S GIO DC-O TO BNH NH THI TH LN 2TRNG THPT QUC HC QUY NHNTUYN SINH I HC NM 2011 Mn thi: TON KHI A-B-D Thi gian lm bi: 180 pht(Khng k thi gain pht )I: PHN CHUNG:( 7im)CuI(2im): Cho hm s y =f(x) =(x + 2)(x2 mx + m2 -3)( 1) 1. Kho st s bin thin v v th ca hm s (1) khi m = 22. Tm m th hm s (1) tip xc vi trc honh.Cu II (2 im): 1: Gii phng trnh: 4sin2x + 1 = 8sin2xcosx + 4cos22x 2: Gii bt phng trnh:x2 + 4x + 1 > 3 x (x + 1)Cu III (1im): Tnh tchphn 14 24 2022 1xI dxx x +Cu IV (1im): Cho hnh hnh chp S.ABCD c cnh SA = 34, tt c cc cnh cn li u bng 1. Chng minh rng tam gic SAC vung v tnhth tch khi chp S.ABCD.Cu V(1im):Gii h phng trnh: 32 1 0(3 ) 2 2 2 1 0x yx x y y + ' PHN RING: Th sinh ch c lm mt trong hai phn A hoc BA.Theo chng trnh chun Cu VI/a: (2im)1 . Trong mpOxy cho tam gicABCcn ti A. ng thng AB v BC ln lt c phng trnh: 7x+ 6y 24 = 0; x 2y 2 = 0. Vit phng trnh ng cao k t B ca tam gic ABC.2. Trong kgOxyzvit phng trnh mt phng (P) i qua giao tuyn ca hai mt phng ( ) : 2x y 1 = 0;( ) : 2x z = 0 v to vi mt phng (Q): x 2y + 2z 1 = 0gc m 2 2os =9c Cu VII/a:(1 im) Tm s phc z tha mn ng thi:( ) 1 2 5 . 34 z i va z z + B. Theo chng trnh nng cao Cu VI/b.(2im)1.Trong mpOxy cho tam gicABCcn ti A. ng thng AB v BC ln lt c phng trnh: 7x+ 6y 24 = 0; x 2y 2 = 0. Vit phng trnh ng trung tuynk t B ca tam gic ABC2. Trg kgOxyzvit phng trnh ng thng d nm trong mt phng (P): x + y z + 1= 0, ct cc ng thng ( ) ( )1 3: ; ' : 12 2 1 2x t x tD y t D y tz t z t + +' ' + v to vi (D) mt gc 300Cu VII/b: (1im)Gii phng trnh:3 3log 1 log4.15 5 0x xx++ -------------------- Ht--------------------H ng dn gii :CuI:1. bn c t gii2. th hm s (1) tip xc vi trc honh khi h sau c nghim:

( )( )2 22 2( 2) 3 0 (1)3 2 4 2 3 0 (2)x x mx mx m x m m+ + ' + (1) 2 223 0 (3)xx mx m + 1 thi chnh thc*)Vi x = - 2thay vo (2):m =1*) (3) c nghim khi v ch khi2 m , (3) c hai ngim x = 212 32m m t Thay vo (2) ta c: 212 3 0 m 2 m t Cu II : 1.4sin2x + 1 = 8sin2xcosx + 4cos22x 5 4cos2x = 8cosx 8cos3x + 16cos4x 16cos2x + 4 16cos4x 8cos3x 12cos2x+ 8cosx- 1 = 0 (2cosx 1)(8cos3x 6cosx + 1) = 0 (2cosx 1)(2cos3x + 1)= 02. x2 + 4x + 1 > 3 x (x + 1) iu kin x 0tt x , t 0Bt phng trnh tr thnh t4 + 4t2+1 > 3t3 + 3t t4 3t3 + 4t2 3t +1 > 0 (t 1)2(t2 t + 1) > 0 t 1Vy nghim ca bt phng trnh x 0 v x 1Cu III:. 14 24 2022 1xI dxx x + = ( ) ( )12 22 204 221 1xdxx x _+ + , = 1 + ( ) ( )122 201 3 1 3 12 1 11 1dxx xx x _+ + + + , =11 1 11 3ln 1 3ln 122 1 10x xx x _+ + + , = Cu VI:ABCD l hnh thoi , gi O l tm , P l trung im caSCTa c BD (SAC), SC (PBD), 1 32 8OP SA ==> SC OPOP l ng TB ca tam gic SAC, vySC SA ==> SAC vung ti A ==> SA = 54Gi H l chn ng cao==> H AC, . 35SASCSHAC Ta c:BD = 22 2BP OP = 394

