Tuyển Tập Đề Dự Bị Môn Toán Thi Đh 2002 - 2008 Và Hd Giảichi Tiet

Trn Xun Bang - Trng THPT Chuyn Qung Bnh

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

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D B THI 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 B THI 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 trnh x 2x + 1 x1 1

2 2

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

2. Xc nh m phng trnh 2(sin4x + cos4x) + cos4xx + 2sin2x + m = 0 c t

nht mt nghim thuc on 0; 2

.

Cu III: 1. Cho hnh chp S.ABC c y ABC l tam gic u cnh a v cnh bn SA vung 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 phn 1 3

20

xI = dxx + 1 .

Cu IV: 1. Trong mt phng Oxy cho hai ng trn: (C1): x2 + y2 - 10x = 0 (C2): x2 + y2+ 4x - 2y - 20 = 0 Vit phng trnh ng trn i qua cc giao im ca (C1) , (C2) v c tm nm trn ng thng x + 6y - 6 = 0. 3. Vit phng trrnh ng tip tuyn chung ca hai ng trn (C1) v (C2). Cu V: 1. Gii phng trnh 24 4 2 12 2 16x x x x . 2. i tuyn hc sinh gii ca mt trng gm 18 em, trong c 7 hc sinh khi 12, 6 hc sinh khi 11 v 5 hc sinh khi 10. Hi c bao nhiu cch c 8 hc sinh 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 ba gc nhn n cc cnh BC, CA, AB. Chng minh rng:

2 2 2a + b + cx + y + z

2R ; a, b, c l cnh tam gic, R l bn knh

ng trn ngoi tip. Du ng thc xy ra khi no ?





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S 2 Cu I: 1. Tm s nguyn dng tho mn bt phng trnh: 3 n-2n nA + 2C 9n, trong

k kn nA , C ln lt l s chnh hp v s t hp chp k ca n.

2. Gii phng trnh 84 221 1log (4x + 3) + log (x - 1) log (4 )2 4

x

Cu II:

Cho hm s 2x - 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 trnh sau c nghim: 2 21 + 1 - t 1 + 1 - t9 - (a + 2).3 + 2a + 1 = 0 Cu III:

1. Gii phng trnh 4 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 = 0 v 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 trn mf(P), tm gi tr nh nht ca MA + MB. Cu V:

Tnh ln3 x

x 30

I = .(e 1)e dx





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S 3

Cu I: Cho hm s y = 13

x3 + mx2 -2x - 2m - 13

(1)(m l tham s)

1. Cho m = 12

: a) Kho st s bin thin v v th (C) hm s (1) .

b) Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn song song vi ng thng y = 4x + 2.

2. Tm m thuc khong 50; 6

sao cho hnh phng gii hn bi th hm s (1)

v cc ng x = 0, x = 2, y = 0 c din tch bng 4. Cu II:

1. Gii h phng trnh 4 2

4 3 0

log log 0

x y

x y

2. Gii phng trnh 2

44

(2 - sin 2x)sin3xtan x + 1 = cos x

.

Cu III: 1. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung gc vi mt phng (ABCD) v SA = a. Gi E l trung im ca cnh CD. Tnh theo a khong cch t im S n ng thng BE. 2. Trong khng gian Oxyz cho ng thng v mt phng (P).

2x + y + z + 1 = 0

: (P): 4x - 2y + z - 1 = 0x + y + z + 2 = 0

Vit phng trnh hnh chiu vung gc ca ng thng v mf(P). Cu IV:

1. Tm gii hn 3

x 0

x + 1 + x - 1L = limx

2. Trong mt phng Oxy cho hai ng trn: (C1): x2 + y2 - 4y - 5 = 0 (C2): x2 + y2 - 6x + 8y + 16 = 0 Vit phng trrnh ng tip tuyn chung ca hai ng trn (C1) v (C2). Cu V:

Cho x, y l hai s dng thay i tho mn iu kin x + y = 54

.

Tm gi tr nh nht ca biu thc 4 14

Sx y

.





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S 4 Cu I: 1. Gii bt phng trnh: x + 12 x - 3 + 2x + 1

2. Gii phng trnh tanx + cosx - cos2x = sinx(1 + tanx.tan x2

).

Cu II: Cho hm s y = (x - m)3 - 3x (m l tham s). 1. Xc nh m hm s cho t 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 trnh sau c nghim:

3

2 32 2

x - 1 - 3x - k < 01 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 trnh mt phng(P) cha d2 v song song d1 v tnh khong cch gia d1 v d2. Cu IV: 1. Gi s n l s nguyn dng v (1 + x)n = a0 + a1x + a2x2 + ...+akx2k +...+anxn.

Bit rng tn ti s nguyn k( 0 k n - 1 sao cho 1 12 9 24k k ka a a . Hy tnh n ?

2. Tnh tch phn 0

2x 3

- 1

I = x(e + x + 1)dx

Cu V: Gi A, B, C l ba gc ca tam gicABC. Chng minh rng tam gic ABC u th iu 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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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 trnh 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 hai im cc tr ca th hm s (1) bng 10. Cu II: 1. Gii phng trnh 3 232716 log 3log 0xx x x .

2. Cho phng trnh 2sinx + cosx+1sinx-2cosx+3

a (2)(a l tham s)

a) Gii phng trnh (2) khi a = 13

. b) Tm a phng trnh (2) c nghim.

b) Tm a phng trnh (2) c nghim. Cu III: 1. Trong mt phng Oxy cho ng thng d: x - y + 1 = 0 v ng trn (C): x2 + y2 + 2x - 4y = 0. Tm ta im M thuc ng thng d m qua k c hai ng thng tip xc vi ng trn (C) ti A v B sao cho gc AMB bng 600.

2. Trong khng gian Oxyz cho ng thng 2x - 2y - z + 1 = 0

d: 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 gc BAC, CAD, DAB u bng 600. Cu IV:

1. Tnh tch phn

2

6 3 5

0

I = 1 - cos x .sinxcos xdx .

2. Tm gii hn 3 2 2

x 0

3x - 1 2 1L = lim1 - 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 thc 2a c b + b + 50 +

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

thc a c + b d

.





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S 6 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 trc honh. Cu II:

1. Gii phng trnh 21 s inx

8 osc x .


3 2

log ( 2 3 5 ) 3

log ( 2 3 5 ) 3x

y

x x x yy y y x

Cu III: 1. Cho hnh t din u ABCD, cnh a = 6 2 cm. Hy xc nh v tnh di on vung gc chung ca ng thng AD v ng thng BC.

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

9 4y

v ng thng dm: mx - y - 1 = 0 a) Chng minh rng vi mi gi tr ca m, ng thng dm lun ct elip (E) ti hai im phn bit. b) Vit phng trnh 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 hn 6

2x 1

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

.

