 l s nguyn t. 0,5 7 Vy cc c nguyn t ca M l: 17; 271;323030,5 SKI720MHDABCSE KIMHTa c: 1 2345103 3(mod10); 103 9(mod10);103 3 9 27 7(mod10);103 21 1(mod10);103 3(mod10); = Nh vy cc lu tha ca 103 c ch s tn cng lin tip l: 3,9,7,1 (chu k 4). 2006 2(mod10) , nn c ch s hng n v l 9. 20061030,5 0,5 8 1 23 45 629 29( 1000); 29 841(mod1000);29 389(mod1000); 29 281(mod1000);29 149(mod1000); 29 321(mod1000);Mod ( )210 5 220 240 8029 29 149 201(mod1000);29 201 401(mod1000);29 801(mod1000); 29 601(mod1000);= 100 20 8029 29 29 401 601 1(mod1000); = ( )202000 100 202006 2000 629 29 1 1(mod1000);29 29 29 1 321(mod1000);= = Ch s hng trm ca P l 3. 1,0 2 Gii thut: 1 STO A, 0 STO D, ALPHA D, ALPHA =, ALPHA D + 1, ALPHA : , ALPHA A, ALPHA =, ALPHA A + (-1)D-1 x ((D-1)D2. Sau bm = lin tip, theo di s m D ng vi ch s ca uD, ta c: 4 5 6113 3401 967; ;144 3600 1200u u u = = = ; 1,0 9 200, 8474920248; u u25 0,8895124152; u30 0.8548281618 1,0 2 u10 = 28595 ;u15 = 8725987 ;u21 = 98848794231,0 S10 = 40149 ;S15 = 13088980 ;S20 = 49424397110,5 10 1STOA,2STOB,3STOM,2STOD,ALPHAD, ALPHA=, ALPHA D+1, ALPHA : , ALPHA C, ALPHA =,ALPHA3ALPHAA,+,2ALPHAB,ALPHA:, ALPHA M, ALPHA =, ALPHA M + ALPHA C, ALPHA : ALPHA A, ALPHA =, ALPHA B, ALPHA : , ALPHA B, ALPHA =, ALPHA C, ALPHA : ,ALPHA D, ALPHA=, ALPHA D+1, ALPHA : , ALPHA C,ALPHA=,ALPHA2ALPHAA,+,3ALPHAB, ALPHA : , ALPHA M, ALPHA =, ALPHA M + ALPHA C, ALPHA : ALPHA A, ALPHA =, ALPHA B, ALPHA : ,ALPHAB,ALPHA=,ALPHAC,saubm=lin tip, D l ch s, C l uD , M l SD 0,5 2 Bi 2: TX: R. Y' = 13*x^2-14*x-2/(3*x^2-x+1)^2 ( )22213 14 2'3 1x xyx x = +, 1 2' 0 1.204634926; 0.1277118491 y x x = = = 1 20.02913709779; 3.120046189 y y = =1 23.41943026 d M M = =Y"=-6*(13*x^3-21*x^2-6*x+3)/(3*x^2-x+1)^3 Bi 3:0.4196433776 x ( )3 2326(13 21 6 3)"3 1x x xyx x += +, 1 2 3" 0 1.800535877; 0.2772043294; 0.4623555914 y x x x = = = = 1 2 30.05391214491; 1.854213065; 2.728237897 y y y = = =Bi 4: 83 17;13 13C 16.07692308; 9.5ADC ABCS S Din tch hnh trn ngoi tip ABCD: ( )58.6590174ABCDS Bi 5: Sau 4 nm, bn Chu n ngn hng: A=4 3 22000000(1.03 1.03 1.03 1.03) 8618271.62 + + + Nm th nht bn Chu phi gp 12m (ng). Gi1 0.03 1.03 q = + =Sau nm th nht, Chu cn n: 112 x Aq m = Sau nm th hai, Chu cn n:( )2212 12 12 ( 1) x Aq mq m Aq mq = = +... Sau nm th nm, Chu cn n 5 4 3 2512 ( 1) x Bq m q q q q = + + + + . Gii phng trnh, ta c 5 4 3 2512 ( 1) 0 x Bq m q q q q = + + + + = 156819 m =Bi 6: .27.29018628; 4.992806526SH MHSH IHMH MS= = =+: bn knh mt cu ngoi tip. Th tch hnh cu (S1):.521.342129 V =Bn knh ng trn giao tuyn: 24.866027997 74.38734859IHr SSH IH= = = HT S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 12 THPT - Nm hc 2007-2008 Thi gian lm bi: 150 phtNgy thi: 01/12/2007 Ch : - thi gm 4 trang - Th sinh lm bi trc tip vo bn thi ny im ca ton bi thi Bng sBng ch Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng chm thi ghi) Gim kho 1: Gim kho 2: Qui nh: Hc sinh trnh by vn tt cch gii, cng thc p dng, kt qu tnh ton vo trng lin k bi ton. Cc kt qu tnh gn ng, nu khng c ch nh c th, c ngm nh chnh xc ti 4 ch s phn thp phn sau du phy Bi 1. (5 im) Cho cc hm s v 2( ) 3 2, ( 0) f x ax x x= + ( ) sin 2 g x a x =. Gi tr no ca a tho mn h thc [ ][ ( 1)] (2) 2 f f g f = Cch giiKt qu Bi 2. (5 im) Tnh gn ng ta cc im un ca th hm s 222 5( )3 4xf xx x+=+ + . Cch giiKt qu MTBT12THPT-Trang 1 Bi 3. (5 im) Tm nghim gn ng (, pht, giy) ca phng trnh:

