WWW.VNMATH.COM THI TH I HC S 1PHN CHUNG CHO TT C CC TH SINH (7 im)Cu I (2 im) Cho hm s 4 2( ) 8x 9x 1 y f x +1. Kho st s bin thin v v th (C) ca hm s.2. Da vo th (C) hy bin lun theo m s nghim ca phng trnh4 28 os 9 os 0 c x c x m + vi [0; ] x .Cu II (2 im) 1. Gii phng trnh: ( )3log12 22xx x x _ ,2. Gii h phng trnh: 2 22 21212x y x yy x y+ + ' Cu III (1 im) Tnh din tch ca min phng gii hn bi cc ng2| 4 | y x x v 2 y x .Cu IV (1 im) Cho hnh chp ct tam gic u ngoi tip mt hnh cu bn knh r cho trc. Tnh th tch hnh chp ct bit rng cnh y ln gp i cnh y nh.Cu V (1 im) nh m phng trnh sau c nghim24sin3xsinx + 4cos 3x -os x +os 2x +04 4 4c c 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. ChoABC c nh A(1;2), ng trung tuyn BM: 2 1 0 x y + + v phn gic trong CD: 1 0 x y + .Vit phng trnh ng thng BC.2. Trong khng gian vi h ta Oxyz, cho ng thng (D) c phng trnh tham s 222 2x ty tz t + ' +.Gi lngthngquaimA(4;0;-1) song song vi (D) v I(-2;0;2) l hnh chiu vung gc ca A trn (D). Trong cc mt phng qua, hy vitphngtrnhcamt phngckhongcchn(D) lln nht.Cu VII.a (1 im) Cho x, y, z l 3 s thc thuc (0;1]. Chng minh rng1 1 1 51 1 1 xy yz zx x y z+ + + + + + +2. Theo chng trnh nng cao.Cu VI.b (2 im) 1. Cho hnh bnh hnh ABCD c din tch bng 4. Bit A(1;0), B(0;2) v giao im I ca hai ng cho nm trn ng thng y = x. Tm ta nh C v D.1WWW.VNMATH.COM2. Trong khng gian vi h ta Oxyz, cho hai im A(1;5;0), B(3;3;6) v ng thng c phng trnh tham s 1 212x ty tz t + '.Mt im M thay i trn ng thng, xc nh v tr ca im M chu vi tam gic MAB t gi tr nh nht.Cu VII.b (1 im) Cho a, b, c l ba cnh tam gic. Chng minh1 1 223 3 2 3 3b caa b a c a b c a c a b _+ + + + < + + + + + + ,----------------------Ht----------------------P N THI TH S 1Cu Ni dung imI 2,001 1,00+ Tp xc nh:D 0,25+ S bin thin: Gii hn: lim ; limx xy y + + + ( )3 2' 32x 18x = 2x 16x 9 y 0' 034xyx

t0,25 Bng bin thin.( )3 49 3 49; ; 0 14 32 4 32CT CTy y y y y y _ _ , ,C0,25 th0,252 1,002WWW.VNMATH.COMXt phng trnh 4 28 os 9 os 0 c x c x m + vi [0; ] x (1)t osx t c , phng trnh (1) tr thnh: 4 28 9 0 (2) t t m + V[0; ] x nn [ 1;1] t , gia x v t c s tng ng mt i mt, do s nghim ca phng trnh (1) v (2) bng nhau.0,25Ta c: 4 2(2) 8 9 1 1 (3) t t m + Gi (C1): 4 28 9 1 y t t +vi [ 1;1] t v (D): y = 1 m.Phng trnh (3) l phng trnh honh giao im ca (C1) v (D).Ch rng (C1) ging nh th (C) trong min1 1 t .0,25Da vo th ta c kt lun sau:8132m >: Phng trnh cho v nghim.1.8132m : Phng trnh cho c 2 nghim.81132m >> 0,502 1,00iu kin: | | | | x y t 2 2; 0 u x y uv x y ' +; x y khng tha h nn xt x y ta c 212uy vv _ ,. H phng trnh cho c dng:212122u vu uvv+ _' , 0,25 3WWW.VNMATH.COM48uv ' hoc 39uv '+ 2 24488ux yvx y ' '+ (I)+ 2 23399ux yvx y ' '+ (II)0,25Gii h (I), (II).0,25Sau hp cc kt qu li, ta c tp nghim ca h phng trnh ban u l ( ) ( ) { }5;3 , 5; 4 S 0,25Sau hp cc kt qu li, ta c tp nghim ca h phng trnh ban u l ( ) ( ) { }5;3 , 5; 4 S 1,00III 0,25Din tch min phng gii hn bi: 2| 4 | ( ) y x x C v( ) : 2 d y x Phng trnh honh giao im ca (C) v (d):2 2 22 20 00| 4 | 2 2 4 2 6 064 2 2 0x xxx x x x x x x x xxx x x x x