2. Cho tam gic ABC c din tch bng 32

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

BC, CA, AB v ha, hb, 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 3a b ca b c h h h





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S 7 Cu I:

1. Kho st v v thi hm s 22x - 4x - 3y =

2(x - 1).

2. Tm m phng trnh 2x2 - 4x - 3 + 2m 1x = 0 c hai nghim phn bit. Cu II: 1. Gii phng trnh 3 - tanx(tanx + 2sinx) + 6cosx = 0 .

2. Gii h phng trnh y xx y

log xy = log y

2 + 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 cho IM = 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 ABC l tam gic cn vi AB = AC = a v gc 0BAC = 120 , cnh bn BB' = a. Gi I l trung im CC' . Chng minh rng tam 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 + 3 cosx.





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S 8 Cu I:

Cho hm s 2 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 trnh cos2x + cosx(2tan2x - 1) = 2 . 2. Gii bt phng trnh 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 090BDC . Xc nh tm v bn knh mt cu ngoi tip t din ABCD theo a v b. 2. Trong khng gian Oxyz cho hai ng thng :

11:

1 2 1x y zd 2

3 1 0:

2 1 0x z

dx y

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

song vi ng thng : 4 7 31 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

0

I = x 1 - x dx

Cu V:

Tnh cc gc ca tam gic ABC bit rng 4 ( )

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 +cp = 2

.





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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 trnh 3cos4x - 8cos6x + 2cos2x + 3 = 0. 2. Tm m phng trnh 22 1

2

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 trnh ng trn 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 cho mt phng (BD'M) ct hnh lp phng theo mt thit din c din tch nh nht. 3. Trong khng gian Oxyz cho t din OABC vi A(0; 0; a 3 ), B(a; 0; 0), C(0; a 3 ; 0) (a > 0). Gi M l trung im BC. Tnh khong cch gia hai ng thng 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 mn iu 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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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). Tm im M thuc (C) sao cho tip tuyn ca (C) ti M vung gc vi ng thng IM. Cu II:

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

2cosx - 1

.

2. Gii bt phng trnh 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

phng trnh cc ng thng d1, d2 i qua M v tip xc vi (E). Tm n trong s cc tip tuyn ca (E) qua N c mt tip tuyn song song vi d1 hoc d2. 2. Cho hnh chp u S.ABC, y ABC c cnh bng a, mt bn to vi y mt gc bng 0 0(0 90 ) . Tnh th tch khi chp S.ABC v khong cch t nh A n mt phng (SBC). 3. Trong khng gian Oxyz cho hai im I(0; 0; 1), K(3; 0; 0). Vit phng trnh mt phng i qua hai im I, K v to vi mt phng Oxy 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 v 1

0

(x)dx = 5f

Cu V:

Chng minh rng 2

x xe + cosx 2 + x - 2

, x .





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S 11 Cu I:

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

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

1. Kho st s bin thin v v th ca hm s (1) khi m = 1. 2. Tm m hm s (1) ng bin trn khong (1; + ) . Cu II:

1. Gii phng trnh 2cos 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 trnh 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 trnh 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 tm c, hy xc nh to tip im ca mt phng (P) v mt cu (S). 3. Cho hnh chp S.ABC c y 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 rng tam gic AMB cn ti M v tnh din tch tam 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

0

x 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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S 12 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 trnh 2cos4x cotx = tanx + sin2x

2. Gii phng trnh 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 trnh mt phng (P) i qua trung im I ca AB v vung gc vi AB. Gi K l giao im ca ng thng d v mt phng (P), chng minh rng d vung gc vi IK. b) Vit phng trnh tng qut ca hnh chiu vung gc ca d trn mt phng c phng trnh x + y - z + 1 = 0. 2. Cho t din ABCD c AD vung gc vi mt phng (ABC) v tam gic ABC vung ti A, AD = a, AC = b, AB = c. Tnh din tch ca tam gic BCD theo a, b, c v chng minh 2S 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 knC l s t hp cp k ca n.


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 + cp =2

.





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S 13 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 trnh 4(sin3 x + cos3 x) = cosx + 3sinx.

2. Gii bt phng trnh log 4

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

Cu III: 1. Trong mt phng Oxy cho ng thng d: x - y + 1 - 2 = 0 v im A(-1; 1). Vit phng trnh ng trn i qua A, qua gc to O v tip xc vi ng thng d. 2. Trong khng gian Oxyz cho hnh hp ch nht ABCD.A 1B 1C 1D 1 c A trng vi gc to O, B(1; 0; 0), D(0; 1; 0), A 1(0; 0; 2 ). a) Vit phng trnh mt phng (P) i qua ba im A 1, B, C v vit phng trnh 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 din ca hnh chp A 1ABCD vi mt phng (Q). Cu IV: 1. Tnh th tch ca vt th trn 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 trnh x - my = 2 - 4mmx + y = 3m + 1

(m l tham s). Tm

gi tr ln nht ca biu thc A = x2 + y2 - 2x, khi m thay i.





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S 14 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 trnh 2 2 cos(x + 4 ) + 1

sinx = 1

cosx .

2. Gii bt phng trnh x - 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 trnh 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. Tm im C 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 ABC c AB = BC = 2a, gc B bng 1200. Tnh khong cch t nh A n mt phng (SBC). Cu IV:


31

dxx + x .

2. Bit rng (2 + x)100 = a 0 + a 1x + a 2x2 + ... + a 100 x100 . Chng minh a 2 < a 3. Vi gi 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

rng phng trnh f(x) = 3 c ng hai nghim.





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S 15 Cu I:

Cho hm s 2x - 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 trnh sin4xsin7x = cos3xcos6x. 2. Gii bt phng trnh log 3x > log x3. Cu III:

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

8 + y2 4 = 1. Vit phng trnh cc tip

tuyn ca (E) song song vi ng thng d: x + 2 y - 1 = 0 2. 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 minh

rng 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. Bit rng 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 thc A1 - sin2S =

sinB.





18

S 16 Cu I:

Cho hm s 2x + x + 4y = x + 1

(1) c th (C).

1. Kho st hm s (1) . 2. Vit phng trnh tip tuyn ca (C), bit tip tuyn vung gc vi ng thng d: x - 3y + 3 = 0. Cu II: 1. Gii phng trnh 2sinxcos2x + sin2xcosx = sin4xcosx.


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 trn trc 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:


0

xsin 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 v chng minh rng tng ny l mt s chnh phng. Cu V:

Cho phng trnh x2 + ( m2 - 53 )

2x + 4 + 2 - m2 = 0.

Tm tt c cc gi tr m phng trnh c nghim.





19

S 17 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 trnh 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 trn d 1 v C trn d 2 sao cho tam gic ABC c trng tm l G(2; 0). 2. Cho hnh vung ABCD c cnh AB = a. trn cc na ng thng Ax, By vung gc vi mf(ABCD) v nm v cng mt pha i vi mf(ABCD), ln lt ly 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 thang ABNM.

