6dedutruToan_DH_da

8/9/2019 6dedutruToan_DH_da

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THI TUYN SINH I HC (D TR ) MN TON

D B 1 KHI A:

Cu I:(2 )Gi (Cm) l th ca hm s : y =2 22 1 3x mx m

x m

(*) (m l tham s)

1. Kho st s bin thin v v th ca hm s (*) ng vi m = 1.2. Tm m hm s (*) c hai im cc tr nm v hai pha trc tung.

Cu II: ( 2 im) 1. Gii h phng trnh :2 2 4( 1) ( 1) 2

x y x y

x x y y y

2. Tm nghim trn khang (0; ) ca phng trnh :2 2 34sin 3 cos 2 1 2cos ( )

2 4

xx x

Cu III: (3 im) 1.Trong mt phng vi h ta Oxy cho tam gic ABC cn ti nh A c trng

tm G4 1

( ; )3 3

, phng trnh ng thng BC l 2 4 0x y v phng trnh ng thng BG l

7 4 8 0x y .Tm ta cc nh A, B, C.2.Trong khng gian vi h ta Oxyz cho 3 im A(1;1;0),B(0; 2; 0),C(0; 0; 2) .

a)

Vit phng trnh mt phng (P) qua gc ta O v vung gc vi BC.Tm ta giao im ca AC vi mt phng (P).b) Chng minh tam gic ABC l tam gic vung. Vit phng trnh mt cu ngai tipt din OABC.

Cu IV: ( 2 im). 1.Tnh tch phn3

2

0

sin .I x tgxdx

.

2. T cc ch s 1, 2, 3, 4, 5, 6, 7, 8, 9 c th lp c bao nhiu s t nhin, mi s gm 6 ch skhc nhau v tng cc ch s hng chc, hng trm hng ngn bng 8.

Cu V: (1 im) Cho x, y, z l ba s tha x + y + z = 0. Cmrng :

3 4 3 4 3 4 6x y z

Bi gii CU I

1/ Khi m = 1 th2x 2x 2

yx 1

(1)

MX: D = R\ {1}

2

2

x 2xy '

x 1

, y ' 0 x 0hay x 2

BBTx 0 1 2 y ' + 0 - - 0 +

y 2 6


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Tim cn:x 1 l pt t/c ngy = x + 3 l pt t/c xin2/ Tm m

Ta c

2 2

2

x 2mx m 1y '

x m

Hm s (*) c 2 cc tr nm v 2 phatrc tung

y ' 0 c 2 nghim tri du2

1 2x x P m 1 0 1 m 1

CU II: 1/ Gii h phng trnh

2 2x y x y 4I

x x y 1 y y 1 2

(I)

2 2

2 2x y x y 4x y x y xy 2 xy 2

Ta c 2 2 2 2 2 2S x y;P xy S x y 2xy x y S 2P

Vy

2

2

S 2P S 4 P 2I

S 0 hay S 1S P S 2

1S x y 0

TH :P xy 2

vy x, y l nghim ca phng trnh 2X 0X 2 0

Vy h c 2 nghimx 2

x 2

hayx 2

y 2

2S x y 1

TH :P xy 2

vy x,y l nghim ca phng trnh 2X X 2 0

X 1hay X 2 . Vy h c 2 nghimx 1

y 2

V

x 2

y 1

Tm li h Pt (I) c 4 nghimx 2

y 2

V

x 2

y 2

V

x 1

y 2

V

x 2

y 1

CCH KHC (I)

2 2

2 2

x y x y 4

x y x y xy 2

2 2x y x y 4

xy 2

2(x y) x y 0

xy 2

x y 0hay x y 1

xy 2

x y 0hay x y 1

xy 2

2

x y

x 2hay

2

x y 1

x x 2 0

x 2

y 2

V

x 2

y 2

V

x 1

y 2

V

x 2

y 1

2/ Tm nghim 0,


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Ta c 2 2x 3

4sin 3 cos2x 1 2 cos x2 4

(1)

(1) 3

2 1 cosx 3cos2x 1 1 cos 2x2

(1) 2 2cosx 3cos2x 2 sin2x

(1) 2cosx 3 cos2x sin2x . Chia hai v cho 2:

(1) 3 1

cosx cos2x sin2x2 2

cos 2x cos x6

5 2 7

x k a hay x h2 b18 3 6

Do x 0, nn h nghim (a) ch chn k=0, k=1, h nghim (b) ch chn h = 1. Do ta c ba

nghim x thuc 0, l 1 2 35 17 5

x ,x ,x18 18 6

CU III. 1/ Ta nh B l nghim ca h pt

x 2y 4 0B 0, 2

7x 4y 8 0

V ABC cn ti A nn AG l ng cao ca ABC

V GA BC pt GA: 4 1

2(x ) 1(y ) 0 2x y 3 03 3

2x y 3 0

GA BC = H

2x y 3 0H 2, 1

x 2y 4 0

Ta c AG 2GH

vi A(x,y).4 1 4 1

AG x, y ;GH 2 , 13 3 3 3

x 0

1 8y

3 3

A 0,3

Ta c :

A B C A B CG Gx x x y y y

x vay3 3

C 4,0

Vy A 0,3 ,C 4,0 ,B 0, 2

2a/ Ta c BC 0, 2,2

mp (P) qua O 0,0,0 v vung gc vi BC c phng trnh l 0.x 2y 2z 0 y z 0

Ta c AC 1, 1,2 , phng trnh tham s ca AC lx 1 t

y 1 t

z 2t

.