1. .6V AC DB SH Cu V:32 1 0 (1)(3 ) 2 2 2 1 0 (2)x yx x y y + ' iu kin122x va y (2)( ) ( ) 1 2 2 1 2 1 2 1 x x y y + + 11 ] ]Xt hm s f(t) = (1 + t2)t= t3 + t f(t)= 3t2 + 1 > 0t R. Vy hm s tng trn R(2)( ) ( )2 2 1 2 2 1 f x f y x y 2 x = 2y 1 2y = 3 x Thay vo (1): x3 + x 2 = 0x = 1. Nghim ca h (1;1)Cu VI.a:1.B = ABAC,B13;2 _ , Theo yu cu bi ton ta c v s tam gictha mn bi ton m cc cnh AC nm trn cc ng thng // vi nhau.Chn M(4;1) BC, M l trung im ca BC ==> C35;2 _ ,Tam gic ABC cn ti A, Vy AM BC ==> AM: 2x + y 9 = 022PPOOHHDDCCBBAASA = AM AB ==> A(6;-3)ng cao BH i qua B c VTPT ACuuur==> pt2. Gi d l giao tuyn ca ( ) v ( ) ==> d: 2 1 02 0x yx z ' Ly A(0;-1;0), B(1;1;2) d(P) qua A, (P) c dng phng trnh:Ax + By + Cz + B = 0(P) qua B nn: A + B + 2C + B = 0 ==> A = - (2B + 2C)Vy(P): - (2B + 2C)x + By + Cz + B = 0 2 2 22 2 2 22 2os93 (2 2 )B C B CcB C B C + + + + 13B2 -8BC 5C2 = 0, ChnC = 1 ==> B = 1; B = - 5/13 +. Vi B = C = 1; (P): - 4x + y + z + 1 = 0 +. Vi B = 5/13 v C = 1; (P): - 16x- 5y + 13z - 5 = 0Cu VII.a:Gi z = x + yi (x;y R)Ta c: 2 22 2( 1) ( 2) 2534x yx y + + '+ 22 75 28 15 0x yy y ' + 3529/ 53/ 5xyxy '

'

==> zCu VI.b:1.Cch gii nh cu VI.a , ng trung tuyn xut pht t B v qua trung im N ca AC 2. Ta c (D) nm trong (P) Gi A = (D)(P) , gii h ta c A(5;-1;5)Ly B(1+t;t;2+2t) (D);( 4; 1; 2 3) AB t t t + uuur lVTCP ca d Ta c cos300 = ( ) ( )2 226 9326 ( 4) 1 2 3tt t t + + + 14tt *) Vi t = - 1 th ABuuur = ( -5;0;-5) ==> d: 515x tyz t + ' +*) Vi t = 4 thABuuur = (0; 5;5) ==> d: 515xy tz t +' +Cu VII.b:3 3log 1 log4.15 5 0x xx++ 33 31loglog log23 4.15 5.5 0xx x + 33loglog3 34 5 05 5xx _ _ + , , 3log31 15xx _ ,---------------------------Ht------------------------- THI TH H- L QU N QUY NHN 2011( KHI A-B)I: PHN CHUNG CuI(2im): Cho hm s y = - 4x3 - 12x2 - 9x(C)1. Kho st s bin thin v v th (C) ca hm s.2. Xc nh tham sk (d): y = kx ct (C) ti ba im phn bit O; M; N v M l trung im ca ONCu II (2 im): 1: Gii phng trnh:cot2x cotx.cot3x = 23 2: Gii h phng trnh: t t= tan