2sin 2 4(sin cos ) 3 x x x + + = Cch giiKt qu Bi 4. (5 im) Cho 2 dy s{v { }nu }nv vi :1 1111; 222 1517 12n nn nu vu vv v++= = = = nnuuvi n = 1, 2, 3, , k, .. 1.Tnh 5 10 15 18 19 5 10 15 18 19, , , , ; , , , , u u u u u v v v v v2.Vit quy trnh n phm lin tc tnh1 nu + v 1 nv + theo nuv nv . 3.Lp cng thc truy hi tnh un+1 theo un v un-1; tnh vn+1 theo vn v vn-1. Cch giiKt qu Bi 5.(5 im) Xc nh cc h s a, b, c ca hm s f(x) = ax3 + bx2 + cx 2007 bit rng f(x)chiacho(x16)csdl29938vchiacho(x210x+21)cathcsdl 10873375016x (Kt qu ly chnh xc). Tm khong cch gia im cc i v im cc tiu ca th hm s f(x) vi cc gi tr a, b, c va tm c. Cch giiKt qu MTBT12THPT-Trang 2 Bi6.(5im)TheochnhschtndngmicaChnhphchohcsinh,sinhvinvayvn trang tri chi ph hc i hc, cao ng, THCN: Mi sinh vin c vay ti a 800.000 ng/thng (8.000.000 ng/nm hc) vi li sut 0,5%/thng. Mi nm lp th tc vay hai ln ng vi hai hc k v c nhn tin vay u mi hc k (mi ln c nhn tin vay l 4 triu ng). Mt nm sau khi tt nghip c vic lm n nh mi bt u tr n. Gi s sinh vin A trong thi gian hc i hc 4 nm vay ti a theo chnh sch v sau khi tt nghip mt nm c vic lm n nh v bt u tr n. 1.Nu phi tr xong n c vn ln li trong 5 nm th mi thng sinh vin A phi tr bao nhiu tin ? 2.Nu tr mi thng 300.000 ng th sinh vin A phi tr my nmmi ht n ? Cch giiKt qu Bi 7. (5 im) Tmchiudibnhtcacithangncth tavotngvmtt,ngangquactcao4m, songsongvcchtng0,5mkttimcact (hnh v) Cch giiKt qu Bi 8. (5 im) Cho tam gic ABC vung ti nh A(-1; 3) c nh, cn cc nh B v C di chuyn trn ng thng i qua 2 im M(-3 ; 1), N(4 ; 1). Bit rng gc n030 ABC = . Hy tnh ta nh B. MTBT12THPT-Trang 3 Cch giiKt qu Bi9.(5im)Chohnhnggicunitiptrongngtrn (O)cbnknhR=3,65cm.Tnhdintch(ctmu)giihn binangtrnngknhABlcnhcanggicuv ng trn (O) (hnh v). Cch giiKt qu ASBMO Bi10.(5im)Chohnhchpthpdinucynitiptrong ng trn c bn knh r =3,5 cm, chiu cao h = 8 cm a)Tnh din tch xung quanh v th tch ca hnh chp. b)Tm th tch phn gia hnh cu ni tip v hnh cu ngoi tip hnh chp u cho. Cch giiKt qu --------------HT------------- MTBT12THPT-Trang 4 S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh CasioKhi 12 THPT - Nm hc 2007-2008 S LC CCH GII V HNG DN CHM BiCch giiKt quim1

2( ( 1)) ( ) 3 2af f f t tt = = +vi ( 1) 5 t f a = = + [ ] (2) ( ) g f g u =vi(2) 44au f = = - Gii phng trnh tm a (dng chc nng SOLVE): [ ] [ ]( )2( 1) (2) 23 13 sin 8 225f f g fa aa aa = = + ( )2( ( 1)) 3 13 ( 5)5af f a aa = + [ ] (2) sin 82ag f a = 5, 8122 a 1,5 1,5 2,0 2Tnh o hm cp 2 tm im un ca th hm s. Gii phng trnh tm honh cc im un "( ) 0 f x = ( )( )2223 2 2 5'( )3 4x xf xx x+ =+ + ( )( )3 2326 2 3 15 19"( )3 4x x xf xx x + =+ +

12, 6607 x , 11,0051 y 22, 9507 x , 25,8148 y 31, 2101 x , 34,3231 y

1,0 1,0 3,0 3 Theo cch gii phng trnh lng gic t ( )0sin cos 2 cos 45 t x x x = + = Dng chc nng SOLVE , ly gi tr u ca X l2; 2 ta c 2 nghim t, loi bt nghim2, 090657851 2 < Gii pt002 cos( 45 ) 0, 6764442880, 676444288cos( 45 )2xx = = 2sin 2 1 x t = Phng trnh tng ng: ( )4 22 4 2 0 | | 2 t t t t + = Gii pt c 1 nghim:0, 676444288 t 0 01106 25' 28" 360 x k +02106 25' 28" 360ox k + 1,0 2,0 2,0 MTBT12THPT-Trang 5 4 a) 5 10 15 18 19 5 10 15 18 19, , , , ; , , , , u u u u u v v v v vb)Qui trnh bm phm: 1 Shift STO A, 2 Shift STO B, 1 Shift STO D, Alpha D Alpha = Alpha D +1, Alpha :,C Alpha = Alpha A, Alpha :, Alpha A Alpha = 22 Alpha B - 15 Alpha A, Alpha :, Alpha B, Alpha =, 17 Alpha B - 12 Alpha A, = = =... c) Cng thc truy hi: u5 = -767 v v5 = -526;u10 = -192547 v v10 = -135434 u15 = -47517071 v v15 = -34219414 u18 = 1055662493 v v18 = 673575382 u19 = -1016278991 v v19 = -1217168422 2 12 9n nu u+ + nu = v 2 12 9n nv v+ + nv = 2,5 1,5 1,0 5 Tm cc h s ca hm s bc 3:

( )3 2( ) 2007, 0 f x ax bx c x a = + + Tmccimcctr,tmkhongcch gia chng a = 7;b = 13 c = 5516 11, 4210 kc 3,0 2,0 6 a) Sau na nm hc H, s tin vay (c vn ln li): Sau 4 nm (8 HK), s tin vay (c vn ln li): Sau mt nm tm vic, vn v li tng thm: + Gi x l s tin hng thng phi tr sau 5 nm vay, sau n thng, cn n (L = 1,005): + Sau 5 nm (60 thng) tr ht n th P = 0 b) Nu mi thng tr 300000 ng, th phi gii phng trnh:0 Shift STO A, 0 Shift STO D, D Alpha = Alpha D + 1, Alpha : Alpha AAlpha = (Alpha A + 4000000) 1.0056. n phm = nhiu ln cho n khi D = 8 ta c A = 36698986 Alpha A Alpha = Alpha A 1.00512 A = 38962499 ( )2 111 ...1nn n nLP AL xL L L L AL xLL= + + + + = ( )596010 7495071AL LP xL= = 0,0051,005x-1A-300000(1.005x - 1) = 0 Dng chc nng SOLVE, gii c x = 208,29, tc phi tr trong 209 thng (17 nm v 5 thng) mi ht n vay. 1,0 1,0 1,0 2,0 BiCch giiKt quim7 Cho AB = l l chiu di ca thang, HC = 4 ml ct , C l giao im ca ct vthang,xlgchpbimttvthang (hnh v). Ta c:

sin cosCH CIAB AC CBx x= + = + 4 1( ) 0;sin 2cos 2f x AB xx x = = + 32 2 24cos sin 8cos sin'( )sin 2cos 2sin cos32x x xf xxx x x = + =x+ 3 3'( ) 0 sin 8cos 2 f x x x tgx = = =1 00tan (2) 63 26' 6" x= ( )min 0( ) 5, 5902( ) AB Min f x f x m = =

1,0 1,0 1,0 1,0 1,0 MTBT12THPT-Trang 6 8 Pt ng thng MN

2 12 7 1 07 7x y y x = = H s gc ca ng thng AB l:( ) ( )( ) ( )1 01 02tan tan 30 1, 033672tan tan 150 0, 25037kk= + = + Gn gi tr k cho bin A. V ng thng AB i qua im A(-1; 3) nn: b = 3 + A, gn gi tr cho bin B.. Gii h pt: 2 7 1 x yAx y B = + = ta c ta im B: ( )15, 5846; 1, 7385 B v( )25, 3959;1, 3988 B 1,0 2,0 2,0 9 + Tnh bn knh ca na ng trn + Tnh din tch vin phn gii hn bi AB v (O) + Hiu din tch ca na ng trn v vin phn: 0sin36 2,1454( ) r AI R cm = = = , gn cho A 22 01sin 72 2, 03555 2vpRS R= =2cm , gn cho B. 225,19452vprS S= = cm2,0 2,0 1,0 10 a) Tnh di cnh v trung on ca hnh chp

b) Phn gic gc SMO ct SO ti I, l mt cu ni tip hnh chp u c tmI, bn knh IO. Trung trc on SA trong mt phng SAO ct SO ti J. Mt cu ngoi tip hnh chp u c tm J, bn knh SJ . Lu : gn cc kt qu trung gian cho cc bin kt qu cui cng khng c sai s ln.a), gn cho A 02 sin18 2,1631( ) a AB r cm = = =0cos18 3, 3287( ) OM r cm = = , gn cho B SM = 2 28, 6649( ) d OM h cm = + = , gn cho C. 2110 93, 71592xqS ad = = cm31 110 96, 00493 2chopV AB OM h = = cm b)111 8tan tan 2, 2203( )2r IO OM cmOM = = = 24, 76562SK SO SAR SJSJ SA SO= = = = (cm )Hiu th tch: (3 32 1 143V V V R r = = )= 407,5157 cm3 0,5 0,5 0,5 0,5 1,0 1,0 1,0 MTBT12THPT-Trang 7 S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 11 THPT - Nm hc 2007-2008 Thi gian lm bi: 150 phtNgy thi: 01/12/2007 Ch : - thi gm 4 trang - Th sinh lm bi trc tip vo bn thi ny im ca ton bi thi Bng sBng ch Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng chm thi ghi) Gim kho 1: Gim kho 2: Qui nh: Hc sinh trnh by vn tt cch gii, cng thc p dng, kt qu tnh ton vo trng lin k bi ton. Cc kt qu tnh gn ng, nu khng c ch nh c th, c ngm nh chnh xc ti 4 ch s phn thp phn sau du phy Bi 1.( 5 im)Chocchms v 2( ) 3 2, ( 0) f x ax x x= + ( ) sin 2 g x a x =. Gi tr no ca a tho mn h thc: [ ][ ( 1)] (2) 2 f f g f = Cch giiKt qu Bi 2. ( 5 im) 1)Tm hai s nguyn dng x sao cho khi lp phngmi s ta c mt s c 2 ch s u (bn phi) v 2 ch s cui (bn tri) u bng 4, ngha l 344......44 x = . Nu qui trnh bm phm. x = MTBT11-Trang 1 2) Tnh tng 1 2 99 100...2 3 3 4 100 101 101 102S = + + .Ly nguyn kt qu hin trn mn hnh. . Cch giiKt qu Bi 3. ( 5 im)Tm nghim gn ng (, pht, giy) ca phng trnh