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

Vit phng trnh 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) trn mt phng (P). Cu IV:

1. Tnh tch phn I = ln8

2x x

ln3

e e 1dx .

2. C bao nhiu s t nhin tho mn ng thi ba im kin sau: gm ng 4 ch 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 0

3x - mx x + 16 = 0





20

S 18

Cau I:

Goi (Cm) la o th cua ham so : y = 2 22 1 3x mx m

x m

(*) (m la tham so)

1. Khao sat s bien thien va ve o th cua ham so (*) ng vi m = 1. 2. Tm m e ham so (*) co hai iem cc tr nam ve hai pha truc tung. Cau II:

1. Giai he phng trnh : 2 2 4( 1) ( 1) 2

x y x yx x y y y

2. Tm nghiem tren khong (0; ) cua phng trnh :

2 2 34sin 3 cos 2 1 2cos ( )2 4x x x

Cau III: 1. Trong mat phang vi he toa o Oxy cho tam giac ABC can tai nh A co

trong tam G 4 1( ; )3 3

, phng trnh ng thang BC la 2 4 0x y va phng

trnh ng thang BG la 7 4 8 0x y .Tm toa o cac nh A, B, C. 2. Trong khong gian vi he toa o Oxyz 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 o O va vuong goc vi BC. Tm toa o giao iem cua AC vi mat phang (P). b) Chng minh tam giac ABC la tam giac vuong. Viet phng trnh mat cau ngoai tiep t dien OABC. Cau IV:

1. Tnh tch phan 3

2

0

sin .I x tgxdx

.

2. T cac ch so 1, 2, 3, 4, 5, 6, 7, 8, 9 co the lap c bao nhieu so t nhien, moi so gom 6 ch so khac nhau va tong cac ch so hang chuc, hang tram hang ngan bang 8.

Cau V: Cho x, y, z la ba so thoa x + y + z = 0. Chng minh rang : 3 4 3 4 3 4 6x y z





21

S 19 Cau I:

1. Khao sat s bien thien va ve o th ( C ) cua ham so 2 1

1x xy

x

.

2. Viet phng trnh ng thang i qua iem M (- 1; 0) va tiep xuc vi o th ( C ) .

Cau II:

1. Giai he phng trnh : 2 1 13 2 4

x y x yx y

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

x x x

Cau III: 1. Trong mat phang vi he toa o Oxy cho ng tron (C): x2 + y2 12 4 36 0x y . Viet phng trnh ng tron (C1) tiep xuc vi hai truc toa o Ox, Oy ong thi tiep xuc ngoai vi ng tron (C). 2. Trong khong gian vi he toa o ecac vuong goc Oxyz cho 3 iem A(2;0;0), C(0; 4; 0), S(0; 0; 4). a) Tm toa o iem B thuoc mat phang Oxy sao cho t giac OABC la hnh ch nhat. Viet phng trnh mat cau qua 4 iem O, B, C, S. b) Tm toa o iem A1 oi xng vi iem A qua ng thang SC.

Cau IV: 1. Tnh tch phan 7

30

21

xI dxx

.

2. Tm he so cua x7 trong khai trien a thc 2(2 3 ) nx , trong o n la so nguyen dng thoa man: 1 3 5 2 12 1 2 1 2 1 2 1... nn n n nC C C C = 1024. ( knC la so to hp chap k cua n phan t)

Cau V: Cm rang vi moi x, y > 0 ta co : 29(1 )(1 )(1 ) 256yx

x y . ang thc xay ra khi nao?





22

S 20 Cau I: 1. Khao sat s bien thien va ve o th ( C ) cua ham so 4 26 5y x x 2. Tm m e phng trnh sau co 4 nghiem phan biet : 4 2 26 log 0x x m . Cau II:

1. Giai he phng trnh : 2 1 13 2 4

x y x yx y

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

x x x

Cau III:

1. Trong mat phang vi he toa o Oxy cho elip (E) : 2 2

64 9x y

= 1. Viet phng

trnh tiep tuyen d cua (E) biet d cat hai hai truc toa o Ox, Oy lan lt tai A, B sao cho AO = 2BO.

2. Trong khong gian vi he toa o Oxyz cho hai ng thang 1x y z: 1 1 2

d va

2

1 2:

1

x td y t

z t

( t la tham so )

a) Xet v tr tng oi cua d1 va d2 . b) Tm toa o cac iem M thuoc d1 va N thuoc d2 sao cho ng thang MN song song vi mat phang (P) : 0x y z va o dai oan MN = 2 . Cau IV:

1. Tnh tch phan 20

lne

x xdx .

2. Mot i van nghe co 15 ngi gom 10 nam va 5 n. Hoi co bao nhieu cach lap mot nhom ong ca gom 8 ngi biet rang trong nhom o phai co t nhat 3 n.

Cau V: Cho a, b, c la ba so dng thoa man : a + b + c = 34

. Chng minh rang :

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





23

S 21

Cau I: Cho ham so : y = 2 2 2

1x x

x

(*)

1. Khao sat s bien thien va ve o th ( C ) cua ham so (*) . 2. Goi I la giao iem cua hai tiem can cua ( C ). Chng minh rang khong co tiep tuyen nao cua (C ) i qua iem I . Cau II: 1. Giai bat phng trnh : 28 6 1 4 1 0x x x

2. Giai phng trnh : 2 2cos 2 1( ) 3

2 cosxtg x tg x

x

Cau III: 1. Trong mat phang vi he toa o Oxy cho 2 ng tron : (C1 ): x2 + y2 9 va (C2 ): x2 + y2 2 2 23 0x y . Viet phng trnh truc ang phng d cua 2 ng tron (C1) va (C2). Chng minh rang neu K thuoc d th khoang cach t K en tam cua (C1) nho hn khong cach t K en tam cua (C2 ). 2. Trong khong gian vi he toa o Oxyz cho iem M(5;2; - 3) va mat phang (P): 2 2 1 0x y z . a) Goi M1 la hnh chieu cua M len mat phang ( P ). Xac nh toa o iem M1

va tnh o dai oan MM1. b) Viet phng trnh mat phang ( Q ) i qua M va cha ng thang :

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

Cau IV:

1.Tnh tch phan 4

sin

0

(tan cos )xx e x dx

.

2. T cac ch so 1, 2, 3, 4, 5, 6, 7 co the lap c bao nhieu so t nhien, moi so gom 5 ch so khac nhau va nhat thiet phai co 2 ch 1, 5 ?