' '

Suy ra din tch cn tnh:( ) ( )2 62 20 24 2 4 2 S x x x dx x x x dx + 0,25Tnh:( )220| 4 | 2 I x x x dx V[ ]20; 2 , 4 0 x x x nn 2 2| 4 | 4 x x x x + ( )22044 23I x x x dx + 0,254WWW.VNMATH.COMTnh( )622| 4 | 2 K x x x dx V[ ]22; 4 , 4 0 x x x v[ ]24; 6 , 4 0 x x x nn ( ) ( )4 62 22 44 2 4 2 16 K x x x dx x x x dx + .0,25Vy 4 52163 3S + 1,00IV 0,25Gi H, H l tm ca cc tam gic u ABC, ABC. Gi I, I l trung im ca AB, AB. Ta c: ( ) ( ) ( ) ' ' ' ' ''AB ICAB CHH ABBA CII CAB HH 'Suy ra hnh cu ni tip hnh chp ct ny tip xc vi hai y ti H, H v tip xc vi mt bn (ABBA) ti im' K II .0,25Gi x l cnh y nh, theo gi thit 2x l cnh y ln. Ta c:1 3 1 3' ' ' ' ' ;3 6 3 3x xI K I H I C IK IH IC Tam gic IOI vung O nn: 2 2 2 23 3' . . 6r6 3x xI K IK OK r x 0,25Th tch hnh chp ct tnh bi: ( )' . '3hV B B B B + +Trong : 2 2 22 24x 3 3 3r 33 6r 3; ' ; 2r4 4 2xB x B h 0,25T , ta c: 2 2 32 22r 3r 3 3r 3 21r . 36r 3 6r 3.3 2 2 3V _ + + ,0,25V 1,00Ta c:+/( ) 4sin3xsinx = 2 cos2x - cos4x;+/( ) 4 os 3x -os x +2 os 2x -os4x 2 sin 2x + cos4x4 4 2c c c c 1 _ _ _ + 1 , , , ]+/( )21 1os 2x +1 os 4x +1 sin 4x4 2 2 2c c _ _ _ + , , ,Do phng trnh cho tng ng:0,255WWW.VNMATH.COM( )1 12 os2x + sin2x sin 4x + m -0 (1)2 2c + t os2x + sin2x =2 os 2x - 4t c c _ , (iu kin: 2 2 t ). Khi 2sin 4x = 2sin2xcos2x = t 1 . Phng trnh (1) tr thnh:24 2 2 0 t t m + + (2)vi 2 2 t 2(2) 4 2 2 t t m + y l phung trnh honh giao im ca 2 ng ( ) : 2 2 D y m (l ng song song vi Ox v ct trc tung ti im c tung 2 2m) v (P): 24 y t t +vi 2 2 t .0,25Trong on 2; 2 1 ], hm s 24 y t t +t gi tr nh nht l 2 4 2 ti 2 t v t gi tr ln nht l 2 4 2 + ti 2 t . 0,25Do yu cu ca bi ton tha mn khi v ch khi 2 4 2 2 2 2 4 2 m +2 2 2 2 m .0,25VIa 2,001 1,00im( ) : 1 0 ;1 C CDx y Ct t + . Suy ra trung im M ca AC l 1 3;2 2t tM+ _ ,. 0,25im( )1 3: 2 1 0 2 1 0 7 7;82 2t tM BM x y t C+ _ + + + + ,0,250,25T A(1;2), k : 1 0 AK CDx y + ti I (imK BC ). Suy ra( ) ( ) : 1 2 0 1 0 AK x y x y + . Ta im I tha h:( )1 00;11 0x yIx y+ ' + . Tam gic ACK cn ti C nn I l trung im ca AK ta ca( ) 1; 0 K.ng thng BC i qua C, K nn c phng trnh: 14 3 4 07 1 8x yx y+ + + + 26WWW.VNMATH.COMGi (P) l mt phng i qua ng thng, th ( ) //( ) P D hoc ( ) ( ) P D . Gi H l hnh chiu vung gc ca I trn (P). Ta lun c IH IA v IH AH . Mt khc( ) ( ) ( ) ( ) ( )( ), , d D P d I P IHH P 'Trong mt phng( ) P,IH IA ; do axIH = IA H A m . Lc ny (P) v tr (P0) vung gc viIA ti A.Vect php tuyn ca (P0) l( ) 6; 0; 3 n IA r uur, cng phng vi( ) 2; 0; 1 v r.Phng trnh ca mt phng (P0) l:( ) ( ) 2 4 1. 1 2x - z - 9 = 0 x z + .VIIa rng( ) ( ) ( ) ( ) 1 1 1 0 xy x y x y + + ; v tng t ta cng c11yz y zzx z x+ + '+ +0,25V vy ta c:( )1 1 11 1 11 1 1 1 1 131 zx+y1511 55x y zx y zxy yz zx yz zx xyx y zyz xy zz yxyz zx y xy zz yxz y y z _+ + + + + + + + + + + + + + + , + + ++ + _ + + + + , _ + + + ,vv1,00Ta c:( ) 1; 2 5 AB AB uuur. Phng trnh ca AB l: 2 2 0 x y + .( ) ( ) : ; I d y x I t t . I l trung im ca AC v BD nn ta c: ( ) ( ) 2 1; 2 , 2 ; 2 2 C t t D t t . 0,25Mt khc: D. 4ABCS ABCH (CH: chiu cao) 45CH . 0,257WWW.VNMATH.COMNgoi ra:( )( ) ( )4 5 8 8 2; , ;| 6 4 | 43 3 3 3 3 ;5 50 1; 0 , 0; 2t C Dtd CAB CHt C D _ _ , ,