Th pt (AC) vo pt mp (P). Ta c1

1 t 2t 0 t3

. Th1

t3

vo pt (AC) ta c

2 2 2M , ,

3 3 3

l giao im ca AC vi mp (P)

2b/ Vi A 1,1,0 B 0,2,0 C 0,0,2 .Ta c: AB 1,1,0

, AC 1, 1,2

AB.AC 1 1 0 AB AC ABC vung ti A


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Ta d thy BOC cng vung ti O. Do A, O cng nhn on BC di 1 gc vung. Do A, O nm trn mt cu ng knh BC, s c tm I l trung im ca BC. Ta d dng

tm dc I 0,1,1 2 2R 1 1 2

Vy pt mt cu ngoi tip t din OABC l : 2 22x y 1 z 1 2

CU IV.

1/ Tnh

/ 3 / 3

2 2

0 0

sinxI sin xtgxdx sin x. dxcosx

2/ 3

0

1 cos x sinxI dx

cosx

, t u cosx du sinxdx

i cn 1

u ,u 0 13 2

21/2

1

1 u duI

u

=

11 2

1/ 2 1/ 2

1 u 3u du lnu ln2

u 2 8

2/ Gi 1 2 3 4 5 6n a a a a a a l s cn lp

3 4 5ycbt: a a a 8 3 4 5 3 4 5a ,a ,a 1,2,5 hay a ,a ,a 1,3,4

a) Khi 3 4 5a ,a ,a 1,2,5 C 6 cch chn 1a C 5 cch chn 2a C 3! cch chn 3 4 5a ,a ,a C 4 cch chn 6a

Vy ta c 6.5.6.4 = 720 s nb) Khi 3 4 5a ,a ,a 1,3,4 tng t ta cng c 720 s nTheo qui tc cng ta c 720 + 720 = 1440 s n

Cch khc Khi 3 4 5a ,a ,a 1,2,5

C 3! = 6 cch chn 3 4 5a a a

C 36A cch chn 1 2 6a ,a ,a

Vy ta c 6. 4.5.6 = 720 s n

Khi 3 4 5a ,a ,a 1,3,4 tng t ta cng c 720 s n

Theo qui tc cng ta c 720 + 720 = 1440 s n

CU V: Ta c: 4x x x3 4 1 1 1 4 4 4

84x x x3 4 2 4 2. 4 . Tng t 84y y x3 4 2 4 2. 4 8z z3 4 2 4

Vy

8 8 8x y z x y z3 4 3 4 3 4 2 4 4 4

3 8 x y z6 4 .4 .4 24 x y z6 4 6


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D B 2 KHI A:

Cu I:(2 im) 1. Kho st s bin thin v v th ( C ) ca hm s2 1

1

x xy

x

.

2. Vit phng trnh ng thng i qua im M (- 1; 0) v tip xc vi th ( C ) .

Cu II:( 2 im). 1. Gii h phng trnh :2 1 1

3 2 4

x y x y

x y

2. Gii phng trnh : 32 2 cos ( ) 3cos sin 04x x x

Cu III: (3 im). 1. Trong mt phng vi h ta Oxy cho ng trn(C): x2 + y2 12 4 36 0x y . Vit phng trnh ng trn (C1) tip xc vi hai trc ta Ox,Oy ng thi tip xc ngai vi ng trn (C).2. Trong khng gian vi h ta cac vung gc Oxyz cho 3 im A(2;0;0), C(0; 4; 0), S(0; 0;4) a) Tm ta im B thuc mt phng Oxy sao cho t gic OABC l hnh ch nht. Vit

phng trnh mt cu qua 4 im O, B, C, S.b) Tm ta im A1 i xng vi im A qua ng thng SC.


30

2

1

xI dx

x

.

2. Tm h s ca x7 trong khai trin a thc 2(2 3 ) nx , trong n l s nguyn dng tha mn:1 3 5 2 12 1 2 1 2 1 2 1...

n

n n n nC C C C = 1024. (

k

nC l s t hp chp k ca n phn t)

Cu V: (1 im) Cmrng vi mi x, y > 0 ta c :29(1 )(1 )(1 ) 256

yx

x y . ng thc xy ra khi no?

Bi gii:CU I.

1/ Kho st v v th

2x x 1y (C)

x 1

MX: D R \ 1 .