Cau V: Chng minh rang neu 0 1y x th

14

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





24

S 22 Cau I: Goi (Cm) la o th cua ham so y= x3+ ( 2m + 1) x2 m 1 (1) (m la tham so). 1) Khao sat s bien thien va ve o th cua ham so (1) khi m 1 . 2) Tm m e o th (Cm) tiep xuc vi ng thang y= 2mx m 1. Cau II: 1. Giai bat phng trnh : 2 7 5 3 2x x x

2. Giai phng trnh : 3 sin( ) 22 1 cos

xtg xx

Cau III: 1. Trong mat phang vi he toa o Oxy cho ng tron (C): x2 + y2 4 6 12 0x y . Tm toa o iem M thuoc ng thang d : 2 3 0x y sao cho MI = 2R, trong o I la tam va R la ban knh cua ng tron (C). 2. Trong khong gian vi he toa o Oxyz cho lang tru ng OAB.O1A1B1 vi A(2;0;0), B(0; 4; 0), O1(0; 0; 4) a) Tm toa o cac iem A1, B1. Viet phng trnh mat cau qua 4 iem O, A, B, O1. b) Goi M la trung iem cua AB.Mat phang ( P ) qua M vuong goc vi O1A va cat OA, OA1 lan lt tai N, K. Tnh o dai oan KN. Cau IV:

1. Tnh tch phan 3 2

1

lnln 1

e xI dxx x

.

2. Tm k 0;1;2;...;2005 sao cho 2005kC at gia tr ln nhat. ( knC la so to hp chap k cua n phan t)

Cau V: Tm m e he phng trnh sau co nghiem:

2 1 2 1

2

7 7 2005 2005( 2) 2 3 0

x x x xx m x m





25

S 23 Cau I:

1. Khao sat s bien thien va ve o th cua ham so 2 3 3

1x xy

x

.

2. Tm m e phng trnh 2 3 3

1x x m

x

co 4 nghiem phan biet

Cau II:

1. Giai bat phng trnh : 2

22

2 19 2 33

x xx x

.

2. Giai phng trnh :sin 2 cos 2 3sin cos 2 0x x x x Cau III: 1. Trong mat phang vi he toa o Oxy cho 2 iem A(0;5), B(2; 3) . Viet phng trnh ng tron i qua hai iem A, B va co ban 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 o cac iem con lai cua hnh lap phng ABCD.A1B1C1D1. Goi M la trung iem cua BC. Chng minh rang hai mat phang ( AB1D1) va ( AMB1) vuong goc nhau. b) Chng minh rang t so khong cach t iem N thuoc ng thang AC1 ( N A ) ti 2 mat phang ( AB1D1) va ( AMB1) khong phu thuoc vao v tr cua iem N. Cau IV:

1. Tnh tch phan

2

2

0

I = ( 2x - 1)cos xdx .

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

( Pn la so hoan v cua n phan t va knA la so chnh hp chap k cua n phan t)

Cau V: Cho x, y, z la ba so dng va xyz = 1. Chng minh rang :

2 2 2 3

1 1 1 2x y z

y z x

.





26

S 24 Cu I:

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

x 1

(C)

2. Da vo th (C), tm m phng trnh sau c hai nghim dng phn bit 2 22 5 ( 2 5)( 1)x x m m x Cu II:

1. Gii phng trnh: 3 3 2 3 2cos3xcos x sin 3x sin x8

2. Gii h phng trnh: 2

2

( 1) ( ) 4 ( , )

( 1)( 2)x y y x y

x y Rx 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 trnh mt phng (ABC'). 2. Vit phng trnh hnh chiu vung gc ca ng thng B'C' trn mf(ABC') Cu IV:

1. Tnh 6

2

dxI2x 1 4x 1

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

1. Trong mt phng Oxy cho elip (E) 2 2

112 2x y

. Vit phng trnh ca hypebol

(H) c hai dng tim cn l 2y x v c hai tiu im l hai tiu im ca (E).

2. p dng khai trin ca nh thc Newton ca 1002x x , chng minh rng:

99 100 198 199

0 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 trnh: 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 = 600 , 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 chp A.BDMN.





27

S 25 Cu I:

1. Kho st v v th (C) ca hm s 4

2xy 2(x 1)2

(C)

2. Vit phng trnh cc ng thng i qua im A(0; 2) v tip xc vi (C). Cu II:

1. Gii phng trnh: 2sin 2x 4s inx +1=06

2. Gii h phng trnh: 3 3

2 2

8 2 ( , )

3 3( 1)x x y y

x y Rx 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 trung im ca on thng AB. 1. Tm giao im 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 thng d: y = 2x + 1 2. Cho x, y, z tho mn cc iu kn 3 3 3 1x 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 trnh ng cao BH: x + y + 3 = 0 v trung 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 s khc nhau ? Tnh tng ca tt c cc s t nhin . Cu Vb: 1. Gii bt phng trnh: x 2x 2xlog 2 2log 4 log 8

2. Cho hnh chp S.ABCD c y 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. Trn

cnh SA ly im M sao cho AM = 33

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

Tnh th tch khi chp S.BCNM.





28

S 26 Cu I:

1. Kho st v v th (C) ca hm s 2x x 1yx 1

(C)

2. Vit phng trnh tip tuyn ca (C) i qua A(0; - 5). Cu II: 1. Gii phng trnh: 2 2 2(2sin x 1) tan 2x 3(2cos x -1) = 0 2. Gii phng trnh: 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:

11

: 12

x ty tz

v 23 1:

1 2 1x y z

1. Vit phng trnh mt phng cha 1 v song song 2 . 2. Xc nh to im A trn 1 v im B trn 2 sao cho on thng AB c di nh nht. Cu IV:

1. Tnh 10

5

dxIx 2 x 1

2. Tm gi tr nh nht ca hm s: 211 74 1 , 02

y x xx x

Cu Va: 1. Trong mt phng Oxy cho tam gic ABC cn ti B, vi A(1; -1), C(3; 5). im B thuc ng thng d: 2x - y = 0. Vit phng trnh cc ng 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 5 ch s khc nhau trong c ng hai ch s l v hai ch s l ng cnh nhau. Cu Vb: 1. Gii phng trnh: 31 82

2

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

2. Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh a. 060BAD . SA vung gc vi mf(ABCD), SA = a. Gi C' trung im ca SC. Mt phng (P) di qua 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'.





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 thi honh ca im cc tiu nh hn 1. Cu II: 1. Gii phng trnh: cos2x + (1 + 2cosx)(sinx - cosx) = 0


2 2

( )( ) 13 ( , )

( )( ) 25x y x y

x y Rx y x y

Cu III: Trong khng gian Oxyz cho mt phng (P): 2x + y - z + 5 = 0 v cc im A(0; 0; 4), B(2; 0; 0) 1. Vit phng trnh hnh chiu vung gc ca ng thng AB trn mf(P). 2. Vit phng trnh mt cu i qua O, A, B v tip xc vi mt phng (P). Cu IV:

1. Tnh e

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

2

3 4 24

x yx y

Cu Va: 1. Trong mt phng Oxy cho tam gic ABC c A(2; 1), ng cao qua nh B c phng trnh x - 3y - 7 = 0 v ng trung tuyn qua nh C c phng trnh 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 gic c nh l cc im cho. Tm n. Cu Vb:

1. Gii phng trnh: 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 gic u, cnh y AB = 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.