Vy ta ca C v D l 5 8 8 2; , ;3 3 3 3C D _ _ , , hoc( ) ( ) 1; 0 , 0; 2 C D 0,502 1,00Gi P l chu vi ca tam gic MAB th P= AB + AM + BM.V AB khng i nn P nh nht khi v ch khi AM + BM nh nht.ng thng c phng trnh tham s: 1 212x ty tz t + '.imM nn( ) 1 2 ;1 ; 2 M t t t + .( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( )( )( )( )22 2 2 2222 2 2 222 22 22 2 4 2 9 20 3 2 54 2 2 6 2 9 36 56 3 6 2 53 2 5 3 6 2 5AM t t t t tBM t t t t t tAM BM t t + + + + + + + + + + ++ + + +0,25Trong mt phng ta Oxy, ta xt hai vect( )3 ; 2 5 u t r v( )3 6; 2 5 v t +r.Ta c ( )( )( )( )2222| | 3 2 5| | 3 6 2 5u tv t +' +rrSuy ra| | | | AM BM u v + +r r v( )6; 4 5 | | 2 29 u v u v + + r r r rMt khc, vi hai vect, uvrr ta lun c| | | | | | u v u v + +r r r rNh vy2 29 AM BM + 0,25ng thc xy ra khi v ch khi, uvrr cng hng 3 2 513 6 2 5ttt +( ) 1; 0; 2 M v( ) min 2 29 AM BM + .0,25Vy khi M(1;0;2) th minP =( )2 11 29 +0,25VIIb 1,008WWW.VNMATH.COMV a, b, c l ba cnh tam gic nn:a b cb c ac a b+ > + >'+ >. t( ) , , , , 0 , ,2 2a b c ax y a z xyz x y z y z x z x y+ + > + > + > + > .V tri vit li:23 3 2a b a c aVTa c a b a b cx y zy z z x x y+ + + ++ + + + + ++ + +0,50Ta c:( ) ( )22z zx y z z x y z z x yx y z x y+ > + + < + >+ + +.Tng t: 2 2; .x x y yy z x y z z x x y z< +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.9WWW.VNMATH.COMPHN 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 + + . Tm im M trn sao cho t M v c vi (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). Tm ta 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 v tr 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----------------------P N THI S 2Cu 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 0,2510WWW.VNMATH.COM 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 ak 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 2 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 sin 1 cos .sin 22 sinsin cos 2 coscos1cos sin 2 sinx x x xxx x xxx x x + 0,252sin .cos 2 sin x x x 0,2511WWW.VNMATH.COM( )224cos224x kx kx k

