ĐỀ dỰ bỊ thi Đh 2002 - 2008 vÀ hd giẢichi tiet

8/13/2019 D B THI H 2002 - 2008 V HD GIIchi tiet

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T r n Xun Bang - Tr ng THPT Chuyn Qu ng Bnh

D b thi i Hc 2002- 2008 ra v Hng dn gii. 6/2010

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D BTHI I HC 2002- 2008 RA V HNG DN GII

http://www.VNMATH.com

http://www.vnmath.com


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PHN TH NHT

D BTHI I HC 2002- 2008




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S 1

Cu I: Cho hm s y= x4 - mx2 + m - 1 (1)(m l tham s)

1. Kho st s bin thin v v th hm s (1) khi m = 8.2. Xc nh m sao cho th hm s (1) ct trc honh ti 4 im phn bit.

Cu II: 1. Gii bt phng tr nh x 2x + 1 x1 1

2 2log (4 + 4) log (2 - 3.2 )

2. Xc nh m phng tr nh 2(sin4x + cos4x) + cos4xx + 2sin2x + m = 0 c tnht mt nghim thuc on 0;

2

.

Cu III:

1. Cho hnh chp S.ABC cy ABC l tam gic u cnh a v cnh bn SAvung gc vi mt phng y (ABC). Tnh khong cch t im A n mt phng

(SBC) theo a, bit rng a 6SA =2

.

2. Tnh tch phn1 3

20

xI = dxx + 1

.

Cu IV: 1. Trong mt phng Oxy cho hai ng tr n:

(C1): x2 + y2 - 10x = 0 (C2): x2 + y2+ 4x - 2y - 20 = 0Vit phng tr nhng tr ni qua cc giao im ca (C1) , (C2) v c tm nm

trn ng thng x + 6y- 6 = 0.3. Vit phng trr nhng tip tuyn chung ca hai ng tr n (C1) v (C2).

Cu V: 1. Gii phng tr nh 24 4 2 12 2 16 x x x x .2. i tuyn hc sinh gii camt trng gm 18 em, trong c 7 hc sinh

khi 12, 6 hc sinh k hi 11 v 5 hc sinh khi 10. Hi c bao nhiu cch c 8 hcsinh i d tri h sao cho mi khi c t nht mt em c chn.

Cu VI: Gi x, y, z l khong cch t im M thuc min trong ca tam gic ABC c bagc nhn n cc cnh BC, CA, AB. Chng minh rng:

2 2 2a + b + cx + y + z2R

; a, b, c l cnh tam gic, R l bn knh

ng tr n ngoi tip. Du ng thc xy ra khi no ?




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4

S 2

Cu I: 1. Tm s nguyn dng tho mn bt phng tr nh: 3 n-2n nA + 2C 9n, trong k k n nA , C ln lt l s chnh hp v s t hp chp k ca n. 2. Gii phng tr nh 84 22

1 1log (4x + 3) + log (x - 1) log (4 )2 4

x

Cu II:

Cho hm s2x - 2x + my =

x - 2 (1)(m l tham s).

1. Xc nh m hm s (1) nghch bin trn on [- 1; 0].2. Kho st s bin thin v v th hm s (1) khi m = 1.3. Tm a phng tr nh sau c nghim:

2 2

1 + 1 - t 1 + 1 - t9 - (a + 2).3 + 2a + 1 = 0 Cu III:

1. Gii phng tr nh4 4sin x + cos x 1 1 = cotx -5sin2x 2 8sin2x

2. Xt tam gic ABC c di cc cnh AB = c, BC = a, CA = b. Tnh din tch

tam gic ABC, bit rng: bsinC(b.cosC + c.cosB) = 20.

Cu IV: 1. Cho t din OABC c cc cnh OA, OB v OC i mt vung gc. Gi

, , ln lt lcc gc gia mt phng (ABC) vi cc mt phng (OBC), (OCA)v (OAB), chng minh rng:cos + cos + cos 3.

2.Trong khng gian Oxyz cho mf(P): x - y + z + 3 = 0v hai im A(- 1; - 3; -2), B( - 5; 7; 12).

a) Tm to im A' i xng im A qua mf(P). b) Gi s M l mt im chy tr n mf(P), tm gi tr nh nht ca MA + MB.

Cu V:

Tnhln3 x

x 30

I = .(e 1)e dx




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6

S 4

Cu I: 1. Gii bt phng tr nh: x + 12 x - 3 + 2x + 1

2. Gii phng tr nh tanx + cosx - cos2x = sinx(1 + tanx.tanx2 ).

Cu II: Cho hm s y = (x- m)3 - 3x (m l tham s). 1. Xc nh m hm s chot cc tiu tai im c honh x = 0. 2. Kho st s bin thin v v th hm s cho khi m = 1.3. Tm k h bt phng tr nh sau c nghim:

3

2 32 2

x - 1 - 3x - k < 0

1 1log x + log (x - 1) 12 3

Cu III: 1. Cho tam gic ABC vung cn c cnh huyn BC = a. Trn ng thng

vung gc vi mt phng(ABC) ti A ly im S sao cho gc gia hai mt phng(ABC) v (SBC) bng 600. Tnh di SA theo a.

2. Trong khng gian Oxyz cho hai ng thng:

d1: 2x - az - a = 0 ax + 3y - 3 = 0

d :

y - z + 1 = 0 x - 3z - 6 = 0

a) Tm a hai ng thng d1 v d2 ct nhau. b) Vi a = 2, vit phng tr nh mt phng(P) cha d2 v song song d1 v tnh

khong cch gia d1 v d2.

Cu IV: 1. Gi s nl s nguyn dng v

(1 + x)n = a0 + a1x + a2x2 + ...+ak x2k +...+anxn.Bit rng tn ti s nguyn k(0 k n - 1 sao cho 1 1

2 9 24k k k a a a . Hy tnh n ?

2. Tnh tch phn0

2x 3

- 1I = x(e + x + 1)dx

Cu V: Gi A, B, C l ba gc ca tam gicABC. Chng minh rng tam gic ABC u

thiu kn cn v l:2 2 2A B C 1 A - B B - C C - Acos + cos + cos - 2 = cos cos cos

2 2 2 4 2 2 2




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7

S 5

Cu I: Cho hm s y =2x + mx1 - x

(1)(m l tham s)

1. Cho m =12 . a) Kho st s bin thin v v th hm s (1) khi m = 0. b) Vit phng tr nh tip tuyn ca th (C), bit rng tip tuyn

song song vi ng thng y = 4x + 2. 2. Tm m hm s (1) cc tr. Vi gi tr no ca m th khong cch gia haiim cc tr ca th hm s (1) bng 10. Cu II:

1. Gii phng tr nh 3 232716log 3log 0 x x x x .

2. Cho phng tr nh 2sinx + cosx+1sinx-2cosx+3

a (2)(a l tham s)

a) Gii phng tr nh (2) khi a =13

. b) Tm a phng tr nh (2) c nghim.

b) Tm a phng tr nh (2) c nghim. Cu III:

1. Trong mt phng Oxy cho ng thng d: x- y + 1 = 0 v ng tr n (C):x2 + y2 + 2x - 4y = 0. Tm ta im M thuc ng thng d m qua kc hai ng thng tip xc vi ng tr n (C) ti A v B sao cho gc AMB bng 600.

2. Trong khng gian Oxyz cho ng thng2x - 2y - z + 1 = 0d:x + 2y - 2z - 4 = 0

v mt

cu (S): x2 + y2 + z2 + 4x - 6y + m = 0. Tm m ng thng d ct mt cu ti hai im M, N sao cho MN = 9.

3. Tnh th tch ca khi t din ABCD, bit AB = a, AC = b, AD = c v cc gcBAC, CAD, DAB u bng 600.Cu IV:

1. Tnh tch phn2

6 3 5

0I = 1 - cos x .sinxcos xdx .

2. Tm gii hn

3 2 2

x 0

3x - 1 2 1L = lim 1 - cosx

x

Cu V: Gi s a, b, c l bn s nguyn thay i tho mn 1 a < b < c < d 50 .

Chng minh bt ng thc2a c b + b + 50 +

b d 50b v tm gi tr nh nht ca biu

thc a c + b d

.




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8

S6

Cu I:

1. Kho st v v thi hm s y = 3 21 2 33

x x x (1)

2. Tnh din tch hnh phng gii hn bi th hm s (1) v tr c honh.

Cu II:

1. Gii phng tr nh 21 sinx

8 osc x.

2. Gii h phng trnh3 2

3 2

log ( 2 3 5 ) 3log ( 2 3 5 ) 3

x

y

x x x y

y y y x

Cu III:

1. Cho hnh t din u ABCD, cnh a =6 2cm. Hy xcnh v tnh dion vung gc chung ca ng thng AD v ng thng BC.

2. Trong mt phng Oxy cho elip (E) :2 2x + = 1

9 4 y

v ng thng dm: mx - y - 1 = 0a) Chng minh rng vi mi gi tr ca m, ng thng dm lun ct elip (E) ti

hai im phn bit. b) Vit phng tr nh tip tuyn ca (E) , bit rng tip tuyn i qua im

N(1; - 3).Cu IV:

Gi a1, a2, ..., a11 l cc h s trong khai trin (x + 1)10 (x + 2) = x11 + a1x10+ ...+a11. Hy tnh h s a5.Cu V:

1. Tm gii hn6

2x 1

x - 6x + 5L = lim(x - 1)

.

2. Cho tam gic ABC c din tch bng32

. Gi a, b, c ln lt l di cc cnh

BC, CA, AB v ha, h b, hc tng ng l di cc ng cao k t cc nh A, B,

C ca tam gic. Chng minh rng: 1 1 1 1 1 1 3

a b ca b c h h h




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9

S 7

Cu I:

1. Kho st v v thi hm s22x - 4x - 3y =2(x - 1)

.

2. Tm m phng tr nh 2x2 - 4x - 3 + 2m 1 x = 0 c hai nghim phn bit.

Cu II: 1. Gii phng tr nh 3 - tanx(tanx + 2sinx) + 6cosx = 0 .

2. Gii h phng tr nh y xx y

log xy = log y2 + 2 = 3

Cu III: 1. Trong mt phng Oxy cho parabol (P):2y x v im I(0; 2). Tm to hai

im M, N thuc (P) sao choIM = 4IN

.2. Trong khng gian Oxyz cho t din ABCD vi A(2; 3; 2), B(6;- 1; - 2),

C( - 1; - 4; 3), D(1; 6; -5). Tnh gc gia hai ng thng AB v CD. Tm to im M thuc ng thng CD sao cho tam gic ABM c chu vi nh nht.

3. Cho lng tr ng ABCA'B'C' c y ABCl tam gic cn vi AB = AC = av gc 0BAC = 120, cnh bn BB' = a. Gi I l trung imCC'. Chng minh rngtam gic AB'I vung A. Tnh cosin ca gc gia hai mt phng (ABC) v(AB'I).

Cu IV: 1. C bao nhiu s t nhin chia ht cho 5 m mi s c 4 ch s khc nhau.

2. Tnh tch phn:4

0

xdxI =1 + cos2x

Cu V: Tm gi tr ln nht v gi tr nh nht ca hm s: y = sin5x + 3cosx.




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10

S 8

Cu I:

Cho hm s2 2x + (2m + 1)x + m 4y =

2(x + m)

m (1)(m l tham s).

1. Tm m hm s (1) c cc tr v tm khong cch gia hai im cc tr ca th hm s (1).

2. Kho st s bin thin v v th hm s (1) khi m = 0.

Cu II: 1. Gii phng tr nh cos2x + cosx(2tan2x - 1) = 2 .2. Gii bt phng tr nh x + 1 x x + 115.2 + 1 2 - 1 + 2 .

Cu III: 1. Cho t din ABCD vi AB = AC = a, BC = b. Hai mt phng (BCD) v

(ABC) vung gc nhau v gc 090 BDC . Xc nh tm v bn knh mt cungoi tip t din ABCD theo a v b.

2. Trong khng gian Oxyz cho hai ng thng :

11:

1 2 1 x y z

d 2

3 1 0:

2 1 0 x z

d x y

a) Chng minh rng, d1 v d2 cho nhau v vung gc nhau. b) Vit phng tr nh tng qut ca ng thng d ct c hai ng v song

song vi ng thng: 4 7 3

1 4 2 x y z

.

Cu IV: 1. 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 2 ng cnh ch s ba.

2. Tnh tch phn:1

3 2

0I = x 1 - x dx

Cu V:

Tnh cc gc ca tam gic ABC bit rng4 ( )

2 3 3sin sin sin2 2 2 8

p p a bc

A B C

trong BC = a, CA = b, AB = c v a + b +c p =2

.