2sin 2 4(sin cos ) 3 x x x + + = Cch giiKt qu Bi 4. ( 5 im)Cho 2 dy s{v { }nu }nv vi :1 1111; 222 1517 12n nn nu vu vv v++= = = = nnuuvi n = 1, 2, 3, , k, .. 1.Tnh 5 10 15 18 19 5 10 15 18 19, , , , ; , , , , u u u u u v v v v v2.Vit quy trnh n phm lin tc tnh1 nu + v 1 nv + theo nuv nv . 3.Lp cng thc truy hi tnh un+1 theo un v un-1; tnh vn+1 theo vn v vn-1. Cch giiKt qu Bi 5.( 5 im) 1) Xc nh cc h s a, b, c ca hm s f(x) = ax3 + bx2 + cx 2007 bit rng f(x) chia cho(x16)csdl29938vchiacho(x210x+21)cbiuthcsdl 10873375016x (Kt qu ly chnh xc).2) Tnh chnh xc gi tr ca biu thc s: P = 3 + 33 + 333 + ... +33.....33 13 ch s 3 MTBT11-Trang 2 Nu qui trnh bm phm. Cch giiKt qu Bi 6. ( 5 im)Theo chnh sch tn dng mi ca Chnh ph cho hc sinh, sinh vin vay vn trangtrichiphhcihc,caong,THCN:Misinhvincvaytia800.000 ng/thng (8.000.000 ng/nm hc) vi li sut 0,5%/thng. Mi nm lp th tc vay hai ln ng vi hai hc k v c nhn tin vay u mi hc k (mi ln c nhn tin vay 4 triu ng). Mt nm sau khi tt nghip c vic lm n nh mi bt u tr n. Gi s sinh vin A trong thi gian hc i hc 4 nm vay ti a theo chnh sch v sau khi tt nghip mt nm c vic lm n nh v bt u tr n. 1.Nu phi tr xong n c vn ln li trong 5 nm th mi thng sinh vin A phi tr bao nhiu tin ? 2.Nu tr mi thng 300.000 ng th sinh vin A phi tr my nmmi ht n ? Cch giiKt qu Bi 7. ( 5 im) 1) Tm s nguyn dng nh nht c ba ch s labcsao cho 3 3abc a b c3= + + . C cn s nguyn no tha mn iu kin trn na khng ? Nu s lc cch tm. 2) Cho dy s c s hng tng qutsin(2 sin(2 sin(2 sin 2)nu = (n ln ch sin) Tm vi mi th gn nh khng thay i (ch xt n 10 ch s thp phn), cho bit gi tr. Nu qui trnh bm phm. 0n0n n nu0nu Cch giiKt qu abc =Bi 8.( 5 im)Cho tam gic ABC vung ti nh A(-1; 3) c nh, cn cc nh B v C di chuyntrnngthngiqua2imM(-3;-1),N(4;1).Bitrnggc n030 ABC = .Hy tnh ta nh B. MTBT11-Trang 3 Cch giiKt qu Bi9.(5im)Chohnhnggicunitiptrongng trn(O)cbnknhR=3,65cm.Tnhdintch(ctmu) giihnbinangtrnngknhABlcnhcang gic u v ng trn (O) (hnh v). Cch giiKt qu Bi 10. ( 5 im)Cho tam gic ABC c cc nh) 3 ; 9 ( A , 3 1;7 7B v.( ) 1; 7 C 1) Vit phng trnh ng trn ngoi tip tam gic ABC.2) Vit phng trnh tip tuyn ca ng trn, bit tip tuyn i qua im.( ) 4;1 M Cch giiKt qu --------------HT------------- MTBT11-Trang 4 S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh CasioKhi 11 THPT - Nm hc 2007-2008 S LC CCH GII V HNG DN CHM BiCch giiKt quim 1 2( ( 1)) ( ) 3 2af f f t tt = = +vi ( 1) 5 t f a = = + [ ] (2) ( ) g f g u =vi(2) 44au f = = - Gii phng trnh tm a (dng chc nng SOLVE): [ ] [ ]( )2( 1) (2) 23 13 sin 8 225f f g fa aa aa = = + ( )2( ( 1)) 3 135( 5)af f aaa = + [ ] (2) sin 82ag f a = 5, 8122 a 1,5 1,5 2,0 2 1) Qui trnh bm phm ng. 2) 0 Shift STO D, 0 Shift STO D, Alpha D Alpha =, Alpha D +1, Alpha :, Alpha A Alpha =, Alpha A + (-1)^(D+1) Alpha D (Alpha D +1) (Alpha D +2), Bm = lin tip n khi D = 100. C th dng chc nng1 1001( 1)( 1)( 2X)XX X++ + 164 v 764 0, 074611665 S 2,0 1,0 2,0 3 Theo cch gii phng trnh lng gic t ( )0sin cos 2 cos 45 t x x x = + = Dng chc nng SOLVE , ly gi tr u ca X l 2; 2 ta c 2 nghim t, loi bt nghim 2, 090657851 2 < Gii pt002 cos( 45 ) 0, 6764442880, 676444288cos( 45 )2xx = = 2sin 2 1 x t = Phng trnh tng ng: ( )4 22 4 2 0 | | 2 t t t t + = Gii pt c 1 nghim: 0, 676444288 t 01106 25' 28" 360 x k +0 0216 25' 28" 360ox k + 1,0 2,0 2,0 4 a) 15 18 19v5 10 15 18 19 5 10, , , , ; , , , , u u u u u v v v vb)Qui trnh bm phm: 1 Shift STO A, 2 Shift STO B, 1 Shift STO D, Alpha D Alpha = Alpha D +1, Alpha :,C Alpha = Alpha A, Alpha :, Alpha A Alpha = 22 Alpha B - 15 Alpha A, Alpha :, Alpha B, u5 = -767 v v5 = -526;u10 = -192547 v v10 = -135434 u15 = -47517071 v v15 = -34219414 u18 = 1055662493 v v18 = 673575382 2,5 MTBT11-Trang 5 Alpha =, 17 Alpha B - 12 Alpha A, = = =... c) Cng thc truy hi: u19 = -1016278991 v v19 = -1217168422 2 12 9n nu u+ + nu = v 2 12 9n nv v+ + nv = 1,5 1,0 5 1) Tm cc h s ca hm s bc 3: ( )3 2( ) 2007, 0 f x ax bx c x a = + + 2) Tnh tng P Qui trnh bm phm a = 7;b = 13 c = 5516P = 3703703703699 3,0 1,0 1,0 6 1) Sau na nm hc H, s tin vay (c vn ln li):Sau 4 nm (8 HK), s tin vay (c vn ln li): n phm = nhiu ln cho n khi D = 8 ta c Sau mt nm tm vic, vn v li tng thm: + Gi x l s tin hng thng phi tr sau 5 nm vay, sau n thng, cn n (L = 1,005): + Sau 5 nm (60 thng) tr ht n th P = 0 2) Nu mi thng tr 300000 ng, th phi gii phng trnh:0 Shift STO A, 0 Shift STO D, D Alpha = Alpha D + 1, Alpha : Alpha AAlpha = (Alpha A + 4000000) 1.0056 A = 36698986 Alpha A Alpha = Alpha A 1.00512 A = 38962499 (21 ...n nP AL xL L L L = + + + + ( )596010 74951AL LP xL= = 0,0051,005x-1A-300000(1.005x - 1) = 0 Dng chc nng SOLVE, gii c x = 208,29, tc phi tr trong 209 thng (17 nm v 5 thng) mi ht n vay. 1,0 1,0 1,0 2,0 BiCch giiKt quim 7 1) Tm c s nh nhtS lc cch tm ngTm c thm 3 s na l:2) Tm c 0nTnh c gi tr0nuQui trnh bm phm ng 153 370, 371 v 407 023 n =230,893939842 u =1,0 0,5 1,5 1,0 0,5 0,5 MTBT11-Trang 6 8 Pt ng thng MN