+

+

Giao 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 _ , _ ,0,50

( ) ( )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 1,0012WWW.VNMATH.COMGi M, NtheothtltrungimcaABvCD. Khi OM AB v ' D ON 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 + Vi 11 02x x x Trng hp ny, (1) cng c nghim duy nht.0,2513WWW.VNMATH.COM* 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 . 0,50VIIa 1,00S 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.0,25Mt 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 2,0014WWW.VNMATH.COM1 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 2 2 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 c vectch phng l ( ) 2; 2; 1 P n rv qua I nn c phng trnh l ( )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 _ ,.0,2515WWW.VNMATH.COMTa 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 = 1.0,50 THI TH I HC - CAO NG- S 3Thi 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 cnh AB, BC ln lt l 4x + 3y 4 = 0; x y 1 = 0. Phn gic trong ca gc A nm trn ng thng16WWW.VNMATH.COMx + 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 + + +----------------------Ht----------------------P N THI S 3Cu 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 0,2517WWW.VNMATH.COM 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 mm m m + 104m 0,25II 2,001 1,0018WWW.VNMATH.COM( )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 < > sao choOB 8 =v gc 0AOB 60 = . Xc nh ta im C trn trc Oz th tch t din OABC bng 8.Cu VII.a (1,0 im)T cc ch s 0;1;2;3;4;5 c th lp c bao nhiu s t nhin m mi s c 6 ch s khc nhau v ch s 2ng cnh ch s 3.2. Theo chng trnh Nng cao:Cu VIb (2,0 im)1. Trong mt phng Oxy. Vit phng trnh ng thng ( ) D i qua im M(4;1) v ct cc tia Ox, Oy ln lt ti A v B sao cho gi tr ca tngOA OB +nh nht. 2. Trong khng gian (Oxyz) cho t din ABCD c ba nh A(2;1; 1),B(3;0;1),C(2; 1;3) - -, cn nh D nm trntrc Oy. Tm ta nh D nu t din c th tchV 5 =Cu VII.b (1,0 im)T cc s 0;1;2;3;4;5. Hi c th thnh lp c bao nhiu s c 3 ch s khng chia ht cho 3 m cc ch s trong mi s l khc nhau.------------------------Ht------------------------KT QU 4I. PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I (2,0 im)1. T gii2.m 2 Cu II (2,0 im)1. k2x k2 ;x6 3p p= p = +2.x 2;x 1 33 = =-CuIII (1,0 im)