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11

S 9

Cu I: Cho hm s 2y = (x - 1)(x + mx + m) (1)(m l tham s).

1. Tm m hm s (1) ct trc honh ti ba im phn bit. 2. Kho st s bin thin v v th hm s (1) khi m = 4.

Cu II: 1. Gii phng tr nh 3cos4x - 8cos6x + 2cos2x + 3 = 0.2. Tm m phng tr nh

2

2 12

4 log x - log x + m = 0 c nghim thuc (0; 1).

Cu III: 1. Trong mt phng Oxy cho ng thng d: x - 7y + 10 = 0. Vit phng tr nh

ng tr n c tm thuc ng thng: 2x + y = 0 v tip xc vi ng thng d

ti im A(4; 2) 2. Cho hnh lp phng ABCD.A'B'C'D'. Tm im M thuc cnh AA' sao chomt phng (BD'M) cthnh lp phng theo mt thit din c din tch nh nht.

3. Trong khng gian Oxyz cho t din OABC vi A(0; 0; a3 ), B(a; 0; 0),C(0; a 3 ; 0) (a > 0). Gi M l trung im BC. Tnh khong cch gia hai ngthng AB v OM.

Cu IV: 1. Tm gi tr ln nht v gi tr nh nht ca hm s

y = x6 + 4(1 - x2)3 trn on [- 1; 1].2. Tnh tch phn:

ln5 2x

xln2

eI = dxe 1

Cu V: T cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin, mi s c 6

ch s v tho mniu kin: Su ch s ca mi s l khc nhau v trong mi s tng ca ba ch s u

nh hn tng ca ba ch s cui mt n v?




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12

S 10

Cu I:

Cho hm s 2x - 1y =x - 1

(1)

1. Kho st s bin thin v v th (C) ca hm s (1). 2. Gi I l giao im hai ng tim cn ca (C). Tmim M thuc (C) sao cho

tip tuyn ca (C) ti M vung gc vi ng thng IM.

Cu II:

1. Gii phng tr nh 2 x 2 - 3 cosx - 2sin -2 4 = 1

2cosx - 1

.

2. Gii bt phng tr nh 1 1 22 4

log x + 2log (x - 1) + log 6 0.

Cu III:

1. Trong mt phng Oxy cho elip (E):2 2x y + = 1

4 1, M( - 2; 3), N(5; n). Vit

phngtrnh ccng thng d1, d2 i qua M v tip xc vi (E). Tm n trongs cc tip tuyn ca (E) qua N c mt tip tuyn song song vi d1 hoc d2.

2. Cho hnh chpu S.ABC, y ABC c cnh bng a, mt bn to vi y mtgc bng 0 0(0 90 ) . Tnh th tch khi chp S.ABC v khong cch t nhA n mt phng (SBC).

3. Trong khng gian Oxyz cho hai im I(0; 0; 1), K(3; 0; 0). Vit phng tr nhmt phng i qua hai im I, K v to vi mt phngOxy mt gc 300.

Cu IV: 1. T mt t gm 7 hc sinh n v 5 hc sinh nam cn chn ra 6 em trong s

hc sinh n phi nh hn 4. Hi c bao nhiu cch chn nh vy.

2. Cho hm s x3af(x) = + bxe

(x + 1). Tm a v b bit rng:

f '(0) = - 22 v1

0(x)dx = 5 f

Cu V:

Chng minh rng2

x xe + cosx 2 + x -2

, x .




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13

S 11

Cu I:

Cho hm s2 2x + 5x + m 6y =

x + 3 (1)( m l tham s)

1. Kho st s binthin v v th ca hm s (1) khi m = 1. 2. Tm m hm s (1) ng bin tr n khong(1; + ).

Cu II:

1. Gii phng tr nh2cos x(cosx - 1) = 2(1 + sinx)

sinx + cosx.

2. Cho hm s xf(x) = xlog 2, (x > 0, x 1).Tnh f '(x) v gii bt phng tr nh f '(x) 0.

Cu III: 1. Trong mt phng Oxy cho tam gic ABC c nh A(1; 0) v hai ng thng

ln lt cha cc ng cao v t B v C c phng tr nh tng ng l x - 2y + 1= 0 v 3x + y - 1 = 0.

Tnh din tch tam gic ABC.2. Trong khng gian Oxyz cho mt phng (P): 2x + 2y + z- m2 - 3m = 0(m l

tham s) v mt cu (S): 2 2 2x - 1 + y + 1 + z - 1 = 9.Tm m mt phng (P) tip xc mt cu (S). Vi m va tmc, hy xcnh

to tip im ca mt phng (P) v mt cu (S). 3. Cho hnh chp S.ABC cy ABC l tam gic vung ti B, AB = a, BC =2a, cnh SA vung gc vi y v SA = 2a. Gi M l trung im ca SC. Chng

minh r ng tam gic AMB cn ti M v tnh din tchtam gic AMB theo a.

Cu IV: 1. T 9 ch s 0, 1, 2, 3, 4, 5, 6, 7, 8 c th lp c bao nhiu s t nhin chn

m mi s gm 7 ch s khc nhau?

2. Tnh tch phn I = 21

3 x

0x e dx .

Cu V: Tnh cc gc A, B, C ca tam gic ABC biu thc: 2 2 2Q = sin A + sin B - sin C t gi tr nh nht.




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14

S12

Cu I: 1. Kho st s bin thin v v th (C) ca hm s: y = 2x3 - 3x2 - 1.2. Gi dk l ng thng i qua M(0;- 1) v c h s gc bng k. Tm k

ng thng dk ct (C) ti ba im phn bit.

Cu II: 1. Gii phng tr nh 2cos4xcotx = tanx +

sin2x

2. Gii phng tr nh x5log 5 4 = 1 - x

Cu III: 2. Trong khng gian Oxyz cho hai im A( 2; 1; 1), B(0;- 1; 3) v ng thng

d: 3x - 2y - 11 = 0y + 3z - 8 = 0

a) Vit phng tr nh mt phng (P) i qua trung im I ca AB v vung gcvi AB. Gi K l giao im ca ng thng d v mt phng (P), chng minh rngd vung gc vi IK.

b) Vit phng tr nh tng qut ca hnh chiu vung gc ca d tr n mt phngc phng tr nh x + y - z + 1 = 0.

2. Cho t din ABCD c AD vung gc vi mt phng (ABC) v tam gic ABCvung ti A, AD = a, AC = b, AB = c. Tnh din tch ca tam gic BCD theo a, b,

c v chng minh2S abc(a + b + c).

Cu IV: 1. Tm s t nhin n tho mn: 2 n - 2 2 3 3 n - 3n n n n n nC C + 2C C + C C = 100, trong k nC l

s t hp cp k ca n.

2. Tnh tch phn I =2

1

x + 1lnxdx.x

e

Cu V: Xc nh tam gic ABC bit rng :

2 2(p - a)sin A + (p - b)sin B = csinAsinB .trong BC = a, CA = b, AB = c, a + b + c p =

2.




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15

S13

Cu I: Cho hm s y = x4 - 2m2 x2 + 1 (1)(m l tham s )1. Kho st hm s (1) khi m = 1.2. Tm m th hm s (1) c ba im cc tr l ba nh ca mt tam gic

vung cn.

Cu II: 1. Gii phng tr nh 4(sin3 x + cos3 x) = cosx + 3sinx.

2. Gii bt phng tr nh log 4

[ log 2(x + 22 - x x )] < 0.

Cu III:

1. Trong mt phng Oxy cho ng thng d: x

- y + 1 - 2 = 0 v im A(

-1;1). Vit phng tr nh ng tr n i qua A, qua gc to O v tip xc ving thng d.

2. Trong khng gian Oxyz cho hnh hp ch nht ABCD.A 1B 1C 1D 1 c A trngvi gc to O, B(1; 0; 0), D(0; 1; 0), A 1(0; 0; 2 ).

a) Vit phng tr nh mt phng (P) i qua ba im A 1, B, C v vit phngtrnh hnh chiu vung gc ca ng thng B 1D 1 trn mt phng (P).

b) Gi (Q) l mt phng qua A v vung gc vi A 1C. Tnh din tch thit dinca hnh chp A 1ABCD vi mt phng (Q).

Cu IV: 1. Tnh th tch ca vt th tr n xoay sinh ra bi php quay xung quanh trc Ox

ca hnh phng gii hn bi trc Ox v ng y =x sinx (0 x )2. Cho tp hp A gm n phn t, n 7. Tm n, bit rng s tp con gm 7 phn

t ca tp A bng hai ln s tp con gm ba phn t ca tp A.

Cu V:

Gi (x; y) l nghim ca h phng tr nh x - my = 2 - 4mmx + y = 3m + 1

(m l tham s). Tm

gi tr ln nht ca biu thc A = x2 + y

2 - 2x, khi m thay i.




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16

S14

Cu I: Cho hm s y = 2x3 - 2mx2 + m2 x - 2 (1)(m l tham s).1. Kho st hm s (1) khi m = 1.2. Tm m hm s (1) t cc tiu ti x = 1.

Cu II: 1. Gii phng tr nh 2 2 cos(x +4 ) +

1sinx =

1cosx .

2. Gii bt phng tr nhx - 12 + 6x - 11 > 4

x - 2

Cu III: 1. Trong mt phng Oxy cho im I(- 2; 0) v hai ng thng

d 1 : 2x - y + 5 = 0 v d

2: x + y - 3 = 0.Vit phng tr nh ng thng d i qua im I v ct hai ng thng d 1, d 2 ln

lt ti A, B sao cho

IA = 2.

IB .

2. Trong khng gian Oxyz cho A(4 ; 2; 2), B( 0 ; 0; 7) v ng thng

d: x - 3 y - 6 z - 1 =- 2 2 1

Chng minh rng hai ng thng d v AB cng thuc mt mt phng. TmimC trn ng thng d sao cho tam gic ABC cn ti nh A.

3. Cho hnh chp S.ABC c SA = 3a v vung gc vi y ABC, tam gic ABCc AB = BC = 2a, gc B bng 1200. Tnh khong cch t nh A n mt phng(SBC).

Cu IV:

1. Tnh tch phn I =3

31

dxx + x

.

2. Bit rng (2 + x)100 = a 0 + a 1x + a 2x2 + ... + a 100 x100 . Chng minh a 2 < a 3. Vigi tr no ca k th a k < a k+1 (0 k 99)?

Cu V: Cho hm s f(x) = ex - sinx +

x2 2. Tm gi tr nh nht ca f(x) v chng minh

r ng phng tr nh f(x) = 3 cng hai nghim.




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17

S15

Cu I:

Cho hm s2x - 2mx + 2y =

x - 1 (1)(m l tham s).

1. Kho st hm s (1) khi m = 1.2. Tm m hm s (1) c hai im cc tr A, B. Chng minh rng khi ng

thng AB song song vi ng thng d: 2x- y - 10 = 0.

Cu II: 1. Gii phng tr nh sin4xsin7x = cos3xcos6x.2. Gii bt phng tr nh log 3x > log x3.

Cu III:

1. Trong mt phng Oxy cho elip (E):x2 8 + y

2 4 = 1. Vit phng tr nh cc tip

tuyn ca (E) song song vi ng thng d: x +2 y - 1 = 02. Trong khng gian Oxyz cho A(2 ; 0; 0) v M( 1 ; 1; 1).

a) Tm to O' i xng O qua ng thng AM. b) Gi (P) l mt phng thay i i qua ng thng AM, ct cc trc Oy, Oz

ln lt ti cc im B, C. Gi s B(0; b; 0), C(0; 0; c), b > 0, c > 0. Chng minhr ng b + c = bc2 . Xc nh b, c sao cho din tch tam gic ABC nh nht.

Cu IV:

1. Tnh tch phn I =3

cosx

0

e sin2xdx.

2. Bit rng (1 + 2x)n = a 0 + a 1x + a 2x2 + ... + a n xn . Chng minh a 2 < a 3. Bitr ng a 0 + a 1 + a 2 + ... + a n = 729. Tm n v s ln nht trong cc s a 0, a 1, a 2, ..., a n

Cu V: Cho tam gic ABC tho mn A 900 v sinA = 2sinBsinCtanA2 . Tm gi tr nh

nht ca biu thcA1 - sin2S =

sinB.




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18

S16

Cu I:

Cho hm s2x + x + 4y =x + 1

(1) c th (C).

1. Kho st hm s (1) .2. Vit phng tr nh tip tuyn ca (C), bit tip tuyn vung gc vi ng

thng d: x- 3y + 3 = 0.