30

S 28

Cu I: Cho hm s 3

2 1133 3xy 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 trnh: 3 3 2cos x sin x 2sin x 1


2 2 2

3( ) ( , )

7( )x xy y x y

x y Rx xy y x y

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

ng thng 1 23 1 4 3: , :

1 2 3 1 1 2x y z x y zd d

1. Chng minh rng d1 v d2 cho nhau. 2. Vit phng trnh ng thng nm trn (P), ng thi ct c d1 v d2 . Cu IV:

1. Tnh 2

0

I (x 1) sin 2xdx

2. Gii phng trnh 14 2 2(2 1)sin(2 1) 2 0x 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 trnh ng trn (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 c 10 hc sinh, t 2 c 11 hc sinh v t 3 c 12 hc sinh sao cho trong mi t c t nht 2 hc sinh n. Hi c bao nhiu cch chia nh vy ? Cu Vb: 1. Gii phng trnh: 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 cao ca hnh chp. Khong cch t trung im I ca SH n mt bn (SBC) bng b. Tnh th tch khi chp S.ABCD.





31

S 29

Cu I: Cho hm s 31

xyx

(C)

1. Kho st v v th (C) ca hm s cho. 2. Cho im 0 0 0( ; )M x y thuc (C). Tip tuyn ca (C) ti 0 0 0( ; )M x y ct cc tim cn ca (C) ti cc im A, B. Chng minh 0 0 0( ; )M x y l trung im AB. Cu II: 1. Gii phng trnh: 3 24sin x 4sin x 3sin 2x 6cosx 0 2. Gii phng trnh: 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 trnh ng thng i qua O v vung gc vi mf(ABC). 2. Vit phng trnh mt phng (P) cha OA, sao cho khong cch t B n (P) bng khong cch t C n (P). Cu IV:

1. Tnh 2

1

I (x 2) ln xdx

2. Gii hphng trnh: 2 2ln(1 ) ln(1 )

( , ) 12 20 0x y x y

x y Rx xy y

Cu Va: 1. Trong mt phng Oxy lp phng trnh chnh tc ca elip (E) c di trc ln bng 4 2 , cc nh trn trc nh v cc tiu im ca (E) cng thuc mt ng trn. 2. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin chn c 5 ch s khc nhau v mi s lp nn u nh hn 2500 ? Cu Vb:

1. Gii phng trnh: 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 cnh

CC' sao cho 23

CK 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 .





32

S 30

Cu I: Cho hm s 2x 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 s n cc ng tim cn ca n l hng s. Cu II:

1. Gii phng trnh: 1 1sin 2x sin x 2cot g2x2sin x sin 2x

2. Tm m phng trnh: 2m x 2x 2 1 x(2 x) 0 (2) c nghim x 0;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 = 0 1. 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. Tnh 4

0

2x 1I dx1 2x 1

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

132x2xx1x2

1y2

Cu Va: 1. Trong mt phng Oxy cho ng trn (C) : x2 + y2 = 1. ng trn (C') tm I (2; 2) ct (C) ti cc im A, B sao cho AB 2 . Vit phng trnh ng thng AB. 2. C bao nhiu s t nhin chn ln hn 2007 m mi s gm 4 ch s khc nhau ? Cu Vb: 1. Gii bt phng trnh: 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 tnh khong cch d t im A ti mt phng (A1BM).





33

S 31

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 i qua gc ta O. Cu II: 1. Gii phng trnh: 22 cos x 2 3 sin x cosx 1 3(sin x 3 cosx)

2. Gii h phng trnh 4 3 2 2

3 2

x x y x y 1

x 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) v

ng thng (d) 6x 3y 2z 06x 3y 2z 24 0

1. Chng minh cc ng thng AB v OC cho nhau. 2. Vit phng trnh ng 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 trn trong khi quay (H) quanh trc 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 32 2 2

x 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 phng trnh 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 n im phn bit khc A, B, C, D. Tm n bit s tam gic c ba nh ly t n + 6 im cho l 439. Cu Vb:

1. Gii phng trnh 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 gic u cnh a. Tnh theo a khong cch t nh B n mp(SAC).





34

S 32 Cu I: Cho hm s y = 2x3 + 6x2 5 1. Kho st v v th hm s (C). 2. Vit phng trnh tip tuyn ca (C), bit tip tuyn i qua A(1, 13). Cu II:

1. Gii phng trnh: 2x3cos2

42xcos

42x5sin

2. Tm m phng trnh: 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 giao im I ca ng thng AB vi mt phng (P). 2. Tm im 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

2

y

2

x

c ng 2 nghim tha mn iu kin

x > 0, y > 0 Cu Va:

1. Tm x, y N tha mn h

66CA

22CA2x

3y

3y

2x

2. Cho ng trn (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 trnh 21x2log1xlog 3

23

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





35

S 33

Cu I: Cho hm s x2

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 ct trc Oy ti B m OBA vung cn. Cu II:

1. Gii phng trnh: gxcottgxxsinx2cos

xcosx2sin

2. Tm m phng trnh : 01xmx13x4 4 c ng 1 nghim Cu III: Trong khng 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 trnh mt phng (Q) cha A, M v ct cc trc Oy, Oz ti cc im tng ng B, C sao cho VOABC = 3. Cu IV: 1. Trong mt phng ta Oxy, tnh din tch hnh phng gii hn bi cc ng y = x2 v 2x2y .

2. Gii h phng trnh:

xy9y2y

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 trn (C): x2 + y2 2x + 4y + 2 = 0. Vit phng trnh ng trn (C') tm M(5, 1) bit (C') ct (C) ti cc im A, B sao cho 3AB . Cu Vb:

1. Gii phng trnh: 1xlog1

43logxlog23

x93

2. Trong mt phng (P) cho na ng trn ng knh AB = 2R v im C thuc na ng trn sao cho AC = R. Trn ng thng vung gc vi (P) ti A ly

im S sao cho o60SBC,SAB

. Gi H, K ln lt l hnh chiu ca A trn SB, SC. Chng minh AHK vung v tnh VSABC ?





36

S 34

Cu I: Cho hm s 1x21xy

(C)

1. Kho st v v th hm s. 2. Vit phng trnh tip tuyn vi (C), bit rng tip tuyn i qua giao im ca ng tim cn v trc Ox.

Cu II: 1. Gii phng trnh: 1xcos12

xsin22

2. Tm m phng trnh: m54x6x4x23x c ng 2 nghim

Cu III: Cho ng thng d: 11z

12y

23x

v mt phng (P):

02zyx 1. Tm giao im M ca d v (P). 2. Vit phng trnh ng thng nm trong (P) sao cho d v khong cch t 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: 23ba

baab

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. Tm B, C sao cho din tch ABC ln nht. Cu Vb:

1. Gii bt phng trnh: 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 minh MN l ng vung gc chung ca cc ng thng AA1 v BC1. Tnh 11BCMAV .