22

2x 2xy' ,y ' 0 x 2x 0 x 0hay x 2x 1

BBTx -2 -1 0 y ' + 0 - - 0 +y

-3

1

Tim cn:x 1 l phng trnh tim cn ng

y x l phng trnh tim cn xin

2/ Phng trnh tip tuyn qua M 1,0 ( h s

gc k ) c dng : y k x 1

tip xc vi C h pt sau c nghim


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2

2

2

x x 1k x 1

x 1

x 2xk

x 1

phng trnh honh tip im l

22

2

x 2x x 1x x 1

x 1 x 1

x 1 3

k4

Vy pt tip tuyn vi C qua M 1,0 l: 3

y x 14

CU II. 1/ Gii h pt : 2x y 1 x y 1

I3x 2y 4

2x y 1 x y 1

I 2x y 1 x y 5

t u 2x y 1 0,v x y 0

(I) thnh

1 12 2

2 2

u v 1 u 2 v 1

u 1 v 2 loaiu v 5

Vy 2x y 1 2

Ix y 1

2x y 1 4 x 2

x y 1 y 1

2/ Gii phng trnh 32 2 cos x 3cosx sinx 0 24

(2)3

2 cos x 3cosx sinx 04

3

3 3 2 2

cosx sinx 3cosx sinx 0

cos x sin x 3cos xsinx 3cosxsin x 3cosx sinx 0

3

cosx 0

sin x sinx 0

2 3 2 3

cosx 0hay

1 3tgx 3tg x tg x 3 3tg x tgx tg x 0

2

sin x 1 haytgx 1 x k2

hay

x k4 CU III

1/ 2 22 2C x y 12x 4y 36 0 x 6 y 2 4

Vy (C) c tm I 6,2 v R=2

V ng trn 1C tip xc vi 2 trc Ox, Oy nn tm 1I nm trn 2 ng thng y x

vv (C) c tm I 6,2 ,R = 2

nn tm 1I (x; x) vi x > 0.


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1TH : Tm 1I ng thng y = x I x,x , bn knh 1R x

1C tip xc ngoi vi (C) 1 1I I R R 2 2x 6 x 2 2 x

2 2 2 2x 6 x 2 4 4x x x 16x 4x 36 0

2x 20x 36 0 x 2hay x 18 .ng vi 1 1R 2hay R 18

C 2 ng trn l: 2 2

x 2 y 2 4 ; 2 2

x 18 y 18 18 2TH : Tm 1I ng thng y x I x, x ; 1R x

Tng t nh trn, ta c x= 6

C 1 ng trn l 2 2x 6 y 6 36

Tm li ta c 3 ng trn tha ycbt l:

2 2 2 2

2 2

x 2 y 2 4; x 18 y 18 18;

x 6 y 6 36

2a/ T gic OABC l hnh ch nht

OC AB B(2,4,0)* on OB c trung im l H 1,2,0 . H chnh l tm ng trn ngoi tip tam gic vungOBC. V A, O, C cng nhn SB di mt gc vung nn trung im I ( 1; 2; 2 ) l tm mt cu v

bn knh R = 1 1

SB 4 16 16 32 2

,

Vy phng trnh mt cu l 2 2 2x 1 y 2 (z 2) 9

2b/ SC 0,4, 4

chn 0,1, 1 l vtcp ca SC.

Pt tham s ng thng SC

x 0

y t

z 4 t

Mp (P) qua A 2,0,0 v vung gc vi SC c phng trnh l

O x 2 y z 0 y z 0

Th pt tham s ca SC v pt (P) Ta c t=2 v suy ra M 0,2,2

Gi 1A x,y,z l im i xng vi A qua SC. C M l trung im ca 1AA nn

2 x 2.0 x 2

0 y 2.2 y 4

0 z 2.2 z 4

Vy 1A 2,4,4

CU IV: 1/ Tnh7

30

x 2I dx

x 1

t 3 23t x 1 x t 1 dx 3t dt

3x 2 t 1 .i cn t( 0) = 1 ; t (7 ) = 2.

Vy

23 2 5 22 2 4

1 11

t 1 3t t t 231I dt 3 t t dt 3

t 5 2 10


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2/ Ta c

2n 1 0 1 2 2 3 3 2n 1 2n 1

2n 1 2n 1 2n 1 2n 1 2n 11 x C C x C x C x ... C x

Cho x 1 Ta c 2n 1 0 1 2 3 4 2n 12n 1 2n 1 2n 1 2n 1 2n 1 2n 12 C C C C C ... C

(1)

Cho x 1 Ta c 0 1 2 3 4 2n 12n 1 2n 1 2n 1 2n 1 2n 1 2n 10 C C C C C ... C

(2)

Ly (1) - (2) 2n 1 1 3 5 2n 12n 1 2n 1 2n 1 2n 12 2 C C C ... C

2n 1 3 5 2n 1 10

2n 1 2n 1 2n 1 2n 12 C C C ... C 1024 2

. Vy 2n=10

Ta c 10

10 k kk 10 k10

k 0

2 3x 1 C 2 3x

Suy ra h s ca 7x l 7 7 310C 3 .2 hay 3 7 310C 3 .2

CU V: Ta c:3

43

x x x x1 x 1 4

3 3 3 3

34

3 3

y y y y y1 1 4

x 3x 3x 3x 3 .x

3

4 3

9 3 3 3 31 1 4

y y y y y

2 64

3

9 31 16

y y

Vy

2 3 3 64

3 3 3 3

y 9 x y 31 x 1 1 256 256

x y 3 3 .x y

D B 1 KHI B:Cu I:(2 im). 1. Kho st s bin thin v v th ( C ) ca hm s 4 26 5y x x

2. Tm m phng trnh sau c 4 nghim phn bit : 4 2 26 log 0x x m .

Cu II: 2 im) 1. Gii h phng trnh : 2 1 13 2 4

x y x yx y

2. Gii phng trnh : 32 2 cos ( ) 3cos sin 04

x x x

Cu III: (3 im) 1. Trong mt phng vi h ta Oxy cho elip (E) :2 2

64 9

x y = 1. Vit phng

trnh tip tuyn d ca (E) bit d ct hai hai trc ta Ox, Oy ln lt ti A, B sao cho AO = 2BO.