4I ln3=24WWW.VNMATH.COMCu IV (1,0 im)V 83 =Cu V (1,0 im)minS 5 =II. PHN RING (3 im)Th sinh ch c lm mt trong hai phn (phn 1 hoc 2).1. Theo chng trnh Chun:Cu VIa (2.0 im)1. x 3y 6 0;x y 2 0 + - = - - =2. 1 2C (0;0; 3),C (0;0; 3) -Cu VII.a (1,0 im)192 s2. Theo chng trnh Nng cao:Cu VIb (2,0 im)1. x 2y 6 0 + - = 2. 1 2D (0; 7;0),D (0;8;0) -Cu VII.b (1,0 im)64 s------------------------Ht------------------------ THI TH TUYN SINH I HC- S 5Thi gian lm bi: 180 phtI. PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I (2,0 im)Cho hm s mx 4yx m+=+ (1)1. Kho st s bin thin v v th hm s (1) khim 1 =2. Tm tt c cc gi tr ca tham s m hm s (1) nghch bin trn khong ( ) ;1 - .Cu II (2,0 im)1. Gii phng trnh: 3 3 2cos x 4sinx 3cosxsin x sinx 0 - - + =25WWW.VNMATH.COM2. Gii phng trnh: ( ) ( )23 3log x 1 log 2x 1 2 - + - =CuIII (1,0 im)Tnh tch phn: 460dxIcos xp=Cu IV (1,0 im)Cho lng tr t gic u ABCD.A'B'C'D' c chiu cao bng h. Gc gia hai ng cho ca hai mt bn k nhauk t mt nh bng 0 0(0 90 ) a v lun tha mn 2 2 2a b c 3 + + =. Xc nh a, b, c sao cho khong cch t im O(0;0;0) n mt phng (ABC) ln nht.Cu VII.b (1,0 im)Trong mt phng cho a gic u (H) c 20 cnh. Xt tam gic c ng 3 nh c ly t cc nh ca (H).C tt c bao nhiu tam gic nh vy ? C bao nhiu tam gic c ng 2 cnh l cnh ca (H) ? C bao nhiu tam gic c ng mt cnh l cnh ca (H) ? C bao nhiu tam gic khng c cnh no l cnh ca (H) ?------------------------Ht------------------------KT QU 12I. PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I (2,0 im)1. T gii2.m 3 >-Cu II (2,0 im)1. x k , x k3p=p =- +p2. 3 3 28x log 10, x log27= =CuIII (1,0 im) ( )21I e 14= -42WWW.VNMATH.COMCu IV (1,0 im) 3aV36=Cu V (1,0 im)

minS 12, x y z 1 = = = =II. PHN RING (3 im)Th sinh ch c lm mt trong hai phn (phn 1 hoc 2).1. Theo chng trnh Chun:Cu VIa (2.0 im)x y 1 0; x y 1 0 + += + - =2. x y 5z 1 0;5x 17y 19z 27 0 + - - = - + - =Cu VII.a (1,0 im) 28800 cch 2. Theo chng trnh Nng cao:Cu VIb (2,0 im)1. 2x y 6 0 - - = 2.a b c 1 = = =Cu VII.b (1,0 im)1440, 20, 320, 800 tam gic------------------------Ht------------------------ THI TH TUYN SINH I HC- S 13Thi gian lm bi: 180 phtI. PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I (2,0 im)Cho hm s( )4 2y x 2 m 2 x 2m 3 =- + + - -(1) c th l ( )mC1. Kho st s bin thin v v th hm s (1), khim 0 =43WWW.VNMATH.COM2. nh m th ( )mC ct trc Ox ti bn im phn bit c honh lp thnh cp s cng.Cu II (2,0 im)1. Gii phng trnh: 4 41sinx cos x4 4p + + = 2. Gii phng trnh: ( )20,5log sinx 5sinx 2 149+ +=CuIII (1,0 im)Tnh tch phn: e1I cos(lnx)dxp=Cu IV (1,0 im)y ca hnh chp SABC l tam gic cn ABC cAB AC a = =v B C = =a. Cc cnh bn cng nghing vi y mt gc b. Tnh th tch ca khi chpSABCCu V (1,0 im)Cho x, y, z l cc s thc dng tha mn x y z 1 + + =. Tm gi tr nh nht ca biu thc:

2 2 21 1Px y z xyz= ++ +II. PHN RING (3 im)Th sinh ch c lm mt trong hai phn (phn 1 hoc 2).1. Theo chng trnh Chun:Cu VIa (2.0 im)1. Trong mt phng Oxy , cho im M( 3;1) - v ng trn ( )2 2C: x y 2x 6y 6 0 + - - + = . Gi 1 2T ,T l cc tip im ca cc tip tuyn k t M n (C). Vit phng trnh ng thng 1 1T T.2. Trong khng gian (Oxyz), cho hai ng thng ( ) ( )1 2x 3 2t' x 5 2td : y 1 t ;d : y 3 t'z 5 t z 1 t' = + = + =- =- - = - =- Chng t rng hai ng thng ( )1d v ( )2d song song vi nhau. Vit phng trnh mt phng ( ) a cha hai ng thng .Cu VII.a (1,0 im)Tnh gi tr ca biu thc ( )4 3n1 nA 3AMn 1 !++=+, bit rng 2 2 2 2n1 n 2 n 3 n 4C 2C 2C C 149+ + + ++ + + =2. Theo chng trnh Nng cao:Cu VIb (2,0 im)1. Trong mt phng Oxy, cho ng thng ( ) d : x y 1 0 - += v ng trn ( )2 2C: x y 2x 4y 0 + + - = . Tm ta im M thuc ng thng (d) sao cho t k n (C) c hai tip tuyn to vi nhau mt gc bng 060 .2. Trong khng gian (Oxyz), cho hai im A(2;0;0),M(1;1;1). Gi s (P) l mt phng thay i nhng lun lun i qua ng thng AM v ct cc trc Oy, Oz ln lt ti cc im B(0;b;0),C(0;0;c)(b,c 0) >. Chng minh rng bcb c2+ = v tm b,c sao cho din tch tam gic ABC nh nht.Cu VII.b (1,0 im)Tm s n nguyn dng tha mn bt phng trnh: 3 n2n nA 2C 9n-+ ------------------------Ht------------------------KT QU 1344WWW.VNMATH.COMI. PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I (2,0 im)1. T gii2. 13m 3,m9= =-Cu II (2,0 im)1. x k , x k4p=p = +p2. 1x k , x arctan( ) k2 5p= +p = +pCuIII (1,0 im) ( )1I e 12p= +Cu IV (1,0 im) 3a cos tanV6a b=Cu V (1,0 im)

1minS 30, x y z3= = = =II. PHN RING (3 im)Th sinh ch c lm mt trong hai phn (phn 1 hoc 2).1. Theo chng trnh Chun:Cu VIa (2.0 im)1. 2x y 3 0 + - =2. y z 4 0 - + =Cu VII.a (1,0 im)3n 5,M4= =2. Theo chng trnh Nng cao:Cu VIb (2,0 im)1. 213 21 213 21M(3;4),M'( 3; 2),N( ; ),N'( ; )3 2 3 3- +- - - 2.minS 46, b c 4 = = =Cu VII.b (1,0 im) n 3,n 4 = =------------------------Ht------------------------45WWW.VNMATH.COM THI TH TUYN SINH I HC- S 14Thi gian lm bi: 180 phtI. PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I (2,0 im)Cho hm s( )3 2y 2x 3 m 1 x 6mx 2 = - + + -(1) c th l ( )mC1. Kho st s bin thin v v th hm s (1), khim 1 =2. nh m th ( )mC ct trc trc hong ti duy nht mt im.Cu II (2,0 im)1. Gii phng trnh:9sinx 6cosx 3sin2x+cos2x 8 + - =2. Gii phng trnh: 3 3 3log 4 2 log x log 28x x .2 x3= -CuIII (1,0 im)Tnh tch phn: 2211I x .ln x dxx = + Cu IV (1,0 im)Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh a, 0BAD 60 =, SA vung gc vi mt phng ABCD,SA a = . Gi C' l trung im ca SC. Mt phng (P) i qua AC' v song song vi BD, ct cc cnh SB, SD ca hnh chp ln lt ti B', D'. Tnh th tch ca khi chp S.AB'C'D'.Cu V (1,0 im)Cho x, y l hai s dng v 2 2x y 1 + = . Tm gi tr nh nht ca biu thc:

( ) ( )1 1P 1 x 1 1 y 1y x = + + + + + II. PHN RING (3 im)Th sinh ch c lm mt trong hai phn (phn 1 hoc 2).1. Theo chng trnh Chun:Cu VIa (2.0 im)1. Trong mt phng Oxy , cho im A(3;4) v ng trn ( )2 2C: x y 4x 2y 0 + - - = . Vit phng trnh tip tuyn ( ) D ca(C), bit rng ( ) D i qua im A. Gi s cc tip tuyn tip xc vi (C) ti M, N. Hy tnh di on MN.2. Trong khng gian (Oxyz), cho ng thng( ) D l giao tuyn ca hai mt phng ( )( ) : 2x y z 1 0; : x 2y z 2 0 a - + += b + - - = v mt phng ( ) P: x y z 10 0 - + + =. Vit phng trnh hnh chiu vung gc ca( ) D trn mt phng (P).Cu VII.a (1,0 im)Gii h phng trnh: x xy yx xy y2.A 5.C 905.A 2.C 80 + = - =2. Theo chng trnh Nng cao:Cu VIb (2,0 im)46WWW.VNMATH.COM1. Trong mt phng Oxy, cho ng hai ng trn: ( ) ( )2 2 2 21 2C : x y 2x 2y 2 0, C : x y 8x 2y 16 0 + - - - = + - - + =. Chng minh rng ( )1C tip xc vi ( )2C. Vit phng trnh tip tuyn chung ca ( )1C v ( )2C.2. Trong khng gian (Oxyz), cho im ( ) A 1;2;3 v hai ng thng ( ) ( )1 2x 2 y 2 z 3 x 1 y 1 z 1d : ; d :2 1 1 1 2 1- + - - - += = = =- -. Vit phng trnh ng thng ( ) D i qua A, vung gc vi ( )1d v ct ( )2dCu VII.b (1,0 im)Gii bt phng trnh: 2 2 32x x x1 6A A C 102 x- +------------------------Ht------------------------KT QU 14I. PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I (2,0 im)1. T gii2. 1 3 m 1 3 - < Cu II (2,0 im)1. x k2; x k22p=- + p =p+ p2.x 256 =CuIII (1,0 im) I 2 =p-Cu IV (1,0 im) 310a 3V27=Cu V (1,0 im)