Cu II: 1. Gii phng tr nh 2sinxcos2x + sin2xcosx = sin4xcosx.

2. Gii h phng tr nh2 2

x + y x - 1

x + y = y + x2 - 2 = x - y.

Cu III: 1. Trong mt phng Oxy cho tam gic ABC vung A. Bit A(- 1; 4), B( 1; -

4), ng thng BC i qua im K(73 ; 2). Tm to C.

2. Trong khng gian Oxyz cho A(2 ; 0; 0) , B(2; 2; 0), C(0; 0; 2).a) Tm to O' i xng O qua mf(ABC). b) Cho im S di chuyn tr n tr c Oz, gi H l hnh chiu vung gc ca O

trn ng thng SA. Chng minh rng din tch tam gic OBH nh hn 4.

Cu IV: 1. Tnh tch phn I =

2

0xsin xdx.

2. Bit rng trong khai trin nh thc Niutn ca (x +1x )n tng cc h s ca hai

s hng u tin bng 24,tnh tng cc h s ca cc s hng cha xk vi k > 0 vchng minh rng tng ny l mt s chnh phng.

Cu V: Cho phng tr nh x2 + ( m2 - 53 )

2x + 4 + 2 - m2 = 0.Tm tt c cc gi tr m phng tr nh c nghim.




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19

S17

Cu I:

Cho hm s xy =x + 1

(1) c th (C).

1. Kho st hm s (1) .2. Tm trn (C) nhng im M sao cho khong cch t M n ng thng

d: 3x + 4y = 0 bng 1.

Cu II: 1. Gii phng tr nh sinx + sin2x = 3(cosx + cos2x)2. Tm gi tr ln nht v gi tr nh nht ca hm s y = (x + 1) 21 - x .

Cu III: 1. Trong mt phng Oxy cho im A(2; 3) v hai ng thng

d 1: x + y + 5 = 0 v d2: x + 2y - 7 = 0.Tm to cc im B tr n d 1 v C trn d 2 sao cho tam gic ABC c tr ng tm lG(2; 0).

2. Cho hnh vung ABCD c cnh AB = a. tr n cc na ng thng Ax, Byvung gc vi mf(ABCD) v nm v cng mt pha i vi mf(ABCD), ln ltly cc im M, N sao cho tam gic MNC vung ti M. t AM = m, BN = n.Chng minh rng, m(n- m) = a2 v tm gi tr nh nht ca din tch hnh thangABNM.

3. Trong khng gian Oxyz cho A(0 ; 1; 1) v ng thng d:x + y = 02x - z - 2 = 0

Vit phng tr nh mt phng (P) i qua A v vung gc vi ng thng d.Tm to hnh chiu vung gc B' ca im B(1; 1; 2) tr n mt phng (P). Cu IV:

1. Tnh tch phn I =ln8

2x x

ln3e e 1dx.

2. C bao nhiu s t nhin tho mn ng thi ba im kin sau: gm ng 4ch s i mt khc nhau; l s chn; nh hn 2158 ?

Cu V:

Tm tt c cc gi tr m h sau c nghim:2

2

x - 5x + 4 03x - mx x + 16 = 0




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20

S18

C au I :

Goi (Cm) laothcua ham so : y =2 22 1 3 x mx m

x m

(* ) (m latham so)

1. Khao sat sbien thien vaveoth cua ham so(* ) ng vi m = 1.2. Tm m eham so(* ) cohai iem cc trnam vehai pha truc tung.C au I I :

1. Giai hephng trnh :2 2 4( 1) ( 1) 2

x y x y

x x y y y

2. Tm nghiem trenkhong (0; ) cua phng trnh :2 2 34sin 3 cos 2 1 2cos ( )

2 4 x

x x

C au I I I :1. Trong mat phang vi hetoa oOxy cho tam giac A BC can tai nh A ctrong tam G4 1( ; )

3 3, phng trnh ng thang BC la2 4 0 x y vaphng

trnh ng thang BG la7 4 8 0 x y .T m toa ocac nh A, B , C.2. T rong khong gian vi hetoa oOxyz cho 3 iem A (1;1; 0),B(0; 2; 0),C(0; 0; 2) .

a) Viet phng trnh mat phang (P) qua goc toa oO vavuong goc vi BCTm toa ogiao iem cua A C vi mat phang (P).

b) Chng minh tam giac ABC latam giac vuong. V iet phng trnh mat cngoai tiep tdien OABC.C au I V :

1. Tnh tch phan3

2

0sin . I x tgxdx

.

2. Tcac chso1, 2, 3, 4, 5, 6, 7, 8, 9 cothelap c bao nhieu sotnhien,moi sogom 6 chsokhac nhau vatong cac chsohang chuc, hang tram hangngan bang 8.

C au V : Cho x, y, z laba sothoa x + y + z = 0. Chng minh rang :3 4 3 4 3 4 6 x y z




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21

S19

C au I :

1. Khao sat sbien thien vaveoth ( C ) cua ham so2 1

1

x x y

x

.

2. Viet phng trnh ng thang i qua iem M (- 1; 0) vatiep xuc vi othC ) .

C au I I :

1. Giai hephng trnh : 2 1 13 2 4

x y x y

x y

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

x x x

C au I I I :

1. T rong mat phang vi hetoa oOxy cho ng tron(C): x2 + y2 12 4 36 0 x y . V iet phng trnh ng tron (C1) tiep xuc vihai truc toa oOx, Oy ong thi tiep xuc ngoai vi ng tron (C).2. Trong khong gian vi hetoa oecac vuong goc Oxyz cho 3 iem A(2;0;0C(0; 4; 0), S(0; 0; 4).

a) T m toa oiem B thuoc mat phang Oxy sao cho tgiac OA BC lahnchnhat. Viet phng trnh mat cau qua 4 iem O, B, C, S.

b) Tm toa oiem A1 oi xng vi iem A qua ng thang SC.C au I V : 1. Tnh tch phan

7

30

21

x I dx

x.

2. Tm heso cua x7 trong khai trien a thc 2(2 3 ) n x , trong on laso nguyendng thoa man: 1 3 5 2 12 1 2 1 2 1 2 1... nn n n nC C C C = 1024. ( k nC lasotohp chap kcua n phan t)C au V : Cm rang vi moi x, y > 0 ta co:

29(1 )(1 )(1 ) 256 y x x y

. ang thc xay ra khi nao?




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22

S20

C au I :

1. Khao sat sbien thien vaveoth ( C ) cua ham so4 26 5 y x x

2. Tm m e phng trnh sau co4 nghiem phan biet :4 2 26 log 0 x x m .C au I I :

1. Giai hephng trnh :2 1 13 2 4

x y x y

x y

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

x x x

C au I I I :

1. Trong mat phang vi hetoa oOxy cho el ip (E) :2 2

64 9 x y = 1. V iet phng

trnh tiep tuyen d cua (E) biet d cat hai hai truc toa oOx, Oy lan lt tai Asao cho AO = 2BO.2. Trong khong gian vi hetoa o Oxyz cho hai ng thang1

x y z:1 1 2

d va

2

1 2:

1

x t

d y t

z t

( t l atham so)

a) Xet vtr tng oi cua d1 vad2 .b) Tm toa ocac iem M thuoc d1 vaN thuoc d2 sao cho ng thang M Nsong song vi mat phang (P) : 0 x y z vaodai oan M N =2 .C au I V :

1. Tnh tch phan 20

lne

x xdx .

2. M otivan ngheco15 ngi gom 10 nam va5 n. Hoi cobao nhieucach lap mot nhom ong ca gom 8 ngi biet rang trong nhom ophacot nhat 3 n.

C au V : Cho a, b, c laba sodng thoa man : a + b + c =34 . Chngminh rang :

3 3 33 3 3 3a b b c c a . Khi nao ang thc xay ra ?




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23

S21

C au I : Choham so : y =2 2 2

1 x x

x

(*)

1. Khao sat sbien thien vaveoth( C ) cua ham so(* ) .2. Goi I lagiao iem cua hai tiem can cua ( C ). Chng minh rang khong cotiep tuyen nao cua (C ) i qua iem I .C au I I :

1. Giai bat phng trnh : 28 6 1 4 1 0 x x x 2. Giai phng trnh : 2 2

cos 2 1( ) 32 cos

xtg x tg x

x

C au I I I :

1. Trong mat phang vi hetoa oOxy cho 2 ng tron :

(C1 ): x2

+ y2

9va(C2 ): x2

+ y2

2 2 23 0 x y . V iet phng trnh truc angphng d cua 2 ng tron (C1) va(C2). Chng minh rang neu K thuoc d thkhoang cach tK en tam cua (C1) nhohn khong cach tK en tam cua(C2 ).2. Trong khong gian vi hetoa oOxyz cho iem M (5;2; - 3) vamat phang(P): 2 2 1 0 x y z .a) Goi M1 lahnh chieu cua M len mat phang ( P ). Xac nh toa oiem M1

vatnh odai oan M M1.b) Viet phng tr nh mat phang ( Q ) i qua M vacha ng thang :

x - 1 y - 1 z - 52 1 - 6

C au I V :

1.T nh tch phan4

sin

0(tan cos ) x x e x dx

.

2. Tcac chso1, 2, 3, 4, 5, 6, 7 cothelap c bao nhieu sotnhien, moi sogom 5 chsokhac nhau vanhat thiet phai co2 ch1, 5 ?C au V : Chngminh rang neu0 1 y x th

14

x y y x . ang thc xay ra khi nao?




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24

S22

C au I : Goi (Cm) laothcua ham so y= x3+ ( 2m + 1) x2 m 1 (1) (m l atham so).

1) Khao sat sbien thien vaveothcua ham so(1) khim 1.2) Tm m eoth(Cm) ti ep xuc vi ng thang y= 2mx m 1.C au I I :

1. Giai bat phng trnh : 2 7 5 3 2 x x x 2. Giai phng trnh :3 sin( ) 2

2 1 cos x

tg x x

C au I I I :

1. Trong mat phang vi hetoa oOxy cho ng tron(C): x2 + y2 4 6 12 0 x y .

Tm toa oiem M thuoc ng thang d :2 3 0 x y sao cho MI = 2R, trongoI latam vaR l aban knh cua ng tron (C).2. Trong khong gian vi hetoa o Oxyz cho lang trung OA B.O1A1B1 vi

A(2;0;0), B(0; 4; 0), O1(0; 0; 4)a) Tm toa ocac iem A1, B1. V iet phng trnh mat cau qua 4 iem O, A,

B, O1.b) Goi M latrung iem cua A B.M at phang ( P ) qua M vuong goc vi O1A

vacat OA, OA1 lan lt tai N, K . T nh odai oan K N.C au I V :

1. Tnh tch phan3

2

1

lnln 1

e

x I dx x x

.

2. Tm k 0;1;2;...;2005 sao cho 2005k C at giatrln nhat. ( k nC lasotohpchap k cua n phan t)C au V : Tm m ehephng trnh sau conghiem:

2 1 2 1

2

7 7 2005 2005( 2) 2 3 0

x x x x

x m x m




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25

S23

C au I :

1. Khao sat sbien thien vaveoth cua ham so2 3 3

1

x x y

x

.

2. Tm m e phng trnh2 3 3

1 x x

m x

co4 nghiem phan biet

C au I I :

1. Giai bat phng trnh :2

22

2 19 2 33

x x x x

.

2. Giai phng tr nh :sin 2 cos 2 3sin cos 2 0 x x x x C au I I I :

1. Trong mat phang vi hetoa oOxy cho 2 i em A (0;5), B(2; 3) . V iphng trnh ng tron i qua hai iem A, B vacoban knh R =10.2. Trong khong gian vi he toa o Oxyz cho 3 hnh lap phng

ABCD.A1B1C1D1 vi A(0;0;0), B(2; 0; 0), D1(0; 2; 2)a) Xac nh toa ocac iem con lai cua hnh lap phng ABCD.A1B1C1D1.

Goi M latrung iem cua BC. Chng minh rang hai mat phang ( AB1D1) va(AM B1) vuong goc nhau.

b) Chng minh rang tsokhong cach tiem N thuoc ng thang AC1 (N A ) ti 2 mat phang ( AB1D1) va( AMB1) khong phu thuoc vao vtr cuaiem N.C au I V :

1. Tnh tch phan2

2

0I = (2x - 1)cos xdx.

2. Tm songuyen n ln hn 1 thoa man ang thc : 2 22 6 12n n n nP A P A .

( Pn lasohoan vcua n phan tvak n A laso chnh hp chap k cua n phan t)

C au V : Cho x, y, z laba sodng vaxyz = 1. Chng minhr ang :2 2 2 3

1 1 1 2

x y z

y z x

.