t,cot3x=(1-3tan^2)/3tanx-tan^3x. 2 23823 3 2 1xyx yx yx y x y x+ + +'+ + + Cu III (1im): Tnh tch phn :I = 10(2 1) ln( 1) x x dx +Cu IV (1im): Cho hnh lng tr tam gic ABC.ABC c y ABC l tamgic u cnh a,AA = AB = AC = 2 33a . Tnh th tch khi lng tr v khong cch t A n mt phng(ABC) theo aCu V(1im): Cho a, b, c> 0. Chng minh rng:3 3 3 3 3 3 2 2 22 2 211 11 1124 4 4b a c b a c a b cab b bc c ca a b c a _ + + + + + + + ,PHN RING: Th sinh ch c lm mt trong hai phn A hoc BA.Theo chng trnh chun Cu VI/a: (2im)1 . Trong mpOxy cho tam gic ABC cn ti A(3;1) v 3 1;2 2H _ , l trc tm. Xc nh ta nh B v C, bit B nm trn ng thng d: 3x y= 0.2. Trong kgOxyz cho hai ng thng (d): 1 2 11 2 2x y z , (d):2 5 31 4 1x y z CMR: (d) v (d) cho nhau. Vit phng trnh mt phng (Q) song song vi (d) v (d) ng thi khong cch t (d) n (Q) gp i khong cch t (d) n (Q).Cu VII/a:Trong mt phng phc (Oxy) xc nh tp hi cc im biu din cc s phc (1 + i)z + 1 bit 1 1 z B. Theo chng trnh nng cao Cu VI/b.(2im)1.Trong mpOxy chong thng d: 3x 4y + 22 = 0 v hai im A(6;0), B(1; -5). Vit phng trnh ng trn (S) i qua hai im A, B v tip xc vi d2. Trong kgOxyz cho mt cu (S): ( x 1)2 + y2 + z2 = 9, mt phng (P); y 2z = 0. Vit phng trnh mp(Q) i qua M(1;4;-1), (Q) (P) v (Q) tip xc vi (S).Cu VII/b: Gii h phng trnh 4 22 4log ( 1) log ( 1) 2log ( 1) log ( 1) 2x yx y + ' + + -------------------- Ht--------------------4S GIO DC V O TO THI TH I HC, CAO NG NM 2009THNH PH NNG Mn thi: TON, khi BTRNG THPT CHUYN L QU N Thi gian lm bi: 180 pht, khng k thi gian giao PHN CHUNG CHO TT C CC TH SINH (7 im)Cu I (2 im) Cho hm s 4 2( ) 2 y f x x x 1. Kho st v v th (C) ca hm s.2. Trn (C) ly hai im phn bit A v B c honh ln lt l a v b. Tm iu kin i vi a v b haitip tuyn ca (C) ti A v B song song vi nhau.Cu II (2 im) 1. Gii phng trnh lng gic:( ) 2 cos sin1tan cot 2 cot 1x xx x x+ 2. Gii bt phng trnh: ( )23 1 13 31log 5 6 log 2 log 32x x x x + + > +Cu III (1 im) Tnh tch phn: ( )24 40cos 2 sin cos I x x x dx +Cu IV (1 im) Cho mt hnh tr trn xoay v hnh vung ABCD cnh a c hai nh lin tip A, B nm trn ng trn y th nht ca hnh tr, hai nh cn li nm trn ng trn y th hai ca hnh tr. Mt phng (ABCD) to vi y hnh tr gc 450. Tnh din tch xung quanh v th tch ca hnh tr.Cu V (1 im) Cho phng trnh( ) ( )341 2 1 2 1 x x m x x x x m + + Tm m phng trnh c mt nghim duy nht.PHN RING (3 im): Th sinh ch lm mt trong hai phn (Phn 