2 12 7 1 07 7x y y x = = H s gc ca ng thng AB l:( ) ( )( ) ( )1 01 02tan tan 30 1, 0372tan tan 150 07kk= + = + Gn gi tr k cho bin A. V ng thng AB i qua im A(-1; 3) nn: b = 3 + A, gn gi tr cho bin B.. Gii h pt: 2 7 1 x yAx y B = + = ta c ta im B: ( )15, 5846; 1, 7385 B v ( )25, 3959;1, 3988 B 1,0 2,0 2,0 9 + Tnh bn knh ca na ng trn + Tnh din tch vin phn gii hn bi AB v (O) + Hiu din tch ca na ng trn v vin phn: 0sin36 2,1454( ) r AI R cm = = =, gn cho A 22 01sin 72 2, 0355 2vpRS R= =, gn cho B. 225,19452vprS S= = cm2,0 2,0 1,0 10 + Xc nh tm v tnh bn knh ca ng trn bng cch gii h IA = IB v IA = IC. Phng trnh ng trn dng: ( ) ( )2 22x a y b + = R2 248 34 32507 7x y + = 490) 1 Hoc: thay ta ca A, B, C vo phng trnh:, ta c h pt: 2 22 2 x y ax by c + + =+ Gi tip tuyn ca ng trn l ng thng d: y = ax + b.0 ax y b + =ngthngiqua,nn( 4;1 M 4 b a = +(1) . +ngthngdltiptuyncangtrn 48 34;7 7I 5 1307R = 0,5 0,5 0,5 0,5 1,0 MTBT11-Trang 7 nn: 248 345 130 7 771a ba +=+ (2) T (1) v (2) ta tm c phngtrnh theo a. Giitatmc2gitrcaangvi2tip tuyn 112,10009, 4000ab 220, 47530, 9012ab 1,0 1,0 HT MTBT11-Trang 8 S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 12 BTTH - Nm hc 2007-2008 Thi gian lm bi: 150 phtNgy thi: 01/12/2007 Ch : - thi gm 4 trang - Th sinh lm bi trc tip vo bn thi ny im ca ton bi thi Bng sBng ch Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng chm thi ghi) Gim kho 1: Gim kho 2: Quy c: Khi tnh gn ng ch ly kt qu vi 4 ch s thp phn. Bi 1(5 im).Tnh gn ng nghim (, pht, giy) ca phng trnh: 4cos2x + 3cosx = -1 Cch giiKt qu 01360 k x +

02360 k x + 03360 k x + 04360 k x +

Bi 2(5 im).Tnh gn ng gi tr ln nht v gi tr nh nht ca hm s: 223 4( )1x xf xx+ +=+ Cch giiKt qu MTBT12BTTH- Trang 1 ) ( max x f ) ( min x f Bi 3(5 im).Tnh gi tr ca a, b, c, d nu th hm s3 2( ) y f x a x b x c x d = = + + +i qua cc im AMTBT12BTTH- Trang 2 10;3 , B31;5 ) ; f(x) chia cho( 2 x c s d l 1 v chia cho( 2, 4 x ) c s d l3, 8 . Kt qu l cc phn s hoc hn s. Cch giiKt qu a = b = c = d =