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26

S 24

Cu I:

1. Kho st v v th hm s2x 2x 5y

x 1

(C)

2. Da vo th (C), tm m phng tr nh sau c hai nghim dng phn bit 2 22 5 ( 2 5)( 1) x x m m x

Cu II:

1. Gii phng tr nh: 3 3 2 3 2cos3xcos x sin3xsin x8

2. Gii h phng tr nh:2

2

( 1) ( ) 4 ( , )

( 1)( 2) x y y x y

x y R x y x y

Cu III: Trong khng gian Oxyz cho hnh lng tr ng ABCA'B'C' c A(0; 0;0), B(2; 0; 0), C(0; 2; 0), A'(0; 0; 2).1. Chng minh A'C vung gc vi BC'. Vit phng tr nh mt phng (ABC'). 2. Vit phng tr nh hnh chiu vung gc ca ng thng B'C' tr n mf(ABC')Cu IV:

1. Tnh6

2

dxI2x 1 4x 1

2. Cho x, y l cc s thc tho mniu kin: x2 + xy + y2 3.Chng minh rng: 2 24 3 3 3 4 3 3 x xy y .

Cu Va: 1. Trong mt phng Oxy cho elip (E)

2 2

112 2 x y . Vit phng tr nh ca hypebol

(H) c hai dng tim cn l 2 y x v c hai tiu im l hai tiu im ca (E). 2. p dng khai trin ca nh thc Newton ca 1002 x x , chng minh rng:

99 100 198 1990 1 99 100100 100 100 100

1 1 1 1100 101 ... 199 200 02 2 2 2

C C C C

Cu Vb:

1. Gii bt phng tr nh: x 1log ( 2x) 2 2. Cho hnh hp ng ABCD.A'B'C'D' c y l hnh thoi, cc cnh AB = AD =

a, BAD =60 0 , cnh bn bng 32

a . Gi M, N ln lt l trung im ca cc

cnh A'D' v A'B'. Chng minh AC'mf(BDMN). Tnh th tch khi chpA.BDMN.




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27

S25

Cu I:

1. Kho st v v th (C) ca hm s4

2xy 2(x 1)2

(C)

2. Vit phng tr nh ccng thng i qua im A(0; 2) v tip xc vi (C). Cu II:

1. Gii phng tr nh: 2sin 2x 4sinx +1=06

2. Gii h phng tr nh:3 3

2 2

8 2 ( , )

3 3( 1) x x y y

x y R x y

Cu III: Trong khng gian Oxyz cho mf( ): 3x + 2y - z + 4 = 0 v hai im A(4;0; 0), B(0; 4; 0). Gi I l trungim ca on thng AB.

1. Tm giaoim ca ng thng AB vi mf( ).2. Xc nh to im K sao cho KI vung gc vi mf( ) ng thi K cch

u gc to O v mf( ) .Cu IV:1. Tnh din tch hnh phng gii hn bi parabol y = x2 - x + 3 v ng thngd: y = 2x + 12. Cho x, y, z tho mn cciu kn 3 3 3 1 x y z . Chng minh rng:

9 9 9 3 3 33 3 3 3 3 3 4

x y z x y z

x y z y z x z x y

Cu Va: 1. Trong mt phng Oxy cho tam gic ABC c nh A thuc ng thng d: x- 4y- 2 = 0 , cnh BC song song vi d. Phng tr nh ng cao BH: x + y + 3 = 0 vtrung im ca cnh AC l M(1; 1). Tm to A, B, C.2. T cc ch s 0, 1, 2, 3, 4 c th lp c bao nhiu s t nhin c 5 ch skhc nhau ? Tnh tng ca tt c cc s t nhin .Cu Vb: 1. Gii bt phng tr nh: x 2x 2xlog 2 2log 4 log 8 2. Cho hnh chp S.ABCD cy ABCD l hnh ch nht vi AB = a, AD = 2a,cnh SA vung gc vi y, cnh SB to vi mt phng y mt gc 600. Trncnh SA ly im M sao cho AM =3

3a . Mt phng (BCM) ct SD ti N.

Tnh th tch khi chpS.BCNM.




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28

S 26

Cu I:

1. Kho st v v th (C) ca hm s2

x x 1y x 1 (C)2. Vit phng tr nh tip tuyn ca (C) i qua A(0;- 5).Cu II: 1. Gii phng tr nh: 2 2 2(2sin x 1) tan 2x 3(2cos x -1) = 0 2. Gii phng tr nh: 23 2 1 4 9 2 3 5 2 , ) x x x x x x R Cu III: Trong khng gian Oxyz cho hai ng thng:

1

1: 1

2

x t

y t

z

v 2

3 1:1 2 1

x y z

1. Vit phng tr nh mt phng cha1v song song 2 .2. Xc nh to im A tr n 1v im B tr n 2 sao cho on thng AB c

di nh nht. Cu IV:

1. Tnh10

5

dxIx 2 x 1

2. Tm gi tr nh nht ca hm s: 211 7

4 1 , 02 y x x x x

Cu Va: 1. Trong mt phng Oxy cho tam gic ABC cn ti B, vi A(1;-1), C(3; 5). imB thuc ng thng d: 2x- y = 0. Vit phng tr nh ccng thng AB, BC.2. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s chn, mi s c 5ch s khc nhau trong c ng hai ch s l v hai ch s l ng cnh nhau.Cu Vb: 1. Gii phng tr nh: 31 82

2log x 1 log (3 x) log (x 1) 0

2. Cho hnh chp S.ABCD cy ABCD l hnh thoi cnh a. 060 BAD . SAvung gc vi mf(ABCD), SA = a. Gi C' trung imca SC. Mt phng (P) diqua AC' v song song BD, ct cc cnh SB, SD ca hnh chp ln lt ti B', D'.

Tnh th tch khi chp S.AB'C'D'.




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29

S 27

Cu I: Cho hm s y = x3 + (1 - 2m)x2 + (2 - m)x + 2 (1)1. Kho st v v th ca hm s (1) khi m = 2. 2. Tm tt c cc gi tr m th hm s (1) c cc i, cc tiu, ng thihonh ca im cc tiu nh hn 1. Cu II: 1. Gii phng tr nh: cos2x + (1 + 2cosx)(sinx - cosx) = 0

2. Gii h phng tr nh:2 2

2 2

( )( ) 13 ( , )( )( ) 25 x y x y

x y R x y x y

Cu III: Trong khng gian Oxyz cho mt phng (P): 2x + y- z + 5 = 0 v ccimA(0; 0; 4), B(2; 0; 0)

1. Vit phng tr nh hnh chiu vung gc ca ng thng AB tr n mf(P).2. Vit phng tr nh mt cu i qua O, A, B v tip xc vi mt phng (P).

Cu IV:

1. Tnhe

1

3 2 ln xI dxx 1 2 ln x

2. Cho hai s dng x, y thay i th mn iu kin x + y 4. Tm gi tr nh

nht ca biu thc A =2 3

23 4 2

4 x y

x y

Cu Va: 1. Trong mt phng Oxy cho tam gic ABC c A(2; 1), ng cao qua nh B c phng tr nh x - 3y - 7 = 0 v ng trung tuyn qua nh C c phng tr nh x + y+ 1 = 0. Xc nh to cc nh B, C ca tam gic..2. Cho hai ng thng song song d1 v d2. Trn ng thng d1 c 10 im phn bit, trn ng thng d2 c n im phn bit(n 2). Bit rng c 2800 tam gicc nh l cc im cho. Tm n.Cu Vb:

1. Gii phng tr nh:2 2x x 1 x x 29 10.3 1 0

2. Cho hnh lng tr ABC.A'B'C' c A'ABC l hnh chp tam gicu, cnh yAB = a, cnh bn AA' = b. Gi l gc gia hai mt phng (ABC) v (A'BC).Tnh tan v th tch khi chp A'BB'C'C.




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30

S 28

Cu I: Cho hm s3

2 1133 3

x y x x

1. Kho st v v th (C) ca hm s cho.2. Tm trn th (C) hai im phn bit M, N i xng nhau qua trc tung. Cu II: 1. Gii phng tr nh: 3 3 2cos x sin x 2sin x 1 2. Gii h phng tr nh:

2 2

2 2 2

3( ) ( , )

7( ) x xy y x y

x y R x xy y x y

Cu III: Trong khng gian Oxyz cho mt phng (P): 4x- 3y + 11z - 26 = 0 v haing thng1 2

3 1 4 3: , :1 2 3 1 1 2

x y z x y zd d

1. Chng minh rng d1 v d2 cho nhau.2. Vit phng tr nh ng thng nm trn (P), ng thi ct c d1v d2 .

Cu IV:

1. Tnh 2

0I (x 1) sin 2xdx

2. Gii phng tr nh 14 2 2(2 1)sin(2 1) 2 0 x x x x y Cu Va:

1. Trong mt phng Oxy cho ng thng d: x- y + 1 - 2 = 0 v im A(- 1; 1).Vit phng tr nhng tr n (C)i qua A, O v tip xc d.2. Mt lp hc c 33 hc sinh, trong c 7 n. Cn chia lp thnh 3 t, t 1 c10 hc sinh, t 2 c 11 hc sinh v t 3 c 12 hc sinh sao cho trongmi t c tnht 2 hc sinh n. Hi c bao nhiu cch chia nh vy ? Cu Vb: 1. Gii phng tr nh: x x 13 3log (3 1)log (3 3) 6 2. Cho hnh chp t gic u S.ABCD c cnh y bng a. Gi SH l ng caoca hnh chp. Khong cch t trung im I ca SH n mt bn (SBC) bng b.

Tnh th tch khi chp S.ABCD.




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31

S 29

Cu I: Cho hm s 31

x y

x (C)

1. Kho st v v th (C) ca hm s cho.2. Cho im 0 0 0( ; ) M x y thuc (C). Tip tuynca (C) ti 0 0 0( ; ) M x y ct cc timcn ca (C) ti cc im A, B. Chng minh0 0 0( ; ) M x y l trung im AB. Cu II: 1. Gii phng tr nh: 3 24sin x 4sin x 3sin 2x 6cosx 0 2. Gii phng tr nh: 2x + 2 7 - x = 2 x - 1 + - x 8 7 +1 ( ) x x R Cu III: Trong khng gian Oxyz cho A(1; 2; 0), B(0; 4; 0), C(0; 0; 3)

1. Vit phng tr nhng thng i qua O v vung gc vi mf(ABC). 2. Vit phng tr nh mt phng (P) cha OA, sao cho khong cch t B n (P)

bng khong cch t C n (P). Cu IV:

1. Tnh2

1I (x 2) ln xdx

2. Gii hphng tr nh: 2 2ln(1 ) ln(1 )

( , ) 12 20 0 x y x y

x y R x xy y

Cu Va: 1. Trong mt phng Oxy lp phng tr nh chnh tc ca elip (E) c di tr c ln

bng4 2 , cc nhtrn tr c nh v cc tiu im ca (E) cng thuc mt ngtrn.2. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin chn c 5ch s khc nhau v mi s lp nn u nh hn 2500 ? Cu Vb:

1. Gii phng tr nh: 2 4 212(log x 1)log x log 04

2. Cho hnh lp phng ABCD.A'B'C'D' c cnh bng a v im K thuc cnhCC' sao cho 2

3CK a . Mt phng ( ) i qua A, K v song song vi BD, chia

khi lp phng thnh hai khi a din. Tnh th tch ca hai khi a din .




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32

S 30

Cu I: Cho hm s2x 4x 3yx 2

1. Kho st v v th hm s. 2. Chng minh rng tch cc khong cch t mt im bt k trn th hm sn cc ng tim cn ca n l hng s. Cu II:

1. Gii phng tr nh: 1 1sin 2x sin x 2cotg2x2sin x sin 2x

2. Tm m phng tr nh: 2m x 2x 2 1 x(2 x) 0 (2) c nghim x0;1 3

Cu III: Trong khng gian Oxyz cho hai im A (-1;3;-2), B (-3;7;-18) v mt phng (P): 2x- y + z + 1 = 01. Vit phng trnh mt phng cha AB v vung gc vi mp (P). 2. Tm ta im M (P) sao cho MA + MB nh nht. Cu IV:

1. Tnh4

0

2x 1I dx1 2x 1

2. Gii h phng tr nh: )Ry,x( 132y2yy

132x2xx1x2

1y2

Cu Va: 1. Trong mt phng Oxy cho ng tr n (C) : x2 + y2 = 1. ng tr n (C') tm I

(2; 2) ct (C) ti cc im A, B sao choAB 2. Vit phng tr nhng thngAB.