37

S 35

Cu I: Cho hm s 1x

xy

(C)

1. Kho st v v th hm s. 2. Vit phng trnh tip tuyn d ca (C) sao cho d v hai tim cn ca (C) ct nhau to thnh mt tam gic cn. Cu II: 1. Gii phng trnh: (1 tgx)(1 + sin2x) = 1 + tgx

2. Tm m h phng trnh :

1xyx

0myx2 c nghim duy nht

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

2z

33y

21x:d1

v 55z

4y

65x:d2

1. Vit phng trnh mt phng (Q) cha d1 v (Q) (P). 2. Tm cc im M d1, N d2 sao cho MN//(P) v cch (P) mt khong bng 2.

Cu IV:

1. Tnh

2

0

2 xdxcosxI

2. Gii phng trnh: 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 m mi 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 = 0

d2: (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 ln nht Cu Vb: 1. Gii phng trnh: 022.72.72 xx21x3 . 2. Cho lng tr ng ABCA1B1C1 c tt c cc cnh u bng a. M l trung im ca on AA1. Chng minh BM B1C v tnh d(BM, B1C).





38

S 36 PHN CHUNG CHO TT C CC TH SINH Cu I (2im) Cho hm s y = x3 + 3mx2 + (m + 1)x + 1 (1), m l tham s thc 1. Kho st s bin thin v v th ca hm s (1) khi m = - 1. 2. Tm cc gi tr ca m tip tuyn ca th hm s (1) ti im c honh x = -1 i qua A(1; 2). Cu II (2im). 1. Gii phng trnh: tanx cotx = 4cos22x

2. Gii phng trnh : 2(2x - 1)2x + 1 + 3 - 2x =

2

Cu III (2im). Trong khng gian Oxyz, cho hai ng thng:

1 25x - 6y - 6z + 13 = 0x - 3 y - 3 z - 3d : = = , d : x - 6y + 6z - 7 = 02 2 1

1. Chng minh d 1 v d 2 ct nhau. 2. Gi I l giao im ca d 1 v d 2 . Tm cc im A, B ln lt thuc d 1 , d 2 sao

cho tam gic IAB cn ti I v c din tch bng 4142 .

Cu IV (2im)

1. Tnh tch phn 3

312

2 2xdxIx

.

2. Gii phng trnh: sin x - 4e = tanx

PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im) 1. Cho tp hp E = {0, 1, 2, 3, 4, 5, 7}. Hi c bao nhiu s t nhin chn gm 4 ch s khc nhau c thnh lp t cc ch s ca E. 2. Trong mt phng Oxy, cho tam gic ABC vi ng cao k t nh B v ng phn gic trong gc A ln lt c phng trnh 3x + 4y + 10 = 0 v x - y + 1 = 0, im M(0; 2) thuc ng thng AB ng thi cch C mt khong bng

2 . Tm to cc nh ca tam gic ABC. Cu V.b. Theo chng trnh phn ban (2 im)

1. Gii bt phng trnh logarit 1 22

2x + 3log log 0x + 1

.

2. Cho hnh chp S.ABC c y ABC vung cn ti nh B, BA = BC = 2a, hnh chiu vung gc ca S trn mt phng y (ABC) l trung im E ca AB v SE = 2a. Gi I, J ln lt l trung im ca EC, SC; M l im di ng trn tia i ca tia BA sao cho 0ECM = ( < 90 ) v H l hnh chiu vung gc ca S trn MC. Tnh th tch ca khi t din EHIJ theo a, v tm th tch ln nht.





39

S 37 PHN CHUNG CHO TT C CC TH SINH Cu 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 trnh 2sin 2x - = sin x - + 4 4 2

.

2. Gii phng trnh 2 21 31

1 1x

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 trnh mt cu (S) i qua ba im A, B, C v c tm thuc mf(P). 2. Vit phng trnh 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 phn

2

0

sin2xdxI = 3 + 4sinx - cos2x .

2. Chng minh phng trnh 4x(4x2 + 1) = 1c ng ba nghim thc phn bit. PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)

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 trn (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 k c 2 tip tuyn vi (C) sao cho gc gia hai tip tuyn bng 60o. Cu V.b. Theo chng trnh phn ban (2 im)

1. Gii phng trnh 3

1 63 log 9log x

xx 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 i xng ca S qua E; I l giao im ca ng thng AD vi mf(SMN). Chng minh rng AD vung gc vi SI v tnh theo a th tch ca khi t din MBSI.





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 trnh 12sin x + - sin 2x - = 3 6 2

.

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

Trong khng gian Oxyz cho ng thng d 1 c phng trnh : x - 3 y z + 5 = =

2 9 1

v hai im A(5; 4; 3), B(6; 7; 2) 1. Vit phng trnh ng thng d 2 qua 2 im A, B. Chng minh rng hai ng thng d 1 v d 2 cho nhau. 2. Tm im C thuc d 1 sao cho tam gic ABC c din tch nh nht. Tnh gi tr nh nht . Cu IV (2im)

1. Tnh 2

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.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)

1. Cho s nguyn n tho mn ng thc 3 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), ng thng AB c phng trnh x + 2y - 3 = 0 v trng 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 trnh phn ban (2 im)

1. Gii phng trnh 2 12

2log 2 2 log 9 1 1x x

2. Cho hnh chp ABCD c y ABCD l hnh vung c cnh bng a, SA = a 3 v SA vung gc vi mt phng y. Tnh theo a th tch khi t din S.ACD v tnh cosin ca gc gia hai ng thng SB, AC.





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 bin thin v v th ca hm s (1) khi m = 1. 2. Tm cc gi tr m hm s (1) ng bin trn tng khong xc nh ca n. Cu II (2im)

1. Gii phng trnh 2 x3sinx + cos2x + sin2x = 4sinxcos2

2. Gii h phng trnh

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

4x yx y z

1. Tm to im D thuc ng thng d sao cho th tch khi t din ABCD bng 1. 2. Vit phng trnh tham s ca ng thng i qua trc tm H ca tam gic ABc v vung gc vi mf(ABC). Cu IV (2im)

1. Tnh tch phn 1 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.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)

1. Chng minh rng vi n l s nguyn dng

n 0 n - 1 1 0 n n + 1

n 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 ng trn ni tip tam gic OAB tip xc vi ng trn i qua trung im cc cnh ca tam gic OAB. Cu V.b. Theo chng trnh phn ban (2 im)

1. Gii phng trnh 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 din ABCD v tnh s o ca gc gia hai nng thng AD v BC.