Bi 4 (5 im). Cho tam gic ABC c cc nh) 3 ; 9 ( A , 3 1;7 7B v( ) 1; 7 C . a) Tnh din tch tam gic ABC v bn knh ng trn ni tip tam gic ABC. b) Xc nh tm v tnh bn knh ng trn ngoi tip tam gic ABC Cch giiKt qu SABC =r

( ) ; I a b = = R Bi 5 (5 im). Tnh gn ng nghim ca h phng trnh 2 32 22 3log log 5log log 19x yx y+ = + = Cch giiKt qu 11yx

22yx

Bi6(5im).Tnhgitrcaavbnungthngy=ax+bltiptuyncathhms 23 4 3 4 y x x x = + + +ti im ca th c honh 02 3 x = + . Cch giiKt qu ==11ba

==22ba

Bi 7 (5 im).Cho t gic ABCD ni tip trong ng trn (O) bn knh R = 4.20 cm, AB = 7,69 cm, BC = 6,94 cm, CD = 3,85 cm. Tm di cnh cn li v tnh din tch ca t gic ABCD. (Kt qu ly vi 2 ch s phn thp phn) Cch giiKt qu AD ABCDS

Bi8(5im).Giavblhainghimkhcnhaucaphngtrnh.Xtdys: (n l s nguyn dng). 24 6 1 0 x x + =n nnu a b = +a)Tnh u1, u2, u3, u4, u5, u6, u7, u8, u9 b)Lp cng thc truy hi tnh un+1 theo un v un-1. Tnh u10 vi kt qu chnh xc dng phn s hoc hn s. MTBT12BTTH- Trang 3 Cch giiKt qu a)u1 = , u2=,u3= u4 = , u5 = , u6 = u7 = , u8 = , u9 =1 1....... .......n nu unu+ = +

10u = Bi 9 (5 im). Tnh gn ng th tch v din tch ton phn ca hnh chp u S.ABCD vi cnh y AB = 12 dm, gc ca mi cnh bn v mt y l. 067 = Cch giiKt qu

tpS2dm Bi 10 (5 im). Tnh gn ng gi tr ca a v b nu ng thng y = ax + b l tip tuyn ca ng trn( )v i qua im( )2 21 3 16 x y + = ( ) 4; 5 M . Cch giiKt qu