2. C bao nhiu s t nhin chn ln hn 2007 m mi s gm 4 ch s khcnhau ?Cu Vb: 1. Gii bt phng tr nh: 2x 4 2(log 8 log x )log 2x 0 2. Cho lng tr ng ABCA1B1C1 c AB = a, AC = 2a, AA1 2a 5 v

o120BAC . Gi M l trung im ca cnh CC1. Chng minh MBMA1 v tnhkhong cch d t im A ti mt phng (A1BM).




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33

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Cu I: Cho hm s my x m (Cm)x 2

1. Kho st v v th hm s vi m = 1. 2. Tm m th (Cm) c cc tr ti cc im A, B sao cho ng thng AB iqua gc ta O. Cu II: 1. Gii phng tr nh: 22cos x 2 3 sin x cosx 1 3(sin x 3 cosx)

2. Gii h phng tr nh4 3 2 2

3 2x x y x y 1x y x xy 1

Cu III: Trong khng gian Oxyz cho cc im A(2,0,0); B(0,4,0); C(2,4,6) vng thng (d)6x 3y 2z 0

6x 3y 2z 24 0

1. Chng minh cc ng thng AB v OC cho nhau.2. Vit phng tr nhng thng // (d) v ct cc ng AB, OC. Cu IV:1. Trong mt phng Oxy cho hnh phng (H) gii hn bi cc ng 2xy4 v y= x. Tnh th tch vt th tr n trong khi quay (H) quanh tr c Ox trn mt vng.2. Cho x, y, z l cc bin s dng. Tm gi tr nh nht ca biu thc:

33 3 3 3 3 33 3

2 2 2x y zP 4(x y ) 4(y z ) 4(z x ) 2y z x

Cu Va: 1. Trong mt phng Oxy cho tam gic ABC c trng tm G(2, 0) bit phngtrnh cc cnh AB, AC theo th t l 4x + y + 14 = 0; 02y5x2 . Tm ta cc nh A, B, C. 2. Trn cc cnh AB, BC, CD, DA ca hnh vung ABCD ln lt cho 1, 2, 3 v nim phn bit khc A, B, C, D. Tm n bit s tam gic c ba nh ly t n + 6im cho l 439.Cu Vb:

1. Gii phng tr nh 4 22x 1

1 1log (x 1) log x 2log 4 2

2. Cho hnh chp SABC c gc o60ABC,SBC , ABC v SBC l cc tam gicu cnh a. Tnh theo a khong cch t nh B n mp(SAC).




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34

S 32

Cu I: Cho hm s y = 2x3 + 6x2 51. Kho st v v th hm s (C).

2. Vit phng tr nh tip tuyn ca (C), bit tip tuyn i qua A( 1, 13).Cu II:

1. Gii phng tr nh: 2x3cos2

42xcos

42x5sin

2. Tm m phng tr nh: mx1x4 2 c nghim. Cu III: Trong khng gian Oxyz cho cc im A( 3; 5; 5); B(5; 3; 7) v mt phng (P): x + y + z = 0 1. Tm giaoim I ca ng thng AB vi mt phng (P). 2. Tmim M (P) sao cho MA2 + MB2 nh nht.

Cu IV: 1. Tnh din tch hnh phng gii hn bi cc ng thng y = 0 v

1xx1xy 2 .

2. Chng minh rng h

1xx2007e

1yy2007e

2y

2x

c ng 2 nghim tha mn iu kin

x > 0, y > 0

Cu Va:

1. Tm x, y N tha mn h66CA22CA

2x

3y

3y

2x

2. Cho ng tr n (C): x2 + y2 8x + 6y + 21 = 0 v ng thng d: 01yx .Xc nh ta cc nh hnh vung ABCD ngoi tip (C) bit A d

Cu Vb: 1. Gii phng tr nh 21x2log1xlog 3

23

2. Cho hnh chp SABCD cy ABCD l hnh vung tm O, SA vung gc viyhnh chp. Cho AB = a, SA = a2 . Gi H v K ln lt l hnh chiu ca Aln SB, SD. Chng minh SC (AHK) v tnh th tch hnh chp OAHK.




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35

S 33

Cu I: Cho hm sx2

m1xy (Cm)

1. Kho st v v th hm s vi m = 1 2. Tm m th (Cm) c cc i ti im A sao cho tip tuyn vi (Cm) ti A cttr c Oy ti B m OBA vung cn.Cu II:

1. Gii phng tr nh: gxcottgxxsinx2cos

xcosx2sin

2. Tm m phng tr nh : 01xmx13x4 4 c ng 1 nghim Cu III: Trong k hng gian Oxyz cho cc im A(2;0;0); M(0; 3;6)1. Chng minh rng mt phng (P): x + 2y 9 = 0 tip xc vi mt cu tm M, bn knh MO. Tm ta tip im. 2. Vit phng tr nh mt phng (Q) cha A, M v ct cc trc Oy, Oz ti cc imtng ng B, C sao cho VOABC = 3.Cu IV: 1. Trong mt phng ta Oxy, tnh din tch hnh phng gii hn bi cc ngy = x2 v 2x2y .

2. Gii h phng tr nh:xy

9y2y

xy2y

yx9x2x

xy2x

2

3 2

23 2

Cu Va: 1. Tm h s ca x8 trong khai trin (x2 + 2)n, bit: 49CC8A 1n2n3n .2. Cho ng tr n (C): x2 + y2 2x + 4y + 2 = 0. Vit phng tr nh ng tr n(C') tm M(5, 1) bit (C') ct (C) ti cc im A, B sao cho 3AB .Cu Vb:

1. Gii phng tr nh: 1xlog143logxlog2

3x93

2. Trong mt phng (P) cho na ng tr n ng knh AB = 2R v im C thucna ng tr n sao cho AC = R. Trn ng thng vung gc vi (P) ti A lyim S sao cho o60SBC,SAB . Gi H, K ln lt l hnh chiu ca A tr n SB,SC. Chng minh AHK vung v tnh VSABC?




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36

S 34

Cu I: Cho hm s1x21xy (C)

1. Kho st v v th hm s.

2. Vit phngtrnh tip tuyn vi (C), bit rng tip tuyn i qua giao imca ng tim cn v tr c Ox.

Cu II: 1. Gii phng tr nh: 1xcos12xsin22

2. Tm m phng tr nh: m54x6x4x23x c ng2 nghim

Cu III: Cho ng thng d:11z

12y

23x v mt phng (P):

02zyx 1. Tm giaoim M ca d v (P).2. Vit phng tr nhng thng nm trong (P) sao cho d v khong ccht M n bng 42 .Cu IV:

1. Tnh

1

02 dx4x

1xxI

2. Cho a, b l cc s dng tha mn ab + a + b = 3.

Chng minh: 23

babaab

1ab3

1ba3

22 .Cu Va: 1. Chng minh vi mi n nguyn dng lun c

2 10 1 2 11 ... 2 1 1 0n nn nn n n nnC n C C C .

2. Trong mt phng Oxy cho im A(2, 1) ly im B thuc trc Ox c honh x 0 v im C thuc trc Oy c trung y 0 sao cho ABC vung ti A. TmB, C sao cho din tch ABC ln nht. Cu Vb:

1. Gii bt phng tr nh: 221 22

1 1log 2x 3x 1 log x 12 2

.

2. Cho lng tr ng ABCA1B1C1 c y ABC l tam gic vung aACAB ,AA1 = a 2 . Gi M, N ln lt l trung im ca on AA1 v BC1. Chng minhMN l ng vung gc chung ca cc ng thng AA1 v BC1. Tnh 11BCM AV .




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37

S 35

Cu I: Cho hm s1x

xy (C)

1. Kho st v v th hm s.

2. Vit phng tr nh tip tuyn d ca (C) sao cho d v hai tim cn ca (C) ctnhau to thnh mt tam gic cn. Cu II: 1. Gii phng tr nh: (1 tgx)(1 + sin2x) = 1 + tgx

2. Tm m h phng tr nh :1xyx

0myx2 c nghim duy nht

Cu III: Cho mt phng (P): x 2y + 2z 1 = 0 v cc ng thng:

2

z

3

3y

2

1x:d1 v5

5z

4

y

6

5x:d2

1. Vit phng tr nh mt phng (Q) cha d1 v (Q) (P).2. Tm ccim M d1, N d2 sao cho MN//(P) v cch (P) mt khong bng 2. Cu IV:

1. Tnh2

0

2 xdxcosxI

2. Gii phng tr nh: xx

2 2x1x12log .

Cu Va: 1. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin chn mmi s gm 4 ch s khc nhau. 2. Trong mt phng Oxy cho cc im A(0, 1) B(2, 1) v cc ng thng:

d1: (m 1)x + (m 2)y + 2 m = 0d2: (2 m)x + (m 1)y + 3m 5 = 0

Chng minh d1 v d2 lun ct nhau. Gi P = d1 d2. Tm m sao cho PBPA lnnht

Cu Vb: 1. Gii phng tr nh: 022.72.72 xx21x3 .2. Cho lng tr ng ABCA1B1C1 c tt c cc cnh u bng a. M l trung imca on AA1. Chng minh BM B1C v tnh d(BM, B1C).




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39

S 37

PHN CHUNG CHO TT CCC TH SINHCu I (2im). Cho hm s y = x4 - 8x2 + 7 (1)

1. Kho st s bin thin v v th hm s (1). 2. Tm cc gi tr thc ca tham s m ng thng y = mx- 9 tip xc vi th hm s (1).

Cu II (2im)

1. Gii phng tr nh 2sin 2x - = sin x - +4 4 2

.

2. Gii phng tr nh 2 21 31

1 1 x

x x

.

Cu III (2im) Trong khng gian Oxyz cho mt phng (P): 2x + 3y- 3z + 1 = 0, ng thng d:

3 52 9 1 x y z

v 3 im A(4; 0; 3), B(- 1; - 1; 3), C(3; 2; 6).1. Vit phng tr nh mt cu (S) i qua ba im A, B, C v c tm thuc mf(P). 2. Vit phng tr nh mt phng (Q) cha ng thng d v ct mt cu (S) theo

mt ng trn c bn knh ln nht.

Cu IV (2im). 1. Tnh tch phn2

0

sin2xdxI =3 + 4sinx - cos2x

.

2. Chng minh phng tr nh 4x(4x2 + 1) = 1c ng ba nghimthc phn bit.

PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b

Cu V.a. Theo chng tr nh KHNG phn ban (2im)1. Tm h s ca s hng cha x5 trong khai trin nh thc Newton ca (1 + 3x)2n

, bit rng 3 2n nA + 2A = 100(n l s nguyn dng) 2. Trong mt phng Oxy cho ng tr n (C): x2 + y2 = 1. Tm tt c cc gi tr

thc m trn ng thng y = m tn ti ng hai im m t mi im c th kc 2 tip tuyn vi (C) sao cho gc gia hai tip tuyn bng 60o.Cu V.b. Theo chng tr nh phn ban (2im)1. Gii phng tr nh

3

1 63 log 9log x

x x x

.

2. Cho hnh chp S.ABC m mi mt bn l mt tam gic vung, SA = SB = SC= a. Gi M, N, E ln lt l trung im ca cc cnh AB, AC, BC; D l im ixng ca S qua E; I l giao im ca ng thng AD vi mf(SMN). Chng minhr ng AD vung gc vi SI v tnh theo a th tch ca khi t din MBSI.




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40

S 38

PHN CHUNG CHO TT C CC TH SINH Cu I (2im). Cho hm s 3 2y = x - 3x - 3m(m + 2)x - 1 (1), m l tham s thc

1. Kho st s bin thin v v th ca hm s (1) khi m = 0. 2. Tm cc gi tr ca m hm s (1) c hai cc tr cng du.

Cu II (2im)

1. Gii phng tr nh 12sin x + - sin 2x - =3 6 2

.

2. Gii phng tr nh 10x + 1 + 3x - 5 = 9x + 4 + 2x - 2.Cu III (2im)

Trong khng gian Oxyz cho ng thng d 1 c phng tr nh : x - 3 y z + 5 = =2 9 1 v hai im A(5; 4; 3), B(6; 7; 2)

1. Vit phng tr nh ng thng d 2 qua 2 im A, B. Chng minhr ng haing thng d 1 v d 2 cho nhau.

2. Tmim C thuc d 1 sao cho tam gic ABC c din tch nh nht. Tnh gi trnh nht .Cu IV (2im)

1. Tnh2

0

x + 1I = dx4x + 1

.