2. Trong khng gian vi h ta Oxyz cho hai ng thng 1x y z

:1 1 2

d v

2

1 2

:1

x t

d y tz t

( t l tham s )

a) Xt v tr tng i ca d1 v d2 .b) Tm ta cc im M thuc d 1 v N thuc d2 sao cho ng thng MN song song vi mtphng (P) : 0x y z v di an MN = 2 .Cu IV: ( 2 im)

1. Tnh tch phn 20

lne

x xdx .


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2. Mt vn ngh c 15 ngi gm 10 nam v 5 n. Hi c bao nhiu cch lp mt nhmng ca gm 8 ngi bit rng trong nhm phi c t nht 3 n.

Cu V: (1 im) Cho a, b, c l ba s dng tha mn : a + b + c =3

4.. Cmrng :

3 3 33 3 3 3a b b c c a . Khi no ng thc xy ra ?Bi gii: CU I:

1/ Kho st4 2

y x 6x 5 . MX: D=R 3 2y ' 4x 12x 4x x 3 ,y ' 0 x 0hay x 3

2y '' 12x 12,y '' 0 x 1 BBTx 3 -1 0 1 3

y ' - 0 + + 0 - - 0 +y '' + + 0 - - 0 + +y 5

-4 0 0 -4 th

2/ Tm m pt 4 2 2x 6x log m 0 c 4 nghim phn bit.4 2 4 2

2 2x 6x log m 0 x 6x 5 log m 5

t 2k log m 5

Ycbt ng thng y=k ct (C) ti 4 im phn bit4 k 5 24 log m 5 5

2 91

9 log m 0 m 12

CU II 1/ Gii pt 3x 3 5 x 2x 4 1


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iu kin

3x 3 0

5 x 0 2 x 5

2x 4 0

(1) 3x 3 5 x 2x 4 v 2 x 5

3x 3 5 x 2x 4 2 5 x 2x 4 v 2 x 5

x 2 5 x 2x 4 v 2 x 5

x 2 0 hay[ x 2 5 x 2 v 2 x 5]

x 2 hay [x 2 2 5 x va2 x 5]

x 2 hay x 4

2/ Gii pt: 2 2 3sinx cos2x cos x tg x 1 2sin x 0 2

iu kin : cosx 0 x k2

2 2 3

2 sinxcos2x sin x cos x 2sin x 0v cosx 0 2sinx cos2x 2sin x cos2x 0 v cosx 0 sinx cos2x 1 cos2x cos2x 0 v cosx 0

2sin x 1 2sin x 0 v cosx 0 22sin x sinx 1 0 v cosx 0

1

sinx (vsinx 1 loai )2

1 5

sinx sin x k2 hay x k22 6 6 6 CU III.1/ Do tnh i xng ca elp (E). Ta ch cn xt trng hp x 0,y 0

Gi A 2m,0 ;B 0,m l giao im ca tip tuyn ca (E) vi cc trc ta ( m 0 ). Pt

AB:x y

1 x 2y 2m 02m m

AB tip xc vi (E) 264 4.9 4m

2 24m 100 m 25 m 5 m 0

Vy pt tip tuyn l x 2y 10 0 V tnh i xng nn ta c 4 tip tuyn lx 2y 10 0,x 2y 10 0

x 2y 10 0,x 2y 10 0

2/ a/ 1d qua O 0,0,0 , VTCP a 1,1,2

2d qua B 1,0,1 , VTCP b 2,1,1

a,b 1, 5,3

, OB 1,0,1


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1 2a,b OB 1 3 4 0 d ,d

cho nhau

b/ 1M d M t ',t ',2t ' ; 2N d N 1 2t,t,1 t

MN 2t t ' 1,t t ',t 2t ' 1

V MN // (P) pMN n 1, 1,1

pMN.n 0 2t t ' 1 t t ' t 2t ' 1 0 t t '

2 22MN t ' 1 4t ' 1 3t ' 2

24

14t ' 8t ' 2 2 2t ' 7t ' 4 0 t ' 0 hay t '7

* t=0 ta c M 0,0,0 O P loai

*4

t '7

ta c

4 4 8 1 4 3M , , ; N , ,

7 7 7 7 7 7

CU IV. 1/ Tnhe 2

1I x lnxdx

tdx

u lnx dux

; 3

2 xdv x dx chon v3

3e ee2 311 1

x 1 dxI x lnxdx lnx x

3 3 x

3 e3 3

1

x 1 2 1lnx x e

3 9 9 9

2. Ta c trng hp

* 3 n + 5 nam. Ta c 3 55 10C C 2520

* 4 n + 4 nam. Ta c 4 45 10C C 1050

* 5 n + 3 nam. Ta c 5 35 10C C 120

Theo qui tc cng. Ta c 2520 + 1050 + 120 = 3690 cchCU V:

Ta c

3

3

3

a 3b 1 1 1a 3b 1.1 a 3b 2

3 3b 3c 1 1 1

b 3c 1.1 b 3c 23 3

c 3a 1 1 1c 3a 1.1 c 3a 2

3 3

Suy ra 3 3 31

a 3b b 3c c 3a 4 a b c 63

1 34. 6 3

3 4

Du = xy ra

3a b c 1

a b c44

a 3b b 3c c 3a 1

Cch 2: t 33x a 3b x a 3b ; 33y b 3c y b 3c ;

33z c 3a z c 3a


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3 3 33

x y z 4 a b c 4. 34

. BT cn cm x y z 3 .