11ab

22ab

-------------HT-------------- MTBT12BTTH- Trang 4 S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh CasioKhi 12 BTTH - Nm hc 2007-2008 CCH GII, P S V HNG DN CHO IM BiCch giip s im tng phn im ton bi 1 20, 4529; 0, 8279 t t 0 , ,, 01,263 412 360 x k + 2,5 1 t t = cosx th v1 1 t. 2 2cos 2 2cos 1 2 1 x x t = = 0Phngtrnhchochuynthnhphngtrnh . 28 3 3 t t + =Gii phng trnh ny ta c hai nghimv 1t2tSau gii cc phng trnh 1s co x t = v 2s co x t = . 0 , ,, 03,4145 531 360 x k + 2,5 5 ( )( )2223 2'( )1x xf xx1 + =+ '( ) 0 1 2 f x x = = max ( ) 4, 6213 f x R 1,0 1,0 1,5 2 Hm s 223 4( )1x xf xx+ +=+ c tp xc nh:RTnh o hm ca hm s ri tm nghim ca o hm. Tnh gi tr ca hm s ti hai nghim ca o hm. lim ( ) 1xf x=v hm s lin tc trn R, nn: C( ) f Max f x =R v( )CTf Min f x =R min ( ) 0, 3787 f x R 1,5 5 31= d 1 252937 = a 1,5 1401571= b 1,5 3 Thaytacaccimchovophngtrnh ,tac2phngtrnhbc nht 4 n, trong c mt phng trnh cho d x c bx ax y + + + =2 331= d . Ta c:( ) ( )( ) f x q x x a r = + ( ) f a r = , t ta c thm 2 phng trnh bc nht 4 n. Thay 31= d vo3phngtrnhcnli,tac3 phngtrnhbcnhtcaccna,b,c.Giih3 phng trnh , ta tm c a, b, c. 6304559 = c 1 5 4 a) Tm ta cc vectABv ACTnh din tch tam gic ABC theo cng thc =720;760AB( ) 10 ; 10 = AC 0,5 0,5 5 MTBT12BTTH- Trang 5 ( )21 1 2 22 21 1. .2 2a bS AB AC AB ACa b= =JJJG JJJG Bn knh ng trn ni tip tam gic ABC l: Srp=(p l na chu vi ca tam gic) 7200= S1, 8759 r =1,0 1,0 21 7 1102x yx y = = 1,0 48 34;7 7I 0,5 b)Gi( ; ) I x y ltmngtrnngoitiptamgic ABC, ta c: IA = IB v IA = IC, nn tm c h pt. Gii h pt ta c ta tm ca ng trn (ABC) Bn knh ng trn: R = IA 3250 5 13049 7R = = 0,5 114,302775638v 0, 69722436219, 73622,1511uxy 2,5 5 tvth u , v l nghim ca h phng trnh 2log u x =3log v = x= += +1952 2v uv uH phng trnh tng ng vi h phng trnh = = +35v uv uT tm c u, v ri tm c x, y. 110, 697224362v 4,3027756381, 6214112, 9655uxy 2,5 5 ( )022 3'( )3 4 3 41,0178xa y xda x x xdxa= +== + + + 2,5 6 ng thng y = ax + b l tip tuyn ca th hm s nn a = y'(x0) Tnhy0 .Tiptuyny=ax+biquaim (0 0 0; ) M x ynn: 0 0y ax b = + 016, 3222 y 0 012, 5238 b y ax = 2,5 5 MTBT12BTTH- Trang 6 7 n12sin ( / 2/ ) AOB AB R=n0 1 11360 2sin ( / 2/ ) 2sin ( / 2/ )2sin ( / 2/ )AOD AB R BC RCD R = n2 sin 4, 29 DA R AOD cm = =n nn n ncos cos12 22cos cos .2 sin2 2 2ABCDAOB BOCAB BCS RCOD DOA DOACD R+ = + + n0132 32' 49" AOB n061 28' 31 AOD 4, 29 DA cm SABCD = 29,64 cm2 1,0 1,0 1,0 2,0 5 8 Gi a l nghim nh ca phng trnh cho th3 5 3 5, .4 4a b += =Gn gi tr ca a v b cho cc bin A v B. 0STOD,Alpha:,AlphaAD+AlphaBD,n=nhiu ln tm cc gi tr ca u1, ...,u9. Dy s c tnh cht qui hi, nn: 1 1 n nu au bun + = +Thayccbbav,tach phng trnh v gii. 3 2 1, , u u u4 3, 2, u u u Tnh tay: 9 8106 1 2889 220764 4 256 256u uu = = 1 2 34 56 78 93 7, ,2 447 123, ,16 32161 843, ,32 1282207 2889,256 256u u uu uu uu u= = == == == =9,4 3 1;2 4a b = = 1 1113 12 464n n nn nnu u uu uu+ += = 10151271024u = 2,0 2,0 1,0 5 9 Chrngccmt bncahnhchp choultamgic cn.GcSAH(Hl tmcay)lgc camicnbnv y: n067 SAH = . TnhSHtheoa=AB vgc,tnh trungonSM,t tnh V v Stp. 067 =Gn cc kt qu trung gian cho cc bin. Xc nh c gc n067 SAH = = 02 tan(67 ) SH a =224aSM SH = +31919, 0467 V d = mm 21114, 2686tpS d 1,0 1,0 0,5 1,0 1,5 5 SBMCHADMTBT12BTTH- Trang 7 112, 71365,8543ab 2,5 10 ng thng i qua( ) 4;5 M , nn4 b a 5 = +(1) ng trn c tm v bn knh R = 4.(1; 3 I )ng thng d: y = ax + b0 ax y b + =ngthngdltiptuyncangtrnnn khong cch t I n d bng bn knh R: 2341a ba +=+ (2) T (1) v (2) ta tm c phng trnh theo a. Gii ta tm c 2 gi tr ca a ng vi 2 tip tuyn 220, 49146, 9654ab 2,5 5 Cng 50 MTBT12BTTH- Trang 8 UBND TNH Tha Thin Huk thi chn hoc sinh gii tnh S Gio dc v o tolp 12 thPT nm hc 2004 - 2005 Mn :MY TNH B TI CHNH THC Thi gian: 120 pht(khng k thi gian giao ) im ca ton bi thi Cc Gim kho (H, tn v ch k) Bng sBng ch S phch (Do Ch tch Hi ng thi ghi) Hc sinh lm bi trc tip vo bn thi ny, in kt qu ca mi cu hi vo trng tng ng. Nu khng c yu cu g thm, hy tnh chnh xc n 10 ch s. Bi 1: (2 im): Tnh gi tr gn ng ca a v b nu ng thng y = ax + b l tip tuyn ca th ca hm s 224 2 51x xyx+ +=+ ti tip im c honh 1 5 x = Bi 2: (2 im): Tnh gn ng cc nghim (, pht, giy) ca phng trnh: 2sin 2 5(sin cos ) 1 x x x + = Bi 3: (2 im): Cho ba s: A = 1193984; B = 157993 v C = 38743. Tm c s chung ln nht ca ba s A, B, C. Tm bi s chung nh nht ca ba s A, B, C vi kt qu ng chnh xc. Ch k ca Gim th 1: ---------------------------- Ch k ca Gim th 2: -------------------- H v tn th sinh: ----------------------------------------------S bo danh: ------------------ Phng thi: ------------------ Hc sinh trng: --------------------------- a = b = a) CLN (A, B, C) =b) BCNN (A, B, C ) = Bi 4: (2 im): Hy rt gn cng thc 2 1( ) 1 2 3 ... .nnS x x x nx = + + + +Tnh tng: 11 121 2 3 3.3 4.3 3 ... 