2. Cho 3 s dng x, y, z tho mn h thc yzx + y + z =3x

. Chng minh rng

2 3 - 3x (y + z)6 .

PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.bCu V.a. Theo chng tr nh KHNG phn ban (2im)

1. Cho s nguyn n tho mnng thc3 3n nA + C = 35

(n - 1)(n - 2)(n 3). Tnh tng

2 2 2 3 n 2 nn n nS =2 C - 3 C + ... + (-1) n C

2. Trong mt phng Oxy, cho tam gic ABC vi AB =5, C(- 1; - 1), ngthng AB c phng tr nh x + 2y - 3 = 0 v tr ng tm ca tam gic ABC thuc

ng thng x + y - 2 = 0. Hy tm to cc nh A v B.Cu V.b. Theo chng tr nh phn ban (2im)1. Gii phng tr nh 2 1

22log 2 2 log 9 1 1 x x

2. Cho hnh chp ABCD cy ABCD l hnh vung c cnh bng a, SA = a3v SA vung gc vi mt phng y. Tnh theo a th tch khi t din S.ACD vtnh cosin ca gc gia hai ng thng SB, AC.




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41

S 39

PHN CHUNG CHO TT C CC TH SINH

Cu I (2im). Cho hm s 2x + (3m - 2)x + 1 - 2my =

x + 2 (1), m l tham s thc

1. Kho st s binthin v v th ca hm s (1) khi m = 1. 2. Tm cc gi tr m hm s (1) ng bin tr n tng khong xc nh ca n.

Cu II (2im) 1. Gii phng tr nh 2 x3sinx + cos2x + sin2x = 4sinxcos

2

2. Gii h phng tr nh

3

4

x - 1 - y = 8 - x

x - 1 = y

Cu III (2im). Trong khng gian Oxyz cho 3 im A(1; 0;- 1), B(2; 3; - 1);,

C(1; 3; 1) v ng thng d:1 0

4 x y

x y z

1. Tm to im D thuc ng thng d sao cho th tch khi t din ABCD bng 1.

2. Vit phng tr nh tham s ca ng thng i qua trc tm H ca tam gicABc v vung gc vi mf(ABC). Cu IV (2im)

1. Tnh tch phn1 3

20

x dxI =4 - x

.

2. Cho s nguyn n( n 2) v hai s thc khng m x, y. Chng minh:n n n + 1 n + 1n n + 1x + y x + y .

PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.bCu V.a. Theo chng tr nh KHNG phn ban (2im)1. Chng minh rng vi n l s nguyn dng

n 0 n - 1 1 0 n n + 1n n n2 C 2 C 2 C 3 - 1 + + ... + =

n + 1 n 1 2(n + 1).

2. Trong mt phng Oxycho hai im A(3; 0), B(0; 4). Chng minh rng ngtrn ni tip tam gic OAB tip xc vi ng tr n i qua trung im cc cnhca tam gic OAB.

Cu V.b. Theo chng tr nh phn ban (2im)1. Gii phng tr nh 2x + 1 2x + 1 x3 - 2 - 5.6 0.2. Cho t din ABCD c cc mt ABC v ABD l cc tam gic u cnh a, cc

m ACD v BCD vung gc vi nhau. Hy tnh theo a th tch ca khi t dinABCD v tnh s o ca gc gia hai nng thng AD v BC.




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42

S 40

PHN CHUNG CHO TT C CC TH SINH Cu I (2im). Cho hm s 3 1

1 x

y x

(1)

1. Kho st s bin thin v v th ca hm s 3 11

x y x

(1)

2. Tnh din tch ca tam gic to bi cc trc to v tip tuyn vi thhm s (1) ti im M(-2; 5).Cu II (2im)

1. Gii phng tr nh 4 44(sin x + cos x) + cos4x + sin2x = 0.2. Gii phng tr nh 2 2( 1)( 3) 2 3 2 ( 1) x x x x x .

Cu III (2im). Trong khng gian Oxyz cho mt phng () : 2x - y + 2z + 1 = 0v ng thng

d: x - 1 y - 1 z= =1 2 - 2 1. Tm to giao im ca d vi () ; tnh sin ca gc gia d v ( ) .2. Vit phng tr nh mt cu c tm thuc d v tip xc vi hai mt phng( ) v

Oxy.Cu IV (2im)

1. Tnh1

2x2

0

xI = xe - dx4 - x

2. Cho cc s thc x, y tha0 ,3

x y . Chng minh rng:

cosx + cosy 1 + cos(xy)PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.bCu V.a. Theo chng tr nh KHNG phn ban (2im)1. Chng minh rng vi n l s nguyn dng, ta c:

n - 1 2 n - 2 n n n - 1n n n2C + 2 C + ... + n.2 C = 2n.3

2. Trong mt phng Oxy cho ng tr n (C): (x - 4)2 + y2 = 4 v im E(4; 1).Tm to im M tr n tr c tung sao cho t M k c hai tip tuyn MA, MBn ng tr n (C) vi A, B l cc tip im sao cho ng thng AB qua E. Cu V.b. Theo chng tr nh phn ban (2im)

1. Gii bt phng tr nh2 22x - 4x - 2 2x - x - 1

2 - 16.2 - 2 0.2. Cho t din ABCD v cc im M, N, P ln lt thuc cc cnh BC, BD, AC saocho BC = 4BM, AC = 3AP, BD = 2BN. Mt phng (MNP) ct AD ti Q. Tnh t sAQAD v t s th tch hai phn ca khi t din ABCD c phn chia bi mt phng

(MNP).




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43

S 41

I. PHN CHUNG CHO TT C CC TH SINH(7, 0 im) Cu I. (2 im)

Cho hm s y = - x3 - 3x2 + mx - 4, trong m l tham s thc, (1).1. Kho st s bin thin v v th ca hm s(1) cho, vi m = 0. 2. Tm tt c cc gi tr ca tham s m hm s(1) chong bintrn

khong (0; 2)Cu II. (2 im)

1. Gii phng tr nh2

2tan t anx 2sin

tan 1 2 4 x

x x

2. Tm tt c cc gi tr ca tham s m phng tr nh 24 2 4 1 x x x m c ng mt nghim thc. Cu III. (2 im)

Trong khng gian Oxyz, cho im A(5; 5; 0) v ng thng d: 1 1 72 3 4

x y z

1. Tm ta im A' i xng vi im A qua ng thng d. 2. Tm ta cc im B, C thuc d sao cho tam gic ABC vung ti C v

BC = 29 Cu IV. (2 im)

1. Tnh tch phn1

2

0( 1) . x I x x e dx

2. Gii h phng tr nh

2 2

2 2

2 2

36 60 25 036 60 25 036 60 25 0

x y x y

y z y z

z y z x

II. PHN RING.Th sinh ch c lm 1 trong 2 cu: V.a hoc V.b Cu V.a. Theo chng tr nh KHNG phn ban (2 im)

1. C bao nhiu s t nhin gm 4 ch s khc nhau m mi s u ln hn2500.

2. Trong mt phngOxy, tm ta cc nh ca tam gic ABC bit rng ngthng AB, ng cao k t A v trung tuyn k t B ln lt c phng tr nh lx + 4y - 2 = 0, 2x - 3y + 7 = 0, 2x + 3y - 9 = 0.

Cu V.b. Theo chng tr nh phn ban (2 im)

1. Gii phng tr nh 5 1 2 5 1 3.2 . x x

x .2. Cho hnh chp S.ABC cy ABC l tam gic vung cn nh B, AB = a, SA

= 2a v SA vung gc vi mt phng y. Mt phng qua A vung gc SC ct SB,SC ln lt ti H, K. Tnh theo a th tch khi t din SAHK.




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44

PHN TH HAI

HNG DN GII




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45

S 1

Cu I:1. Kho st s bin thin v v th ca hm s khi m = 8. (Hc sinh t gii).

2. Xcnh, th hm s ct tr c honh ti 4 im phn bit.Bi ton quy v xcnh m ph ng trnh

x4 - mx2 + m - 1 = 0c 4 nghim phn bit.

t2 - mt + m - 1 = 0 (t = x2) c 2 nghim t1, t2 d ng phn bit. 22 4 1 2 0m m m

2 24( 1) 0 ( 2) 02

0 011 0 1

m m mm

S m S mmP m P m

Cu II:1. Bt ph ng trnh 2 11 1

2 2log 4 4 log 2 3.2 x x x

2 12 3.2 4 4 2.4 3.2 4 4 4 3.2 4 02 1 22 4

x x x x x x x x

x

x x

2. 2(sin4x + cos4x) + cos4x + 2sin2x + m = 0 (1) 2(1 - 2sin2xcos2x) + cos4x + 2sin2x + m = 0 3 + m - 3sin22x + 2sin2x = 0 3t2 - 2t - (m + 3) = 0; t = sin2x (2)

Vi 0 0 2 0 12

x x t

, nhn mi gi tr thuc on

[0;1]. ph ng trnh (1) c t nht mt nghim thuc

on 0;2

, iu kin cn v l ph ng trnh (2)

c nht mt nghim t [0;1] 3t2 - 2t = m + 3 c t nht mt nghim t [0; 1].

t f(t) = 3t2 - 2t (0 t 1)




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46

V d th (C) ca hmy = f(t) (0 t 1).

f 1 13 3

T th suy ra, ph ng trnh3t2 - 2t = m + 3 c t nht mt nghim t [0; 1],iu kin cn v l:

1 103 1 23 3

m m

Cu III:1. K AH BC, AO SH. V SA (ABC), BC AH nn BC SH. T BC (SAH), suy ra BC AO. Do AO (ABC). V vy AO l khong cch t A

n (SAB). ABCu nn AH = 32

a . SA (ABC) nn SA (ABC) nn SA

AH. Do SAH vung ti A. Ta c:

2 2 21 1 1

AO AH SA 2 2 2

4 2 23 3a a a

22

a AO

2. Tnh tch phn31

20 1

x I dx

x

Cch 1. t u = x2 + 1 du = 2xdx31

20 1 x

I dx x

= 2 2

1 1

1 (u - 1) 1 1 1du 1 - du 1 ln 22 u 2 u 2

Cch 2. t x = tant dx = 2osdt

c t

34 4 4

32 2 2

0 0 0

tan t d t 1I = . = tan t.d t = tan t - 1 d t1 + tan t cos t cos t

24 4 440

0 0 0

sin tdt tg t= tan t.d tant - = + ln costcost 2

1 1 1 1 1 1= + 1n - 1n1 = + 1n = 1 - 1n22 2 2 2 22.

Cu IV:

y

1

y = m+323

13

x

O1

BH

C

OA

S

a

Hnh 13




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47

1. To giaoim ca (C1) v (C2) l nghim ca h:2 2

2 22 2

7 10 010 010 04 2 20 0

x y x y x

x y x x y x y

Rt y t ph ng trnhu, th vo ph ng trnh th 2 tac.x2 + (7x - 10)2 - 10x = 0 x2 - 3x + 2 = 0 x = 1, x = 2.

Vi x = 1 ta c y = -3Vi x = 2 ta c y = 4

Vy 2 giaoim ca (C1) v (C2) l A1 (1;-3) A2 (2;4).

Trungim A ca A1A2 c to A3 1;2 2

. Ta c 1 2 1;7 A A . ng thng

qua A vung gc vi A1A2 c ph ng trnh3 1

1.3 7 02 2 x y

hay: x + 7y - 5 = 0

To tm I ca ng trn cn tm l nghim ca h:7 5 06 6 0

x y

x y

(12; 1) I

ng trn cn tm c bn knh:

R = IA2 = 2 22 12 4 1 125

Vy ng trn cn tm c ph ng trnh:(x- 12)2 + (y + 1)2 = 125

2. Ta c:(C1): (x - 5)2 + y2 = 52 (C2): (x + 2)2 + (y - 1)2 = 52 (C1) c tm I1(5; 0) bn knh 5(C2) c tm I2(-2; 1) bn knh 5(C1) v (C2) ct nhau nn ch c hai ti p tuyn ngoi.

Ta c 1 2 7;1 I I . Do dng thng I1I2 c ph ng trnh:1. (x + 2) + 7 (y - 1) = 0

hay: x + 7y - 5 = 0Do tip tuyn chung ca (C1) v (C2) c ph ng trnh dng:

x + 7y + d = 0Khong cch t I1 n ti p tuyn l




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48

2 2

5 5 5 25 2 5 25 2 5 25 21 7

d d d d

Vy cc tip tuyn phi tm l:x + 7y - 5 + 252 = 0x + 7y - 5 - 252 = 0

Cu V:1. Gii ph ng trnh 24 4 2 12 2 16 x x x x

iu kin: x - 4 0 x 4.Vi iu kin x 4, ph ng trnh t ng ng vi

4 4 4 4 12 2 4. 4 x x x x x x

2

4 4 4 4 12 x x x x (1)

t 4 4 0 x x t t

Ph ng trnh (1) t2 - t - 12 = 0 43

t

t

t = 4 4 4 4 x x (1)

22 2 16 16

4 x x

x

2 16 8 x x

2 24 816 64 16 x

x x x x = 5

2. Tng s cch chn 8 hc sinh t 18 em ca i tuyn l818

18.17.16.15.14.13.12.11437588.7.6.5.4.3.2.