1 hoc phn 2)1. Theo chng trnh chun.Cu VI.a (2 im) 1. Trong mt phng vi h ta Oxy, cho ng trn (C) v ng thng nh bi: 2 2( ) : 4 2 0; : 2 12 0 C x y x y x y + + . TmimMtrn saochotMvcvi (C) hai tiptuyn lp vi nhau mt gc 600.2. Trong khng gian vi h ta Oxyz, cho t din ABCD vi A(2;1;0), B(1;1;3), C(2;-1;3), D(1;-1;0). Tmta tm v bn knh ca mt cu ngoi tip t din ABCD.Cu VII.a (1 im) C 10 vin bi c bn knh khc nhau, 5 vin bi xanh c bn knh khc nhau v 3 vin bi vng c bn knh khc nhau. Hi c bao nhiu cch chn ra 9 vin bi c ba mu?2. Theo chng trnh nng cao.Cu VI.b (2 im)1. Trong mt phng ta Oxy, cho hnh ch nht ABCD c din tch bng 12, tm I thuc ng thng ( ) : 3 0 d x y v c honh 92Ix , trung im ca mt cnh l giao im ca (d) v trc Ox. Tm ta cc nh ca hnh ch nht.2. Trong khng gian vi h ta Oxyz, cho mt cu (S) v mt phng (P) c phng trnh l2 2 2( ) : 4 2 6 5 0, ( ) : 2 2 16 0 S x y z x y z P x y z + + + + + + .im M di ng trn (S) v im N di ng trn (P). Tnh di ngn nht ca on thng MN. Xc nh vtr ca M, N tng ng.Cu VII.b (1 im) Cho , , a b c l nhng s dng tha mn: 2 2 23 a b c + + . Chng minh bt ng thc2 2 21 1 1 4 4 47 7 7 a b b c c a a b c+ + + ++ + + + + +----------------------Ht----------------------5p n.Cu Ni dung imI 2,001 1,00+ MX:D 0,25+ S bin thin Gii hn: lim ; limx xy y + + + ( )3 20' 4 4 4 1 ; ' 01xy x x x x yx t0,25 Bng bin thin( ) ( ) ( )1 21 1; 1 1; 0 0CT CTy y y y y y C 0,25 th0,252 1,00Ta c 3'( ) 4 4 f x x x . Gi a, b ln lt l honh ca A v B.H s gc tip tuyn ca (C) ti A v B l3 3'( ) 4 4 , '( ) 4 4A Bk f a a a k f b b b Tip tuyn ti A, B ln lt c phng trnh l:( ) ( ) ( ) ( ) ( ) ' ' ( ) af' a y f a x a f a f a x f a + + ; ( ) ( ) ( ) ( ) ( ) ' ' ( ) f' b y f b x b f b f b x f b b + + Hai tip tuyn ca (C) ti A v B song song hoc trng nhau khi v ch khi:( ) ( )3 3 2 24a 4a = 4b 4 1 0 (1)A Bk k b a b a ab b + + V A v B phn bit nna b , do (1) tng ng vi phng trnh:2 21 0 (2) a ab b + + Mt khc hai tip tuyn ca (C) ti A v B trng nhau ( ) ( ) ( ) ( ) ( )2 2 2 24 2 4 21 0 1 0' ' 3 2 3 2a ab b a ab ba bf a af a f b bf b a a b b + + + + ' ' + + , Gii h ny ta c nghim l (a;b) = (-1;1), hoc (a;b) = (1;-1), hai nghim ny tng ng vi cng mt cp im trn th l( ) 1; 1 v( ) 1; 1 .Vy iu kin cn v hai tip tuyn ca (C) ti A v B song song vi nhau l 62 21 01a ab baa b + + t'II 2,001 1,00iu kin: ( ) cos .sin 2 .sin . tan cot 2 0cot 1x x x x xx+ '0,25T (1) ta c: ( ) 2 cos sin1 cos .sin 22 sinsin cos 2 coscos1cos sin 2 sinx xx xxx x xxx x x + 0,252sin .cos 2 sin x x x ( )224cos224x kx kx k