1Max P 6, x y z3= = = =II. PHN RING (3 im)49WWW.VNMATH.COMTh sinh ch c lm mt trong hai phn (phn 1 hoc 2).1. Theo chng trnh Chun:Cu VIa (2.0 im)1. ( )2 2x 3 y 4,A(1;0),B(3;2) - + =2. 176 19x yz7 73 2 1- -= =- -Cu VII.a (1,0 im)n 3;n 4 = =2. Theo chng trnh Nng cao:Cu VIb (2,0 im)1. ( ) ( ) ( ) ( )2 2 2 2x 4 y 6 18; x 2 y 2 8 + + - = - + + = 2. x 5 y 2 z 5x 3 y 4 z 5;2 3 1 2 3 1- + + + + -= = = =- - - -Cu VII.b (1,0 im) x 4y 5 == ------------------------Ht------------------------ THI TH I HC V CAO NG S 16.(Thi gian lm bi 180)I.PHN CHUNG CHO TT C TH SINH. (7 im)Cu I.(2 im) Cho hm s y = x3 + mx + 2 (1)1. Kho st s bin thin v v th ca hm s (1) khi m = -3.2. Tm m th hm s (1) ct trc hanh ti mt im duy nht.Cu II. (2 im)1. Gii h phng trnh :' + + +2 213 2 23 3y xy y xy x 2. Gii phng trnh: x x x tan sin 2 )4( sin 22 2 .Cu III.(1 im)50WWW.VNMATH.COMTnh tch phn I = 2124dxx xCu IV.(1 im)Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA = h vung gc mt phng (ABCD), M l im thay i trn CD. K SH vung gc BM. Xc nh v tr M th tch t din S.ABH t gi tr ln nht. Tnh gi tr ln nht .Cu V.(1 im)Tm m phng trnh sau c nghim thc:m x x +4 21II. PHN RING. (3 im)Th sinh ch c lm mt trong hai phn (phn a hc phn b)Cu VI a.(2 im) 1.Trong mt phng vi h ta Oxy, cho hai ng thng d1: x 2y + 3 = 0, d2 :4x + 3y 5 = 0. Lp phng trnh ng trn (C) c tm I trn d1, tip xc d2 v c bn knh R = 2.2.Trong khng gian vi h ta Oxyz cho hai ng thng d1: 2 1 1z y x , d2: '+ t zt yt x12 1 v mt phng (P): x y z = 0. Tm ta hai im M1d , N2d sao cho MN song song (P) v MN =6Cu VII a.(1 im) Tm s phc z tha mn :14 ,_

+i zi zCu VI b.(2 im)1. Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD c cnh AB: x 2y 1 = 0, ng cho BD: x 7y + 14 = 0 v ng cho AC qua im M(2 ; 1). Tm ta cc nh ca hnh ch nht.2. Trong khng gian vi h ta Oxyz cho ba im O(0 ; 0 ; 0), A(0 ; 0 ; 4), B(2 ; 0 ; 0) v mt phng (P): 2x + 2y z + 5 = 0. Lp phng trnh mt cu (S) i qua ba im O, A, B v c khang cch t tm I n mt phng (P) bng 35.Cu VII b.(1im)Gii bt phng trnh: 3 log 3 log3x x310xxxBt phng trnh tr thnh : 01 log1log11 log1log13log1log13 3 3 333< < < > < x x x xx x* 1 0 log3< < x x kt hp K : 0 < x < 1* 3 0 log3> > x x Vy tp nghim ca BPT: x) ; 3 ( ) 1 ; 0 ( + THI TH I HC V CAO NG- S 17(Thi gian lm bi 180)Cu I: (2 im) Cho hm s:( )3 23 1 9 2 y x m x x m + + + (1) c th l (Cm)1) Kho st v v th hm s (1) vi m =1.2) Xc nh m (Cm) c cc i, cc tiu v hai im cc i cc tiu i xng vi nhau qua ng thng 12y x .CuII: (2,5 im)1) Gii phng trnh:( )( )3sin 2 cos 3 2 3 os 3 3 os2 8 3 cos sinx 3 3 0 x x c x c x x + + .2) Gii bt phng trnh : ( )22 121 1log 4 5 log2 7x xx _+ > + ,.3) Tnh din tch hnh phng gii hn bi cc ng y=x.sin2x, y=2x, x=2.CuIII: (2 im) 1)Cho hnh lng tr ABC.ABC c y ABC l tam gic u cnh a, cnh bn hp vi y mt gc l 450. Gi P l trung im BC, chn ng vung gc h t A xung (ABC) l H sao cho 12AP AH uuur uuur. gi K l trung im AA, ( ) l mt phng cha HK v song song vi BC ct BB v CC ti M, N. Tnh t s th tch ' ' 'ABCKMNA B CKMNVV.2)Gii h phng trnh sau trong tp s phc:( )222 2 2 2656 0a aa aa b ab ba a+ +'+ + + Cu IV: (2,5 im)1) Cho m bng hng trng v n bng hng nhung khc nhau. Tnh xc sut ly c 5 bng hng trong c t nht 3 bng hng nhung? Bit m, n l nghim ca h sau:56WWW.VNMATH.COM2 2 1319 192 2720mm n mnC C AP++ +