C

Tng s cch trnc phn lm hai b phn r i nhau:B phn I gm cc cch chn t i tuyn ra 8 em sao cho mi khi u

c emc chn (s cch phi tm).B phn II gm cc cch chn t i tuyn ra 8 em ch gm hai khi (lu

l s em thuc mi khi u t h n 8 nn khng c cch chn no c 8 em thuc cng cng mt khi).

Ring b phn II c th phn tch thnh 3 loi: 8 emc chn t khi 12 hoc khi 11: C 813C cch chn 513C 8 emc chn t khi 12 hoc khi 10: C 812C cch chn 412C

(loi)




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49

8 emc chn t khi 11 hoc khi 10: C 811C cch chn 311C S cch phi tm s l:

8 5 4 318 13 12 11 43758 1947 41811C C C C (cch).

Cu VI.Ta c:

2 2 2

2 2 2 2a b c a b c

a b c R R R R

= a sinA + b sinB + c sinC =2 2 2S S S a b cbc ca ab

= 2S a b cbc ca ab

Mt khc 2S = ax + by + xy + cz. Do:

2 2 2

ax2

a b c a b cby cz

R bc ca ab

Ta c, theo bt ng thc Bunhicpski:

2

1 1 1 1 1 1ax ax+by+cz2 2 2

b c c a a bby cz

a c b b a c c b a a b c

x y z

Suy ra:2 2 2

2a b c

x y z R

(1)

Theo di phn chng minh trn, ta thy du "=" trong (1) xy ra khi v ch khi:

2b c c a a bc b a c b a

a x b y c z

a b c

x y z

ABCu M trng vi tr ng tm G ca tam gic ABC.

Nhn x t: Cng c th c lng: 1 1 1a b cbc ca ab a b c

nh p dng

bt ng thc Csi cho mi cp s hng v tri:




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50

2 2 21 1 1 1 1 1 1 1.2 .2 .22 2 2 2 2

1 1 1

a b c a c c abc ca ab bc ab ab bc c a b

a b c

C

ch 2 : C th lm cch khc nh sau:1 1 1. ax . x y z by cza b c

2 2 2

1 1 1 1 1 1 1 1 1(ax ) .22

2 2

abcby cz S

a b c a b c a b c R

ab bc ca a b c R R




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51

S 2 Cu I:

1. Tm n N* tho mn bt ph ng trnh 3 22 9nn n A C n iu kin: n N*, n 3.

Bt ph ng trnh n(n - 1) (n - 2) + 22 9nC n n(n - 1) (n - 2) + n(n - 1) 9 (n - 1)(n - 2) + n - 1 9

2 2 8 03 4 ( *)

n n

n n N

K t hp iu kin n 3 suy ra n = 3 hoc n = 4.

2. Ph ng trnh 84 221 1log 3 log 1 log (4 )2 4

x x x

iu kin: 01

x

x

Ph ng trnh 84 221 1log ( 3) log ( 1) log (4 )2 4

x x x

2 2 2log ( 3) log 1 log (4 )( 3). 1 4

x x x

x x x

i. Nu x > 1, ph ng trnh (x + 3)( x -1) - 4x = 02

2 3 0 31

x x x x

ii. Nu 0 < x < 1, ph ng trnh (x + 3)( 1 - x) - 4x = 02 6 3 0 3 2 2

0 1 x x

x x

p s: ph ng trnh c 2 nghim 3

3 2 3 x

x

Cu II:

1.2 2

2 2 x x m m

y x x x

2

2 24 4' 1

( 2) ( 2)m x x m

y x x

hm s nghch bin trnon 1;0 , iu kin cn v l




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52

' 0 [ 1;0 y x

2

[ 1;0]

( ) 4 4 , 1;0max ( ) ( 1) 9g x x x m x

g x m g m m

2. kho st v v th hm s khi m = 1.( Hc sinh t gii)

3. Ph ng trnh.2 2

2

1 1 1 1

2

1 1

9 ( 2 )3 2 1 02 1 ( 2)

(1 )3

t t

t

a a

X X a X

X

Do 21 1 1 2,t t m 1 - t2 0. Suy ra min gi tr ca 21 13 t X lon

3;9 .

2 2 1(1) 2

3 9

X X a

X X

T th v cu 2, hn ch trongon 3;9 suy ra, ph ng trnh c nghim,

iu kin cn v l: 6447

a * Ch : Bn c th thy th v hn, nu tm a phng tr nh sau c nghim:

2 2t + 1 - t t + 1 - t9 - (a + 2).3 + 2a + 1 = 0

Cu III:1. Gii ph ng trnh

4 4sin os 1 1cot25sin 2 2 8sin 2 x c x

x x x

iu kin: sin2x 0Vi iu ph ng trnh:

2 21 2in cos 1 1os25 2 8 x x

c x

2 9os 2 5 os2 049os2 1

2 2 6 21 3cos22

c x c x

c x x k

x

2. trong tam gic ABC h thc: bsinC (bcosC + ccosB) = 20

(loi)6

x k




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53

4R 2sinBsinC (sinBcosC + sinCcosB) = 20 4R 2sinBsinCsinA = 20

M3

28 sin Asin sin 2 sin Asin sin4 4abc R

S B C R B C R R

Vy ta c: S = 10 ( n v din tch)

Cu IV: 1. cho gn, k hiu BC = a, AC = b,

AC = c, OA = x, OB = y, OC = z.Theo gi thit ta c:

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

a y z

b x z

c x y

a b c

b c a

c a b

Suy ra, ABC l tam gic nhn, nntr c tm H nm trong tam gic.

Gi AM, BK, CN l 3ng cao

ca tam gic ABC.Theo gi thit OA (OBC)

. BC OA

BC OM BC AM

Suy ra OMA = , OKB = , ONC =

Ta c: cos = OM AM

.

Trong tam gic vung BOC ta c: 2 2 2 2 21 1 1 1 1

OM OB OC y z

Suy ra: OM2 =2 2

2 2 y z

y z OM =

2 2

y z

y z

Trong tam gic vung AOM:

AM2 = OA2 + OM2 = x2 +2 2

2 2 y z

y z

B

C

O

A

N

A K

H

Hnh 15

M




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54

2 2 2 2 2 2 2 2 2 2 2 22

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

AM AM y z y z

Vy, cos = 2 2 2 2 2 2OM yz AM x y y z z x

Lp lun t ng t ta cng c:

cos =2 2 2 2 2 2

xz

x y y z z x

cos =2 2 2 2 2 2

xy

x y y z z x

Suy ra, cos + cos + cos =

2 2 2 2 2 2 2

313

xy yz xz xy yz xz x y y z z x xy yz xz

(pcm)

Cch khc: cos + cos + cos =2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2

3( ) 3 x y y z z x xy yz xz x y y z z x x y y z z x

2. a) (P) c vect php 1; 1;1n . ng thng (d) qua A v (P) c ph ngtrnh:

1 3 21 1 1

x y z

Gi I l giaoim ca (d) v (P), khi c to ca I l nghim ca h:3 0

1 3 21 1 1

x y z

x y z

Suy ra I(-2; -2; -3).A' lim i xng ca A qua (P) (d) v I l trungim ca AA'. Do A' c

to (xo; yo; zo) 01 2

2 x , 03 2

2 y , 02 3

2 z

x0 = -3; y 0 = -1, zo = 4. Vy A' (-3; -1; -4). b) ' BA = (2; -8; -16) = 2(1; -4; -8). BA' c ph ng trnh:




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55

3 1 41 4 8

x y z

Gi M l giaoim ca BA' vi (P) th to ca M l nghim ca h

3 0

3 1 4 4;3;41 4 8

x y z

x y z M

Ta c: MA + MB = MA' + MB A'B. Do MA + MB t gi tr nh nht khiMA' + MB t gi trnh nht bng A'B. Vy gi tr nh nht ca biu thc MA +MB l A'B = 2 2 22 8 16 18 , khi M l giaoim ca ng thng A'B vi mt phng (P). Cu V:

Ta c: I =

1 3 1 3 1 3

3233 20 0 0

1 1 11 1

xn n n x x x

x

d ee dx e d eex e

= 1 1 32 1 1

1 3 02 20

12[ 1 1 ]1

2

x n x

n

x

ee e

=1 12 2 1 12 4 2 2 2 1

2 2

.




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56

S 3

Cu I:

1. a) Kho st v v th hm s khi m =1

2

.

(Hc sinh t gii). Gi (C) l th hm s.b) Vit ph ng trnh tip tuyn ca (C) bit tip tuyn y song song vi ng

thng y = 4x + 2.

Ta c hm s y = 3 21 1 423 2 3

x x x

y' = x2 + x - 2Theo gi thit tip tuyn th phi tm c h s gc k =4. Vy c:

2 22 4 6 0 x x x x 1 1

2 2

22; 3

13;6

x y

x y

Vy, c 2 tip tuyn tho mniu kin bi.

Ti p tuyn (d1): 2 264 2 43 3 y x y x

Tip tuyn (d2): 1 734 3 46 6 y x y x

2. Do 0 < m


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57

4 3 223 12 0

12 3 x x x

m x m x

4 8 2 4 104 43 3 3 3 3

mm m

Theo gi thit S = 4 m = 12

(tho mniu kin 0 < m


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58

2 211 sin 2 2 sin 2 sin 32 x x x

2 22 sin 2 2 sin 2 .2sin3 x x x

sin3x =12 (d thy tho mniu kin)

3 2

653 26

x k

x k

218 35 218 3

k x

k x

(k Z)

Cu III:1. K AH BE. Do SA (ABC) nn BE SH. Do SH l khong cch t S

n BE. Ko di BE ct AD ti M. E l trungim ca CD nn ED =2 2

a AB D

l trungim ca AM AM = 2aABM vung ti A ta c:SAH vung ti A ta c:

SH =2

2 2 2 4 3 55 5a a

SA AH a

aB C

E

DM

S

A

a

H

Hnh 16




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59

2. (d) l hnh chiu vung gc ca trn (P) th (d) l giao tuyn ca (P) v mt phng (Q) cha v vung gc vi (P). rng, mt phng (Q) cha c ph ng trnh dng: 2 1 2 0 x y z x y z , 2 2 0 (1)

Tht vy, tt c cc im M(x; y) thuc u c ta tha:2 1 0

2 0 x y z

x y z

v do tha (1). (1) 2 2 x y z , 2 2 0

suy ra (Q) c vct php tuyn 2 ; ;Qn

Do (Q) (P) nn . 0 4. 2 2 1 0 7 3 0Q Pn n Chn 3 ta c 7

Vy, (Q) c ph ng trnh x + 4y + 4z + 11 = 0 Nh th, (d) l giao ca hai mt phng: 4 2 1 0

4 4 11 0 x y z

x y z

(d) c vc t ch phng [ , ]P Qa n n

= (4; 5; -6), trong , Pn =(4; - 2; 1), Qn =(1; 4;4) v i qua im (1; 0;- 3).

T , suy ra phng tr nh (d): 1 34 5 6

x y z

Cch 2. (d) l hnh chiu vung gc ca trn (P) th (d) l giao tuyn ca (P) vmt phng (Q) cha v vung gc vi (P). Mt phng (Q) i qua Mo (1; -3; 0) lmt im thuc v c cp ch phng gm mt vc t l vc t ch phng ca

, mt vc t l vc t php tuyn ca (P).

1 2 1 2(2;1;1), (1;1;1) [ , ] (0; 1;1)n n n n

Vc t ch phng ca l (0, 1;1)a .

Suy ra, vc t php tuyn ca (Q) l [ , ]Q Pn a n

=(1; 4; 4).