Ta c : 33 3x 1 1 3 x .1.1 3x ; 3 33y 1 1 3 y .1.1 3y ;

33 3z 1 1 3 z .1.1 3z 9 3 x y z (V 3 3 3x y z 3 ).Vy x y z 3

Hay 3 3 3a 3b b 3c c 3a 3

Du = xy ra 3 3 33

x y z 1 vaa b c4

a 3b b 3c c 3a 1 v3 1

a b c a b c4 4

D B 2 KHI B:

Cu I:(2 im) Cho hm s : y =2 2 2

1

x x

x

(*)

1. Kho st s bin thin v v th ( C ) ca hm s (*) .2. Gi I l giao im ca hai tim cn ca ( C ).Chng minh rng khng c tip tuyn no ca (C )

i qua im I .Cu II:( 2 im). 1. Gii bt phng trnh : 28 6 1 4 1 0x x x

2. Gii phng trnh : 22

cos 2 1( ) 3

2 cos

xtg x tg x

x

Cu III: (3 im). 1. Trong mt phng vi h ta Oxy cho 2 ng trn :(C1 ): x

2 + y2 9 v (C2 ): x2 + y2 2 2 23 0x y . Vit phng trnh trc ng phng d ca 2

ng trn (C1) v (C2). Chng minh rng nu K thuc d th khang cch t K n tm ca (C1)nh hn khang cch t K n tm ca ( C2 ).2. Trong khng gian vi h ta Oxyz cho im M(5;2; - 3) v mt phng(P) : 2 2 1 0x y z . a) Gi M1 l hnh chiu ca M ln mt phng ( P ). Xc nh ta

im M1 v tnh di an MM1. b) Vit phng trnh mt phng ( Q ) i qua M v cha

ng thngx-1 y-1 z-5

:2 1 -6


sin

0

( cos )xtgx e x dx

.

2. T cc ch s 1, 2, 3, 4, 5, 6, 7 c th lp c bao nhiu s t nhin, mi s gm 5 ch skhc nhau v nht thit phi c 2 ch 1, 5 ?

Cu V: (1 im) Cmrng nu 0 1y x th1

4x y y x . ng thc xy ra khi no?

Bi gii

CU I 1/ Kho st2x 2x 2

yx 1

(C)

MX: D R \ 1

22

2

x 2xy ' ,y ' 0 x 2x 0 x 0hay x 2

x 1

BBTx -2 -1 0


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y ' + 0 - - 0 +y

-2

2

Tim cnx 1 l pt t/c ng. y x 1 l pt t/c xin th :Bn c t v.

2/ Chng minh khng c tip tuyn no ca (C) i qua I 1,0 l giao im ca 2 tim cn.

Gi 2o o

o o o oo

x 2x 2M x ,y C y

x 1

Phng trnh tip tuyn ca (C) ti oM

2o o

o o o o o2o

x 2xy y f ' x x x y y x x

x 1

Tip tuyn i qua I 1,0

2o o o

o 2o

x 2x 1 x

0 y x 1

2 2o o o o

o o

x 2x 2 x 2xx 1 x 1

2 0 V l. Vy khng c tip tuyn no ca (C) i qua I 1,0

CU II 1/ Gii bt phng trnh 28x 6x 1 4x 1 0 (1)

(1) 28x 6x 1 4x 1

2

2 22

1 1x Vx

1 14 28x 6x 1 0 x Vx1 4 24x 1 0 x

14x 0hayx8x 6x 1 (4x 1) 48x 2x 0

1 1

x hay x4 2

2/ Gii phng trnh 22

cos2x 1tg x 3tg x

2 cos x

(2)

(2)

2

2 22sin xcot gx 3tg x cos x

2 3

1tg x 0 tg x 1 tgx 1 x k ,k Z

tgx 4

CU III 1/ ng trn 1C c tm O 0,0 bn knh 1R 3

ng trn 2C c tm I 1,1 , bn knh 2R 5


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Phng trnh trc ng phng ca 2 ng trn 1C , 2C l

2 2 2 2x y 9 x y 2x 2y 23 0 x y 7 0 (d)

Gi k k k kK x ,y d y x 7

2 2 22 2 2 2 2

k k k k k k k kOK x 0 y 0 x y x x 7 2x 14x 49

2 2 2 22 2

k k k k k kIK x 1 y 1 x 1 x 8 2x 14x 65

Ta xt 2 2 2 2k k k kIK OK 2x 14x 65 2x 14x 49 16 0 Vy 2 2IK OK IK OK (pcm)

2/ Tm 1M l h/c ca M ln mp (P)

Mp (P) c PVT n 2,2, 1

Pt tham s 1MM qua M, P l

x 5 2t

y 2 2t

z 3 t

Th vo pt mp (P): 2 5 2t 2 2 2t 3 t 1 0

18 9t 0 t 2 . Vy 1 1MM P M 1, 2, 1

Ta c 2 2 2

1MM 5 1 2 2 3 1 16 16 4 36 6

* ng thng

x 1 y 1 z 5:

2 1 6i qua A(1,1,5) v c VTCP a 2,1, 6

Ta c

AM 4,1, 8

Mt phng (Q) i qua M, cha mp (Q) qua A c PVT l

AM,a 2,8,2 hay 1,4,1 nn pt (Q): x 5 4 y 2 z 3 0

Pt (Q): x 4y z 10 0 Cch khc: Mt phng (Q) cha nn pt mp(Q) c dng:

x 2y 1 0haym(x 2y 1) 6y z 11 0 . Mt phng (Q) i qua M(5;2; - 3) nn ta c 5

4 + 1 = 0 ( loi) hay m( 5 4 + 1) + 12 3 11 = 0 m = 1.Vy Pt (Q): x 4y z 10 0

CU IV: 1/ Tnh

/ 4 sinx

0I tgx e cosx dx

Ta c: / 4 / 4 / 4 / 4

sinx sinx0 0 0 0sinxI tgxdx e cosxdx dx e cosxdxcosx

1

/ 4/ 4 sinx 20 o

ln cosx e ln 2 e 1

2/ Gi 1 2 3 4 5n a a a a a l s cn lp

Trc tin ta c th xp 1, 5 vo 2 trong 5 v tr: ta c: 25A 4.5 20 cchXp 1,5 ri ta c 5 cch chn 1 ch s cho cn li u tin

4 cch chn 1 ch s cho cn li th 2


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15

3 cch chn 1 ch s cho cn li th 3

* Theo qui tc nhn ta c: 25A .5.4.3 20.60 1200 s n.

Cch khc : - Bc 1 : xp 1, 5 vo 2 trong 5 v tr: ta c: 25A 4.5 20 cch

-Bc 2 : c 35A 3.4.5 60 cch bc 3 trong 5 s cn li ri xp vo 3 v tr cn li .Vy c 20.60 = 1200 s n tha ycbt.

CU V. Ta c2

0 x 1 x x Ta c

1 1x y y x x y y x

4 4 (1)

Theo bt ng thc Cauchy ta c

2 21 1 1

y x yx 2 yx . x y4 4 4

1

x y y x4

Du = xy ra

2

2

0 y x 1 x 1x x 1

y

1 4yx 4

D B 1 KHI D:Cu I:(2 im) Gi (Cm) l th ca hm s y= x

3+ ( 2m + 1) x2 m 1 (1)(m l tham s). 1) Kho st s bin thin v v th ca hm s (1) khi m 1.

2) Tm m th (Cm) tip xc vi ng thng y= 2mx m 1.

Cu II:( 2 im). 1. Gii bt phng tr nh : 2 7 5 3 2x x x

2. Gii phng trnh :3 sin

( ) 22 1 cos

xtg x

x

Cu III: (3 im). 1. Trong mt phng vi h ta Oxy cho ng trn(C): x2 + y2 4 6 12 0x y . Tm ta im M thuc ng thng

d : 2 3 0x y sao cho MI = 2R , trong I l tm v R l bn knh ca ng trn (C).2. Trong khng gian vi h ta Oxyz cho lng tr ng OAB.O1A1B1 vi A(2;0;0), B(0; 4; 0),O1(0; 0; 4)

a) Tm ta cc im A1, B1. Vit phng trnh mt cu qua 4 im O, A, B, O1.b) Gi M l trung im ca AB.Mt phng ( P ) qua M vung gc vi O 1A v ct OA,

OA1 ln lt ti N, K . Tnh di an KN.

Cu IV: ( 2 im). 1.Tnh tch phn3 2

1

ln

ln 1

ex

I dxx x

.

2. Tm k 0;1;2;.....;2005 sao cho 2005k

C t gi tr ln nht. ( knC l s t hp chp k ca n

phn t)Cu V: (1 im) Tm m h phng trnh sau c nghim:

2 1 2 1

2

7 7 2005 2005

( 2) 2 3 0

x x xx

x m x m

Bi giiCU I

1/ Kho st 3 2y x 2m 1 x m 1 khi m=1

Khi m = 1 th 3 2y x 3x 2


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16

MX: D=R

2y' 3x 6x 3x x 2 ,y ' 0 x 0hayx 2 y'' 6x 6,y '' 0 x 1 BBT

x 0 1 2 y ' - 0 + + -

y'' + + 0 - -y 2

lm -2 lm 0 li li

2/ Tm m mC tip xc vi y 2mx m 1 d

(d) tip xc vi mC

3 2

2

x 2m 1 x m 1 2mx m 1

3x 2 2m 1 x 2mc nghim

2

2

x 0hay x 2m 1 x 2m

3x 2 2m 1 x 2mc nghim

2

2 2

x 2m 1 x 2mm 0hay

3x 2 2m 1 x x 2m 1 xc nghim

2

2

x 2m 1 x 2mm 0hay

2x 2m 1 x 0c nghim

2x 2m 1 x 2mm 0hay 2m 1

x2

c nghim

222m 1 1

m 0hay 2m 1 2m2 2

1

m 0hay m2

CU II: 1/ Gii bpt 2x 7 5 x 3x 2 (1)


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17

iu kin

2x 7 02

5 x 0 x 53

3x 2 0

(1) 2

2x 7 3x 2 5 x va x 53

2x 7 3x 2 5 x 2 3x 2 5 x 2

va x 53

2 3x 2 5 x 2

va x 53

23x 17x 14 0 2

va x 53

14

(x 1 hay x)3

2

va x 53

2 14

x 1 hay x 53 3

2/ Gii phng trnh3 sinx

tg x 22 1 cosx

(2)

(2)sinx cosx sinx

cotgx 2 21 cosx sinx 1 cosx

2 2cosx cos x sin x 2sinx 2sinx cosx v sinx 0

cosx 1 2sinx cosx 1 v sinx 0

2sinx 1

x k26

hay

5

x k26

.