42

S 40 PHN CHUNG CHO TT C CC TH SINH

Cu I (2im). Cho hm s 3 11

xyx

(1)

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

xyx

(1)

2. Tnh din tch ca tam gic to bi cc trc to v tip tuyn vi th hm s (1) ti im M(-2; 5). Cu II (2im) 1. Gii phng trnh 4 44(sin x + cos x) + cos4x + sin2x = 0 . 2. Gii phng trnh 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 = 0 v 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 trnh mt cu c tm thuc d v tip xc vi hai mt phng ( ) v Oxy. Cu IV (2im)

1. Tnh 1

2x

20

xI = xe - dx4 - x

2. Cho cc s thc x, y tha 0 ,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.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)

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

1. Gii bt phng trnh 2 22x - 4x - 2 2x - x - 12 - 1 6 .2 - 2 0 . 2. Cho t din ABCD v cc im M, N, P ln lt thuc cc cnh BC, BD, AC sao cho BC = 4BM, AC = 3AP, BD = 2BN. Mt phng (MNP) ct AD ti Q. Tnh t s AQAD v t s th tch hai phn ca khi t din ABCD c phn chia bi mt phng

(MNP).





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) cho ng bin trn khong (0; 2) Cu II. (2 im)

1. Gii phng trnh 2

2

tan t anx 2 sintan 1 2 4

x xx

2. Tm tt c cc gi tr ca tham s m phng trnh 24 2 4 1x 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 phn 1

2

0

( 1) .xI x x e dx

2. Gii h phng trnh

2 2

2 2

2 2

36 60 25 036 60 25 036 60 25 0

x y x yy z y zz 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 trnh KHNG phn ban (2 im) 1. C bao nhiu s t nhin gm 4 ch s khc nhau m mi s u ln hn 2500. 2. Trong mt phng Oxy, tm ta cc nh ca tam gic ABC bit rng ng thng AB, ng cao k t A v trung tuyn k t B ln lt c phng trnh l x + 4y - 2 = 0, 2x - 3y + 7 = 0, 2x + 3y - 9 = 0. Cu V.b. Theo chng trnh phn ban (2 im)

1. Gii phng trnh 5 1 2 5 1 3.2 .x x x . 2. Cho hnh chp S.ABC c y 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.





44

PHN TH HAI

HNG DN GII





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. Xc nh, th hm s ct trc honh ti 4 im phn bit. Bi ton quy v xc nh m phng trnh x4 - mx2 + m - 1 = 0 c 4 nghim phn bit. t2 - mt + m - 1 = 0 (t = x2) c 2 nghim t1, t2 dng phn bit.

22 4 1 2 0m m m

2 24( 1) 0 ( 2) 02

0 01

1 0 1

m m mm

S m S mm

P m P m

Cu II: 1. Bt phng trnh 2 11 1

2 2

log 4 4 log 2 3.2x x x

2 12 3.2 4 4 2.4 3.2 4 4 4 3.2 4 0

2 12

2 4

x x x x x x x x

x

xx

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]. phng trnh (1) c t nht mt nghim thuc

on 0;2

, iu kin cn v l phng 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)





46

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

f 1 13 3

T th suy ra, phng trnh 3t2 - 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). ABC u 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

aAO

2. Tnh tch phn 31

20 1xI dx

x

Cch 1. t u = x2 + 1 du = 2xdx 31

20 1xI 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 co s t co s t

24 4 4

40

0 0 0

sin td t tg t= tan t.d tan t - = + ln co stcost 2

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

.

Cu IV:

y

1

y = m+3 23

13

x

O 1

B H

C

O A

S

a

Hnh 13





47

1. To giao im ca (C1) v (C2) l nghim ca h:

2 2

2 22 2

7 10 010 010 04 2 20 0

x yx y xx y xx y x y

Rt y t phng trnh u, th vo phng trnh th 2 ta c. x2 + (7x - 10)2 - 10x = 0 x2 - 3x + 2 = 0 x = 1, x = 2. Vi x = 1 ta c y = -3 Vi x = 2 ta c y = 4 Vy 2 giao im ca (C1) v (C2) l A1 (1;-3) A2 (2;4).

Trung im A ca A1A2 c to A3 1;2 2

. Ta c 1 2 1;7A A

. ng thng

qua A vung gc vi A1A2 c phng trnh

3 11.3 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 yx 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 phng 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 tip tuyn ngoi.

Ta c 1 2 7;1I I

. Do d ng thng I1I2 c phng trnh:

1. (x + 2) + 7 (y - 1) = 0 hay: x + 7y - 5 = 0 Do tip tuyn chung ca (C1) v (C2) c phng trnh dng: x + 7y + d = 0 Khong cch t I1 n tip tuyn l





48

2 2

55 5 25 2 5 25 2 5 25 2

1 7

dd d d

Vy cc tip tuyn phi tm l: x + 7y - 5 + 25 2 = 0 x + 7y - 5 - 25 2 = 0 Cu V:

1. Gii phng trnh 24 4 2 12 2 16x x x x iu kin: x - 4 0 x 4. Vi iu kin x 4, phng trnh tng ng vi

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

24 4 4 4 12x x x x (1) t 4 4 0x x t t

Phng trnh (1) t2 - t - 12 = 0 4

3tt

t = 4 4 4 4x x (1)

22 2 16 16

4x x

x

2 16 8x x

2 2

4 816 64 16x

x x x

x = 5

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

81818.17.16.15.14.13.12.11 43758

8.7.6.5.4.3.2.C

Tng s cch trn c phn lm hai b phn ri nhau: B phn I gm cc cch chn t i tuyn ra 8 em sao cho mi khi u c em c 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 hn 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 em c chn t khi 12 hoc khi 11: C 813C cch chn 513C 8 em c chn t khi 12 hoc khi 10: C 812C cch chn 412C

(loi)





49

8 em c 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 ca b c

R R R R

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

= 2Sa b cbc ca ab

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

2 2 2

ax2

a b c a b cby czR 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 cza c b b a c c b a a b c

x y z

Suy ra: 2 2 2

2a b cx 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 b

c b a c b aa x b y c z

a b cx y z

ABC u M trng vi trng tm G ca tam gic ABC.

Nhn xt: 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:





50

2 2 2

1 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 Sa b c a b c a b c R

ab bc ca a b cR R





51

S 2 Cu I: 1. Tm n N* tho mn bt phng trnh 3 22 9nn nA C n

iu kin: n N*, n 3. Bt phng 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 nn n N

Kt hp iu kin n 3 suy ra n = 3 hoc n = 4.

2. Phng trnh 84 221 1log 3 log 1 log (4 )2 4

x x x

iu kin:01

xx

Phng 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, phng trnh (x + 3)( x -1) - 4x = 0

2 2 3 0

31

x xx

x

ii. Nu 0 < x < 1, phng trnh (x + 3)( 1 - x) - 4x = 0

2 6 3 0

3 2 20 1x x

xx

p s: phng trnh c 2 nghim 3

3 2 3

x

x

Cu II:

1.2 2

2 2x x m my x

x x

2

2 2

4 4' 1( 2) ( 2)

m x x myx x

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





52

' 0 [ 1;0y 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. Phng trnh.