+

+

0,25Giao vi iu kin, ta c h nghim ca phng trnh cho l( ) 24x k k + 0,252 1,00iu kin:3 x >0,25Phng trnh cho tng ng:( ) ( ) ( ) 1 1233 31 1 1log 5 6 log 2 log 32 2 2x x x x + + > +( ) ( ) ( )23 3 31 1 1log 5 6 log 2 log 32 2 2x x x x + > +( ) ( ) ( ) ( )3 3 3log 2 3 log 2 log 3 x x x x > +1 ]0,25( ) ( )3 32log 2 3 log3xx xx _ >1 ]+ ,( ) ( )22 33xx xx >+2109 110xxx

< > > 0,25Giao vi iu kin, ta c nghim ca phng trnh cho l10 x > 0,25III 1,001 1,00( )2202201cos 2 1 sin 221 11 sin 2 sin 22 2I x x dxx d x _ , _ ,0,507

( ) ( )2 220 032 20 01 1sin 2 sin 2 sin 22 41 1sin 2 sin 2 02 12| |d x xd xx x 0,50IV 1,00Gi M, N theo th t l trung im ca AB v CD. Khi OM AB v ' D O N C . Gi s I l giao im ca MN v OO.t R = OA v h = OO. Khi :OM I vung cn ti O nn: 2 2 2.2 2 2 2 2h aOM OI IM h a 0,25Ta c: 222 2 22 2 2 22 3a2 4 4 8 8a a a aR OA AM MO _ _ + + + , ,0,252 323a 2 3 2R . . ,8 2 16a aV h 0,25v 2a 3 2 32 Rh=2 . . .2 2 2 2xqa aS 0,25V 1,00Phng trnh( ) ( )341 2 1 2 1 x x m x x x x m + + (1)iu kin :0 1 x Nu[ ]0;1 x tha mn (1)th 1 x cng tha mn (1) nn (1) c nghim duy nht th cn c iu kin 112x x x . Thay 12x vo (1) ta c:301 12. 2.12 2mm mm + ' t0,25* Vi m = 0; (1) tr thnh:( )24 411 02x x x Phng trnh c nghim duy nht.0,25* Vi m = -1; (1) tr thnh( ) ( )( )( )( )( )( ) ( )442 24 41 2 1 2 1 11 2 1 1 2 1 01 1 0x x x x x xx x x x x x x xx x x x+ + + + + + Vi 4 411 02x x x 0,258+ Vi 11 02x x x Trng hp ny, (1) cng c nghim duy nht.* Vi m = 1 th (1) tr thnh: ( ) ( )( ) ( )2 24 441 2 1 1 2 1 1 1 x x x x x x x x x x + Ta thy phng trnh (1) c 2 nghim 10,2x x nn trong trng hp ny (1) khng c nghim duy nht.Vy phng trnh c nghim duy nht khi m = 0 v m = -1.0,25VIa 2,001 1,00ng trn (C) c tm I(2;1) v bn knh5 R .Gi A, B l hai tip im ca (C) vi hai tip ca (C) k t M. Nu hai tip tuyn ny lp vi nhau mt gc 600 th IAM l na tam gic u suy ra2R=2 5 IM .Nh th im M nm trn ng trn (T) c phng trnh:( ) ( )2 22 1 20 x y + .0,25Mt khc, im M nm trn ng thng, nn ta ca M nghim ng h phng trnh: ( ) ( )2 22 1 20 (1)2 12 0 (2)x yx y + '+ 0,25Kh x gia (1) v (2) tac:( ) ( )2 2232 10 1 20 5 42 81 0275xy y y yx

+ + +