24.3 3 25.3 S = + + + . Bi 5: (2 im): a)BnAngititkimmtstinbanul1000000ngvilisut 0,58%/thng(khngkhn).HibnAnphigibaonhiuthngthcc vn ln li bng hoc vt qu 1300000 ng ? b)Vi cng s tin ban u v cng s thng , nu bn An gi tit kim c k hn 3 thng vi li sut 0,68%/thng, th bn An s nhn c s tin c vn ln li l baonhiu?Bitrngtrongccthngcakhn,chcngthmlichkhng cng vn v li thng trc tnh li thng sau. Ht mt k hn, li s c cng vo vn tnh li trong k hn tip theo (nu cn gi tip), nu cha n k hn m rt tin th s thng d so vi k hn s c tnh theo li sut khng k hn. Bi 6: (2 im):Mt thng hnh tr c ng knh y (bn trong) bng 12,24 cm ng nc cao ln 4,56 cm so vi mt trong ca y. Mt vin bi hnh cu c th vo trong thng th mc nc dng ln st vi im cao nht ca vin bi (ngha l mt nc l tip din ca mt cu). Hy tnh bn knh ca vin bi. Bi 7: (2 im): Trong mt phng Oxy cho tam gic ABC vi cc nh(2; 6), ( 1;1), ( 6;3) A B C . Gi D v E l chn cc ng phn gic ca gc A trn ng thng BC. Tnh din tch tam gic DAE. Ch k ca Gim th 1: ---------------------------- Ch k ca Gim th 2: -------------------- H v tn th sinh: ----------------------------------------------S bo danh: ------------------ Phng thi: ------------------ Hc sinh trng: --------------------------- a) S thng cn gi l: n =b) S tin nhn c l:Rt gn: Sn= Tnh tng S Bn knh ca vin bi l: x1 ; x2 Din tch tam gic DAE l: Bi8:(2im):Mtnhnvingc trm hi ng trn bin (im A) cch bbin16,28km,munvotlin nnginhbnbbin(imB) bng phng tin ca n vn tc 8 km/h cpbsauitipbngxepvi vn tc 12 km/h. Hi ca n phi cp b tiimMnothigiandnhcho ltrnhdichuynlbnht?(Gi thit rng thi tit tt, dt ca ca n khi di chuyn khng ng k). Bi 9: (2 im): Cho dy s sp th t 1 2, 3 1, ,..., , ,...n nu u u u u +bit: 1 2 3 1 2 31, 2, 3; 2 3 ( 4)n n n nu u u u u u u n = = = = + + a)Tnh 4 5 6 7, , , . u u u ub)Vit qui trnh bm phm lin tc tnh gi tr ca nuvi4 n . c)S dng qui trnh trn, tnh gi tr ca 22 25 28 30, , , u u u u . 4u = 5u = 6u = 7u = Bi 10: (2 im): Tm s nguyn t nhin n sao cho 16 192 2 2n+ +l mt s chnh phng. Ch k ca Gim th 1: ---------------------------- Ch k ca Gim th 2: -------------------- H v tn th sinh: ----------------------------------------------S bo danh: ------------------ Phng thi: ------------------ Hc sinh trng: --------------------------- 20u = 22u = 25u = 28u = Qui trnh bm phm lin tc tnh gi tr ca nuvi4 n : 16 192 2 2n+ +l s chnh phng th:n = x = MH ; Thi gian b nhtt UBND TNH Tha Thin Huk thi chn hoc sinh gii tnh S Gio dc v o tolp 12 thPT nm hc 2004 - 2005 Mn :MY TNH B TIp n v thang im: BiCch giip s im TP im ton bi 0, 606264 a 1,0 1 1, 91213278 b 1,0 2 tsin cos 2 sin ; 24t x x x t = = Pt tr thnh: 4 22 5 1 0 (0 2) t t t t + = < Pt c nghim duy nht trong (0; 2 1,0 2 00 0 0 010 0 0 020, 218669211 sin( 45) 0,154622482245 8 53' 41" 53 53' 41" .360216 6'18" .360 45 171 6'18"tt xx x kx k x = + + 1,0 2 D = CLN(A, B) = 5830,5 CLN(A, B, C) = CLN(D, C) = 530,5 ( , ) 323569664( , )A BE BCNNA BUCLNA B= = = 0,5 3 BCNN(A, B, C) = BCNN(E, C) = 236.529.424.3840,5 2 ( )( )'2 31( ) ... '1nnnx xS x x x x xx = + + + + = 1,0 4 ( ) 253 8546323,8 S S = 1,0 2 a)n = 46 (thng) 1,0 5b) 46 thng = 15 qu + 1 thng S tin nhn c sau 46 thng gi c k hn: 1000000(1+0.00683)151,0058 = 1361659,061 ng 1,02 Ta c phng trnh: 2 3 2 3 2 24.2 4 6 3 03(0 )Rh x R x x Rx Rhx R + = + =< < Vi R, x, h ln lt l bn knh y ca hnh tr, hnh cu v chiu cao ban u ca ct nc. 1,0 6 Bm my gii phng trnh:34 224, 7264 512, 376192 0(0 6,12) x x x + = < Ta c: 1 22, 588826692; 5, 857864771 x x 1,0 2 ( ) : 5 3 8 0; ( ) : 3 8 42 0;( ) : 2 5 3 0AB x y AC x yBC x y + = + =+ = 0,5 Pt cc ng phn gic ca gc A: 5 3 3 8 42 8;34 73 34 73 73 345 3 3 8 42 834 73 34 73 73 34x yx y + + = + + = 0,5 Giao im ca cc ng phn gic vi (BC) l: (9, 746112158; 3, 298444863),( 3, 02816344;1, 811265376)DE 0,5 7 1 112,10220354 6, 5443048012 239, 60025435DAEDAES AD AES= 0,5 2 Thi gian ca l trnh: ( )2 216, 26 25, 86( ) 0 25, 868 12x xf x x+ = + < ) 2343452,1938 x l < .S:343452,1938 ( ) x km 0,5 0,25 0,25 0,25 0,25 0,5 2 5 a)1000029 =344.82758620689655172413793103448275862068965517241379310344827586... 1000029lshutcphntchthpphnvhntunhonc chu k 28. 611 1(mod 28) ( )3342007 611 11 = ; Vy ch s l thp phn th 11l: 1. 3 334 311 1 11 (mod 28) 15(mod 28) 2007 0,50 0,25 0,25 2 b)Tac: 4 3 2 4 2 3x y xy x xy y = = +32 99 .Vxvychc2ch s,nnvphitial,nnxtial 3 42 99 38 < , suy ra 10 .38 x