Vy phng tr nh (Q): 1(x - 1) + 4(y + 3) + 4z = 0x + 4y + 4z + 11 = 0. Cch 3. Trn ( ) chn im Mo no chng hn Mo (1; -3; 0). Gi H l hnhchiu ca M0 trn (P). Gi I l giaoim ca () vi (P). Suy ra ph ng trnhng thng IH chnh l phng tr nh hnh chiuca( ) ln mt phng (P):

1 34 5 6

x y z




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60

Cu IV:

1. Tm3

0

1 11 x

x xim

x

3 3

0 00

1 1 1 1 1 11 lim lim x x x

x x x xim x x x

Xt tng s hng:

0011 lim21 1 1 1 x x

x xim

x x x

3

00 23 3

1 11 11 lim[1 1 1 ]

x x

x xim

x x x x

20 3 31 1

1 31 1 xim x x x

3

0

1 1 1 112 3 x

x xim

x 5

6

2. (C1): x2 + y2 - 4y - 5 = 0 x2 (y-2)2 = 9(C2): x2 + y2 - 6x + 8y + 16 = 0 (x-3)2 + (y + 4)2 = 9(C1) c tm I1 (0; 2) bn knh R 1 = 3(C2) c tm I2 (3; -4); bn knh R 2 = 3

V I1I2 = 22

1 23 6 45 3 5 R R nn (C1) v (C2) nm ngoi nhau, do c 4 tip tuyn chung.

V R 1 = R 2 = 3 nn d1 // d2 // I1I2

Ph ng trnhng thng I1I2 = 0 2 2 2 2 03 6 x y

x y

Ph ng trnh d1, d2 c dng 2x + y + c = 0

Khong cch t I1 n d1, d2 bng 2 22 32 1

c

1

2

3 5 2 2 3 5 2 02 3 53 5 2 2 3 5 2 0

c x yc

c x y




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61

Hnh 17

Do tnhi xng, d3 v d4 ct nhai ti trungim I ca on I1I2 c to (3/2;1).

Ph ng trnh d3, d4 c dng y + 1 = k3 31 02 2

k x kx y

Khong cch t I1 ti d3, d4 bng 232 1 2 31

k

k

Gii ra tac :1

2

0 144 333

k y

y xk

Cu V:

Cho, 0

54

x y

x y Tm min S vi S = 4 1

4 x y

Cch 1: S =5

1 1 1 1 1 5 5.5 25 54 4 5. . . .4 x x x x y x x x x y x x x x y

min S = 5

1 14

454

x y

x y

x y

114

x

y

Cch 2: S = 4 1 ( )5 4

f x x x

0 < x


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62

f(x) = 22

4 4 05 4 x x

22 5 4

1504

x x x

x

Lp bng du f '(x) suy ra min S = 5

Cch 3: 1 2 1 4 12 . . .2 42

x y x y x y x y

(3)

Du "=" (3) khi

54

2 1 4 1. 2 . 5 1

4 4

x y x x x y y

x y y x y

(3)25 5 4 1 4 1 5

2 4 4 4 x y x y

Vy min S = 5.




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63

S 4

Cu I: 1. Gii bt ph ng trnh: 12312 x x x

iu kin: x 3.Bt ph ng trnh 12312 x x x

12312 2 x x x (v x +12 > x-3 0)

2

2 9 2 12 3 2 1

12 . 3 49 52 0 13 4

x x x x

x x

x x x

Do iu kin x 3, suy ra3 4 x 2. Gii ph ng trnh

2tan cos cos sin (1 tan tan )2 x

x x x x x

iu kin:1cos

0cos0

2cos

0cos

x

x x

x

Ta c:

sin sin2

1 tan tan 12 cos cos2

x x x

x x x

x x x

x x

x x

x x

x x

cos1

2coscos

2cos

2coscos

2sinsin

2coscos

Ph ng trnh 2 sintan cos coscos

x x x x

x

cosx(1 - cosx) = 0.Doiu kin cosx 0 nn ph ng trnh cosx = 1

x = 2k , (k Z)Cu II:1. y = (x - m)3 - 3x

y' = 3(x-m)2 - 3 = 3[(x - m)2 - 1]




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64

y'(0) = 3(m2 -1)y" = 6(x - m) y"(0) = -6m

iu kin cn v hm s t cc tiu ti x = 0 l y'(0) = 0 m = 1 hoc m = -1.

Vi m = 1 th y"(0) = -6 < 0, hmt cc i ti x = 0Vi m = -1 th y"(0) = 6> 0, hmt cc tiu ti x = 0

p s: m = -1 .2. Kho st v v th hm s khi x = 1( Hc sinh t gii)

th hm s y = (x - 1)3 - 3xc cho trn hnh 18.3. Tm k h bt ph ng trnh c nghim

)2(1)1(log31log

21

)1(0313

22

2

3

x x

k x x

iu kin: (x - 1)3 > 0 x > 1Khi x > 1, bt ph ng trnh (1)

(x - 3)3 - 3k < k (1')Bt ph ng trnh (2) log2x + log2(x - 1) 1 (x > 1).

x(x - 1) 2

1022

x

x x

1< x 2Bi ton quy v xcnh k bt ph ng trnh (1')c nghim thaiu kin 1< x 2.

Da vo th hm s v cu 2, xt ch trn khong 1< x 2, ta suy ratp mi tr s k cn tm l k > -5 (k >

(1;2]min ( ) (2) 5 f x f )

Cu III:1. Gi H l trungim BC.

Do ABC vung cn ti A nnAH BC; AB = AC SB = SC

SH BC. Do AHS = 600.

Hnh 18




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65

Ta c AH = .22a BC

Trong SAH vung ti A nn:

SA =0 3

tan 602 2a a

.2. a)

d1 l giao tuyn ca hai mt phng x - az - a = 0 v y - z + 1 = 0 nn i qua M1(a; - 1; 0) v vung gc vi cc vc t

1u = (1; 0; - a), 1v = ( 0; 1; - 1).

Suy ra mt vc t ch phng ca d1 l 1a = [ 1u , 1v ] = (a; 1; 1).

d2 l giao tuyn ca hai mt phng ax + 3y- 3 = 0 v x - 3z - 6 = 0 nn i quaM2(0; 1; - 2) v vung gc cc vc t2u = (a; 3; 0), 2v = ( 1; 0; - 3). Suy ra mt vct ch phng ca d2 l 2a = [ 2u , 2v ] = (3; - a; 1).

1 1 M M = (- a; 2; - 2), 1 2[ ; ]a a

= (1 + a; 3 - a; - a2 - 3)

Ta c 1 2[ ; ]a a

. 1 1 M M = a2 - 3a + 12 > 0, a. Suy ra, khng c a d1 v d2 ct nhau.Ta cng c kt qu l d1 v d2 cho nhau, vi a.Cch 2.

Phng tr nh tham s ca cc ng thngd1 v d2:

1 2

'x = a + atd : y = - 1 + t d : 1 '

3z = t 12 '3

x t

a y t

z t

Xt h phng tr nh:

a + at ' (1)

1 1 ' (2)3

12 ' (3)3

t

at t

t t

T (2) v (3) suy ra:(1 ) ' 12a t a = - 1 khng tha

Hnh 19

H

S

C

B

A




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66

121 '1

a t a

, 421

t a

. Thay vo (1), ta c a2 - 3a + 12 = 0 v

nghim. Vy, khng c a d1 v d2 ct nhau.

b) Vi a = 2 ta c:

1 22 2 0 2 3 3 0

: ; :1 0 3 6 0

x z x yd d

y z x z

Theo a) d2 l giao tuyn ca hai mt phng 2x + 3y- 3 = 0 v x - 3z - 6 = 0 nn iqua M2(0; 1; - 2) v vung gc cc vc t2u = (2; 3; 0), 2v = ( 1; 0; - 3). Suy ramt vc t ch phng ca d2 l 2a = [ 2u , 2v ] = (3; - 2; 1).

d1 l giao tuyn ca hai mt phng x - 2z - 2 = 0 v y - z + 1 = 0 nn i qua

M1(2; - 1; 0) v vung gc vi cc vc t1u = (1; 0; - 2), 1v = ( 0; 1; - 1). Suy ramt vc t ch phng ca d1 l 1a = [ 1u , 1v ] = (2; 1; 1).

Mt phng (P) cha d2 v song song d1 nni qua M2(0; 1; - 2) v c mt vc t php tuyn Pn = [ 1a , 2a ] = (3; 1; - 7). Suy ra, phng tr nh ca (P):

3(x - 0) + y - 1 - 7(z + 2) = 0hay: 3x + y - 7z - 15 = 0

Cch 2. Mt phng (P) cha d2 nn ph ng trnh c dng:

(2 3 3) ( 3 6) 0 x y x z ,2 2

0 (2 ) 3 3 (3 6 ) 0 x y z , 2 2 0

Vy (P) c vect php 2 2 ;3 ; 3n Trong ph ng trnh xcnh d1 t x = 2t ta c z = t - 1, y = t - 2.Do vect ch ph ng ca d1 l )1;1;2(1 n v (P) // d2 nn 1 2. 0n n

2. 2 1.3 1.3 0 7 0 Chn = 1 tac 7 . Th vo ph ng trnh xcnh (P) tac ph ng

trnh ca (P) cn tm:9x + 3y - 21z - 45= 0

hay 3x + y - 7z - 15 = 0. (P) cha d2 v (P)// d1 nn khong cch gia d1, d2 l khong cch t mt im

ca d1 n (P).




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67

im M(2;- 1; 0) thuc d1 Vy khong cch cn tm l

2 2 2

3.2 1 7(0) 15 17593 1 7

d

Cu IV:1. (1 + x)n = a0 + a1x + a2x2 +....+ak xk + anxn.

H thc 11)1(249211 nk

aaa k k k .

)!1()!1(!

241

)!(!!

91

)!1()!1(!

21

2492

11

k nk n

k nk n

k nk n

C C C k nk n

k n

2.(k - 1)!(n - k + 1)!=9.k!(n - k)! = 24.(k + 1)!(n - k - 1)!

2.(n - k + 1)(n - k) = 9k (n - k) = 24 (k + 1)k

1183

1122

)1(24)(99)1(2

nk

nk

k k n

k k n

tn ti k tho mn h thc (1),iu kin t c v l 3n - 8 = 2n + 2 n = 10

2. Tnh0

121

32 )1(( I I dx xe x I x , trong: I1 =

0

1

0

1

32

2 1; dx x x I dx xe x

Tnh

dxe xedx xe I x x x

0

1

22

0

1

21 1021

= 41

43

41

41

21

10

41

21

22222

eeeee x




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68

Tnh I2 =0 0

3 3

1 1

1 ( 1 1) 1 ( 1) x x dx x x d x

289

1

0

4)1(3

7)1(3

)1()1()1()1(

34

37

0

1

310

1

34

x x

xd x xd x

Vy I = I1 + I2 = 289

41

43

2 e = 7

44

32 e

Cu V: Gi thit:

2cos

2cos

2cos

2sin

2sin

2sin8

2cos

2cos

2cos)1coscos(cos2

2cos

2cos

2cos8)coscoscos3(2

2cos

2cos

2cos

412

2cos

2cos

2cos 222

AC C B B AC B A

AC C B B AC B A

AC C B B AC B A

AC C B B AC B A

8 sinA sinB sinC = (sinA + sinB)(sinB + sinC)(sinC + sinA)

sinA = sinB = sinC A = B = C (ch cn p dng bt ng thc csicho v phi).




69/247



69

S 5

Cu I.:

Hm s y = xmx x

12 (1)

1. Kho st, v th hm s khi m = 0 (Hc sinh t gii).

2. Tm m y c cc i v cc tiu.

y' = 22

2

2

)1(2

)1()1)(2(

m

m x x

x

mx x xm x

y c cc i v cc tiu, iu kin cn v l ph ng trnh y(x) = 0 c hainghim v yi du khi x bin thin qua mi nghim. - x2 + 2x +m = 0 (2) c hai nghim phn bit khc 1.

021

01'm

m 1 m

Gi M1(x1,y1) lim cc i vim cc tiu thuc th hm s.x1,x2 l nghim ca(2), y1,y2 c th tnh bng cch thay x1, x2 vo (1),

nhng cng c th tnh bng cch khc nh sau:

Hm s (1) c dng y =vu , y =

vuvvu

2'' .

Ti x1, x2 ta c y = 0 xuvvu 2,1''

x x x x vu

vu

uvuv 2,12,12,12,1 '''0'

Vy tnh y1,2 ta c th dng t s ''

vu ( n gin h n t s

vu ). Vi nhn xt ny

ta c:y1 = x

m x11

2 = -(2x1 + m); y2 = -(2x2 + m)

M1M2 = 2212

21 )()( y y x x =2

21 )(5 x x

= 5 x x x x 212

21 4 = 5 m44



8/13/2019 D B THI H 2002 - 2008 V H

ĐỀ dỰ bỊ thi Đh 2002 - 2008 vÀ hd giẢichi tiet

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