0,25Vy c hai im tha mn bi l: 93;2M _ , hoc 27 33;5 10M _ ,0,252 1,00 Ta tnh c10, 13, 5 AB CD AC BD AD BC . 0,25Vy t din ABCD c cc cp cnh i i mt bng nhau. T ABCD l mt t din gn u. Do tm ca mt cu ngoi tip ca t din l trng tm G ca t din ny. 0,25Vy mt cu ngoi tip t din ABCD c tm l 3 3; 0;2 2G _ ,, bn knh l 142R GA . 0,50VIIa1,00S cch chn 9 vin bi ty l : 918C . 0,25Nhng trng hp khng c ba vin bi khc mu l:+ Khng c bi : Kh nng ny khng xy ra v tng cc vin bi xanh v vng chl 8.+ Khng c bi xanh: c 913Ccch.+ Khng c bi vng: c 915Ccch.0,259Mt khc trong cc cch chn khng c bi xanh, khng c bi vng th c 910Ccch chn 9 vin bi c tnh hai ln.Vy s cch chn 9 vin bi c c ba mu l: 9 9 9 910 18 13 1542910 C C C C + cch.0,50VIb 2,001 1,00I c honh 92Ix v( )9 3: 3 0 ;2 2I d x y I _ ,Vai tr A, B, C, D l nh nhau nn trung im M ca cnh AD l giao im ca (d) v Ox, suy ra M(3;0)( ) ( )2 2 9 92 2 2 3 24 4I M I MAB IM x x y y + + D12. D = 12AD = 2 2.3 2ABCDABCSS AB AAB ( ) AD dM AD ', suy ra phng trnh AD:( ) ( ) 1. 3 1. 0 0 3 0 x y x y + + .Li c MA = MD = 2.Vy ta A, D l nghim ca h phng trnh:( ) ( ) ( ) ( )2 2 2 22 23 03 33 2 3 3 2 3 2x yy x y xx y x x x y+ + + ' ' ' + + + 3 23 1 1y x xx y ' ' t hoc 41xy ' .Vy A(2;1), D(4;-1), 0,509 3;2 2I _ , l trung im ca AC, suy ra:2 9 2 722 3 1 22A CIC I AA C C I AIx xxx x xy y y y yy+ ' '+ Tng t I cng l trung im BD nn ta c: B(5;4).Vy ta cc nh ca hnh ch nht l (2;1), (5;4), (7;2), (4;-1).0,502 1,00Mt cu (S) tm I(2;-1;3) v c bn knh R = 3.Khong cch t I n mt phng (P):( ) ( )( ) 2.2 2. 1 3 16, 53d d I P d R+ + >.Do (P) v (S) khng c im chung.Do vy, min MN = d R = 5 -3 = 2.0,25Trong trng hp ny, M v tr M0 v N v tr N0. D thy N0 l hnh chiu vung gc ca I trn mt phng (P) v M0 l giao im ca on thng IN0 vi mt cu (S).Gi l ng thng i qua im I v vung gc vi (P), th N0 l giao im ca v (P). ngthng cvectchphngl ( ) 2; 2; 1 P n rvquaInn c phngtrnhl ( )2 21 23x ty t tz t + + ' .0,25Ta ca N0 ng vi t nghim ng phng trnh:( ) ( ) ( )15 52 2 2 2 1 2 3 16 0 9 15 09 3t t t t t + + + + + Suy ra 04 13 14; ;3 3 3N _ ,.0,2510Ta c 0 03.5IM IN uuuur uuur Suy ra M0(0;-3;4)0,25VIIb1,00p dng bt ng thc 1 1 4( 0, 0) x yx y x y+ > >+Ta c: 1 1 4 1 1 4 1 1 4; ;2 2 2a+b+c a b b c a b c b c c a a b c c a a b+ + + + + + + + + + + + +0,50Ta li c:( ) ( ) ( )2 2 22 2 2 22 2 21 2 22 4 4 2 2 02 2 4 72 1 1 1 0a b c a b ca b c a b c aa b c + + + + + + + + + + + Tng t: 2 21 2 1 2;2 7 2 7 b c a b c a b c + + + + + +T suy ra2 2 21 1 1 4 4 47 