Ghi ch:Khi sinx 0 th cos x 1

CU III. 1/ ng trn (C) c tm I 2,3 , R=5

M M M M M MM x ,y d 2x y 3 0 y 2x 3

2 2

M MIM x 2 y 3 10

2 2 2M M M M

M M

M M

x 2 2x 3 3 10 5x 4x 96 0

x 4 y 5 M 4, 5

24 63 24 63x y M ,

5 5 5 5

2/ a/ V 1 1AA Oxy A 2,0,4

1 1BB Oxy B 0,4,4

Vit pt mt cu (S) qua O, A, B, O1

Ptmc (S):2 2 2x y z 2ax 2by 2cz d 0

V O S d 0

V A S 4 4a 0 a 1

V B S 16 8b 0 b 2

V 1O S 16 8c 0 c 2

Vy (S) c tm I(1,2,2)


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18

Ta c 2 2 2 2d a b c R

2R 1 4 4 9 Vy pt mt cu (S) l:

2 2 2x 1 y 2 z 2 9

b/ Tnh KN

Ta c M 1,2,0

, 1O A 2,0, 4

Mp(P) qua M vung gc vi 1O A nn nhn 1O A

hay (1;0; -2) lm PVT

pt (P): 1 x 1 0 y 2 2(z 0) 0 (P): x 2z 1 0

PT tham s OA l

x t

y 0

z 0

Th vo pt (P): t 1 0 t 1 OA P N 1,0,0

Pt tham s 1OA l:

x t

y 0

z 2t

vi 1OA 2,0,4

hay (1;0;2) l vtcp.

Th vo pt (P): 1

t 4t 1 0 t3

11 2

OA P K ,0,3 3

Vy

2 221 2 20 20 2 5

KN 1 0 0 03 3 9 3 3

CU IV: 1/ Tnh3 2e

1

ln xI dx

x l n x 1

t t lnx 1 2dx

t lnx 1 2tdtx

v 2t 1 lnx

i cn: 3t(e ) 2; t(1) 1

3 2 4 2e 2 2 4 2

1 1 1

ln x t 2t 1I dx 2tdt 2 t 2t 1 dt

tx l n x 1

25 3

1

t 2t 762 t

5 3 15

2. k2005C ln nhtk k 12005 2005

k k 12005 2005

C C

C C

k N

2005! 2005!k! 2005 k ! k 1 ! 2004 k ! k 1 2005 k

2005! 2005! 2006 k kk! 2005 k ! k 1 ! 2006 k !


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19

k 10021002 k 1003,k N

k 1003

k 1002hay k 1003

CU V: Tm m h phng trnh sau c nghim:

2x x 1 2 x 1

2

7 7 2005x 2005 (1)

x m 2 x 2m 3 0 (2)

iu kin l x 1.Ta c 2x x 1 2 x 17 7 0, x 1;1

Ta c: (1) x 1 2x 27 7 7 2005 1 x : ung x 1;1 v sai khi x > 1Do (1) 1 x 1 . Vy, h bpt c nghim

2f x x m 2 x 2m 3 0 c nghim 1,1

x 1;1

0 max f( 1), f (1) 0Maxf(x)

max 3m 6,m 2 0 3m 6 0hay m 2 0 m 2

D B 2 KHI D:

Cu I:(2 im) 1. Kho st s bin thin v v th ca hm s2 3 3

1

x xy

x

.

2. Tm m phng trnh2 3 3

1

x xm

x

c 4 nghim phn bit

Cu II:( 2 im). 1. Gii bt phng trnh :

2

22

2 19 2 33

x x

x x

.

2. Gii phng trnh : sin 2 cos 2 3sin cos 2 0x x x x Cu III: (3 im). 1. Trong mt phng vi h ta Oxy cho 2 im A(0;5),B(2; 3) . Vit phng trnh ng trn i qua hai im A, B v c bn knh R = 10 .2. Trong khng gian vi h ta Oxyz cho 3 hnh lp phng ABCD.A1B1C1D1 vi A(0;0;0),B(2; 0; 0), D1(0; 2; 2) a) Xc nh ta cc im cn li ca h nh lp phngABCD.A1B1C1D1.Gi M l trung im ca BC . Chng minh rng hai mt phng ( AB1D1) v (AMB1) vung gc nhau.b) Chng minh rng t s khang cch t im N thuc ng thng AC1 ( N A ) ti 2 mtphng ( AB1D1) v ( AMB1) khng ph thuc vo v tr ca im N.


2

0

( 2 1)cosI x xdx

.

2. Tm s nguyn n ln hn 1 tha mn ng thc : 2 22 6 12n n n nP A P A .

( Pn l s han v ca n phn t v knA l s chnh hp chp k ca n phn t)

Cu V: (1 im) Cho x, y, z l ba s dng v x yz = 1. Cmrng :

2 2 2 3

1 1 1 2

x y z

y z x

.