7 7 a b b c c a a b c+ + + ++ + + + + +ng thc xy ra khi v ch khi a = b = c = 1.0,5011S GIO DC V O TO THI TH I HC, CAO NG NM 2009THNH PH NNG Mn thi: TON, khi DTRNG THPT CHUYN L QU N Thi gian lm bi: 180 pht, khng k thi gian giao PHN CHUNG CHO TT C CC TH SINH (7 im)Cu I (2 im) Cho hm s( )3 2( ) 3 1 1 y f x mx mx m x + , m l tham s1. Kho st s bin thin v v th ca hm s trn khi m = 1.2. Xc nh cc gi tr ca m hm s ( ) y f x khng c cc tr.Cu II (2 im) 1. Gii phng trnh : ( )4 4sin cos 1tan cotsin 2 2x xx xx+ +2. Gii phng trnh:( ) ( )2 34 82log 1 2 log 4 log 4 x x x + + + +Cu III (1 im) Tnh tch phn 322121dxAx xCu IV (1 im) Cho hnh nn c nh S, y l ng trn tm O, SA v SB l hai ng sinh, bit SO = 3, khong cch t O n mt phng SAB bng 1, din tch tam gic SAB bng 18. Tnh th tch v din tch xung quanh ca hnh nn cho.Cu V (1 im) Tm m h bt phng trnh sau c nghim ( )227 6 02 1 3 0x xx m x m + + + 'PHN RING (3 im): Th sinh ch lm mt trong hai phn (Phn 1 hoc phn 2)1. Theo chng trnh chun.Cu VI.a (2 im)1. Trong mt phng vi h ta Oxy, cho tam gic ABC bit phng trnh cc ng thng cha cc cnhAB, BC ln lt l 4x + 3y 4 = 0; x y 1 = 0. Phn gic trong ca gc A nm trn ng thngx + 2y 6 = 0. Tm ta cc nh ca tam gic ABC. 2. Trong khng gian vi h ta Oxyz, cho hai mt phng( ) ( ) : 2 2z + 5 = 0; Q : 2 2z -13 = 0. P x y x y + + Vit phng trnh ca mt cu (S) i qua gc ta O, qua im A(5;2;1) v tip xc vi c hai mt phng(P) v (Q).Cu VII.a (1 im) Tm s nguyn dng n tha mn cc iu kin sau:4 3 21 1 24 31 154715n n nnn nC C AC A + + +12p nCu Ni dung imI 2,001 1,00Khi m = 1 ta c 3 23 1 y x x + + MX:D 0,25+ S bin thin: Gii hn: lim ; limx xy y + +2' 3 6 y x x + ; 2' 00xyx 0,25 Bng bin thin( ) ( ) 2 3; 0 1CTy y y y C0,25 th0,252 1,00+ Khi m = 0 1 y x , nn hm s khng c cc tr. 0,25+ Khi0 m ( )2' 3 6 1 y mx mx m + Hm s khng c cc tr khi v ch khi ' 0 y khng c nghim hoc c nghim kp0,50( )2 2' 9 3 1 12 3 0 m m m m m + 104m 0,2513II 2,001 1,00( )4 4sin cos 1tan cotsin 2 2x xx xx+ +(1)iu kin:sin 2 0 x 0,25211 sin 21 sin cos2(1)sin 2 2 cos sinxx xx x x _ + ,0,252211 sin 21 121 sin 2 1 sin 2 0sin 2 sin 2 2xx xx x Vy phng trnh cho v nghim.0,502 1,00( ) ( )2 34 82log 1 2 log 4 log 4 x x x + + + +(2)iu kin: 1 04 44 014 0xxxxx+ < < > ' ' + >0,25( ) ( ) ( )( )22 2 2 2 22 22 2(2) log 1 2 log 4 log 4 log 1 2 log 16log 4 1 log 16 4 1 16x x x x xx x x x + + + + + + + + 0,25+ Vi1 4 x