Bi gii


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20

CU I:

1/ Kho st 2x 3x 3

y Cx 1

MX: D R \ 1

22

2

x 2xy ' ,y ' 0 x 2x 0 x 0hay x 2

x 1

BBTx -2 -1 0 y ' + 0 - - 0 +y

-1

3

Tim cn: x=-1 l tc ngy = x + 2 l tc xin

2/ Tm m pt2x 3x 3

mx 1

c 4 nghim

phn bit

Ta c

2

2

2

x 3x 3neux 1

x 1x 3x 3y

x 1 x 3x 3neux 1

x 1

Do th

2x 3x 3y

x 1c c bng cch

Gi nguyn phn th (C) c x > -1Ly i xng qua Ox phn th (C) c x 3

CU II. 1/ Gii bt phng trnh 2

22x x

x 2x 19 2 3 13

Ta c (1)2 2x 2x x 2x9 2.3 3 . t

2x 2xt 3 0 , (1) thnh

2t 2t 3 0 1 t 3. Do , (1) 2 2x 2x x 2x 11 3 3 0 3 3

2 2x 2x 1 x 2x 1 0 1 2 x 1 2 2/ Gii phng trnh sin2x cos2x 3sinx cosx 2 0 2


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21

(2) 22sinxcosx 1 2sin x 3sinx cosx 2 0

22sin x 2cosx 3 sinx cosx 1 0

22sin x 2cosx 3 sinx cosx 1 0 ( 3 )(phng trnh bc 2 theo sinx)

C 2 22cosx 3 4 2 cosx 1 2cosx 1

Vy (2)

2cosx 3 2cosx 1 1sinx4 2

2cosx 3 2cosx 1sinx cosx 1

4

1

sinx cosx 1 hay sinx2

2 1sin x sin hay sinx

4 2 4 2

5

x k2 hay x k2 hay x k2 hayx k22 6 6 .Cch khc: (3) (2sinx 1) sinx cosx 1 0

CU III.1/ Gi I a,b l tm ca ng trn (C)

Pt (C), tm I, bn knh R 10 l

2 2

x a y b 10

2 2 2 2A C 0 a 5 b 10 a b 10b 15 0

(1)

2 2 2 2B C 2 a 3 b 10 a b 4a 6b 3 0(2)(1) v ( 2)

2 2 a 1 a 3a b 10b 15 0hay

b 2 b 64a 4b 12 0

Vy ta c 2 ng trn tha ycbt l

2 2

2 2

x 1 y 2 10

x 3 y 6 10

2/ Ta c A 0,0,0 ;B 2,0,0 ;C 2,2,0 ;D(0;2;0)

1 1 1 1A 0,0,2 ;B 2,0,2 ;C 2,2,2 ;D 0,2,2

Mp 1 1AB D c cp VTCP l:

1AB 2,0,2

1AD 0,2,2

mp 1 1AB D c 1 PVT l

1 11

u AB ,AD 1, 1,14


22/23

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22

mp 1AMB c cp VTCP l:

AM 2,1,0

M 2,1,0

1AB 2,0,2

mp 1AMB c 1 PVT l 1v AM,AB 1, 2, 1

2

Ta c:

u.v 1 1 1 2 1 1 0 u v 1 1 1AB D AMB

b/

1AC 2,2,2 Pt tham s

1

x t

AC : y t

z t

, 1N AC N t,t,t

Pt 1 1AB D : x 0 y 0 z 0 0 x y z 0

1 1 1t t t t

d N,AB D d3 3

Pt 1AMB : x 0 2 y 0 z 0 0 x 2y z 0

1 2t 2t t 2td N,AMB d1 4 1 6

1

2

ttd 6 6 23

2 td 2 t 23 2 36

Vy t s khong cch t 1N AC N A t 0 ti 2 mt phng 1 1AB D v 1AMB khng ph thuc vo v tr ca im N.

CU IV: 1/ Tnh / 2 / 22

0 0

1 cos2xI 2x 1 cos xdx 2x 1 dx2

2/ 2/ 2 21 0 0

1 1I 2x 1 dx x x

2 2 8 4

/ 2

2 0

1I (2x 1)cos2xdx

2

1 1

at u (2x 1) du dx,dv cos2xdxchonv sin2x2 2

/ 2 / 2 / 2

2 0 00

1 1 1 1I (2x 1)sin2x sin2xdx cos2x

4 2 4 2

Do 2/ 2 2

0

1I 2x 1 cos x

8 4 2

2/ Tac: 2 2n n n n2P 6A P A 12 n N,n 1

6n! n!

2n! n! 12n 2 ! n 2 !

n!

6 n! 2 6 n! 0n 2 !

n!6 n! 0hay 2 0

(n 2)! n! 6hay n(n 1) 2 0


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2n 3hay n n 2 0 n 3hay n 2(vn 2) CU V. Cho x,y, z l 3 s dng tha mn xyz=1

CMR:2 2 2x y z 3

1 y 1 z 1 x 2

Ta c:2 2x 1 y x 1 y

2 . x

1 y 4 1 y 4

2 2y 1 z y 1 z2 y

1 z 4 1 z 4

2 2z 1 x z 1 x2 z

1 x 4 1 x 4

Cng ba bt ng thc trn v theo v ta c:

2 2 2x 1 y y 1 z z 1 x

x y z1 y 4 1 z 4 1 x 4

2 2 2

x y z 3 x y z x y z1 y 1 z 1 x 4 4

3 x y z 34 4

3 3 9 3 6 3.3

4 4 4 4 4 2 ( v 3x y z 3 xyz 3 )

Vy2 2 2x y z 3

1 y 1 z 1 x 2

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