2 2

2

1 1 1 1

2

1 1

9 ( 2)3 2 1 0

2 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 tX l on 3;9 .

2 2 1

(1) 23 9

X X aXX

T th v cu 2, hn ch trong on 3;9 suy ra, phng trnh c nghim,

iu kin cn v l: 6447

a

* Ch : Bn c th thy th v hn, nu tm a phng trnh sau c nghim: 2 2t + 1 - t t + 1 - t9 - (a + 2).3 + 2a + 1 = 0 Cu III:

1. Gii phng trnh 4 4s in os 1 1cot 25sin 2 2 8sin 2

x c x xx x

iu kin: sin2x 0 Vi iu phng trnh:

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

2 9os 2 5 os2 04

9os2 12 2 6 21 3cos 22

c x c x

c xx k

x

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

(loi)

6x k





53

4R2sinBsinC (sinBcosC + sinCcosB) = 20 4R2sinBsinCsinA = 20

M 3

28 sin Asin sin 2 sin Asin sin4 4abc RS 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 zb x zc x y

a b cb c ac a b

Suy ra, ABC l tam gic nhn, nn trc tm H nm trong tam gic. Gi AM, BK, CN l 3 ng cao ca tam gic ABC. Theo gi thit OA (OBC)

.BC OA

BC OMBC AM

Suy ra OMA = , OKB = ,ONC =

Ta c: cos = OMAM

.

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 zy z

OM = 2 2

yzy z

Trong tam gic vung AOM:

AM2 = OA2 + OM2 = x2 + 2 2

2 2

y zy z

B

C

O

A

N

A K

H

Hnh 15

M





54

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

22 2 2 2

x y y z x z x y y z z xAM AMy z y z

Vy, cos = 2 2 2 2 2 2

OM yzAM x y y z z x

Lp lun tng t ta cng c:

cos = 2 2 2 2 2 2

xzx y y z z x

cos = 2 2 2 2 2 2

xyx y y z z x

Suy ra, cos + cos + cos =

2 2 2 2 2 2 2

313

xy yz xz xy yz xzx 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 xxy yz xzx 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 phng

trnh:

1 3 21 1 1

x y z

Gi I l giao im ca (d) v (P), khi c to ca I l nghim ca h:

3 0

1 3 21 1 1

x y zx y z

Suy ra I(-2; -2; -3). A' l im i xng ca A qua (P) (d) v I l trung im ca AA'. Do A' c to (xo; yo; zo)

01 22

x , 03 2

2y

, 02 32

z

x0 = -3; y 0 = -1, zo = 4. Vy A' (-3; -1; -4). b) 'BA

= (2; -8; -16) = 2(1; -4; -8). BA' c phng trnh:





55

3 1 41 4 8

x y z

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

3 03 1 4 4;3;4

1 4 8

x y zx y z M

Ta c: MA + MB = MA' + MB A'B. Do MA + MB t gi tr nh nht khi MA' + MB t gi tr nh 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 giao im ca ng thng A'B vi mt phng (P). Cu V:

Ta c: I =

1 3 1 3 1 3 3

233 20 0 0

11 1

1 1

xn n nxx x

x

d ee dx e d eex e

= 1

1 32 1 11 3 02 2

0

12[ 1 1 ]1

2

x nxn

x

ee e

= 1 12 2 1 12 4 2 2 2 1

2 2

.





56

S 3

Cu I:

1. a) Kho st v v th hm s khi m = 12

.

(Hc sinh t gii). Gi (C) l th hm s. b) Vit phng 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 - 2 Theo gi thit tip tuyn th phi tm c h s gc k =4. Vy c:

2 22 4 6 0xx x x 1 1

2 2

22;313;6

x y

x y

Vy, c 2 tip tuyn tho mn iu kin bi.

Tip 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 < 56

nn:

1 1(0) 2 03 3

y m

8 1 5(2) 2 4 2 03 3 3

y m m

Li c: y' = x2 + 2mx - 2 y" = 2x + 2m > 0, x [0; 2]

Suy ra th hm s 3 21 12 22 3

y x mx x m lm trn on [0; 2]. Kt hp

vi y(0) < 0; y(2) < 0 suy ra y < 0, x [0; 2]. Do :

2 2 2

2 3

0 0 0

1 12 23 3

S y dx ydx x mx x m dx





57

4 3 2

23 12 012 3 xx xm x m x

4 8 2 4 104 43 3 3 3 3

mm m

Theo gi thit S = 4 m = 12

(tho mn iu kin 0 < m < 56

)

Ch : Khng cn dng tnh "lm" ca th hm s trn [0; 2] chng minh y < 0, x [0; 2] nh sau:

1 1(0) 2 03 3

y m

8 1 5(2) 2 4 2 03 3 3

y m m

Li c: y' = x2 + 2mx - 2 y" = 2x + 2m > 0, x [0; 2] Suy ra y' ng bin, lin tc trn [0; 2], vi tp gi tr [-2; 2 + 2m], nn i du t m sang dng trn [0; 2]. Do hm s cho nghch bin ri chuyn sang ng bin, lin tc trn [0; ]. ng thi vi g(0) < 0 v g(2) < 0, ta c pcm. Cu II: Gii h phng trnh:

4 2

4 3 (1)

log log 0 (2)

x y

x y

iu kin: 42

log 0 1log 0 1

x xy y

Vi iu kin phng trnh (2) log4x = log2y log2x = log2y2 x = y2 Phng trnh (1) y2 - 4y + 3 = 0 (do y 1) y = 1; y = 3 x = 1; x = 9 Vy, phng trnh c hai nghim (1;1) v (9;3)

2. Phng trnh tg4x + 1 = 2

4

2 sin 2 sin 3os

x xc x

iu kin: cos 0 sinx 1 Phng trnh sin4x + cos4x = (2 - sin22x) sin3x





58

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

2 22 sin 2 2 sin 2 .2sin 3x x x

sin3x = 12

(d thy tho mn iu kin)

3 2

653 26

x k

x k

218 35 218 3

kx

kx

(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 trung im ca CD nn ED = 2 2a AB D

l trung im ca AM AM = 2a ABM vung ti A ta c: SAH vung ti A ta c:

SH = 2

2 2 2 4 3 55 5a aSA AH a

a B C

E

D M

S

A

a

H

Hnh 16





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 phng trnh dng: 2 1 2 0x 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 0x y z

x y z

v do tha (1).

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

4 2 1 0

4 4 11 0x 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 trnh (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) v mt phng (Q) cha v vung gc vi (P). Mt phng (Q) i qua Mo (1; -3; 0) l mt 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 trnh (Q): 1(x - 1) + 4(y + 3) + 4z = 0 x + 4y + 4z + 11 = 0. Cch 3. Trn () chn im Mo no chng hn Mo (1; -3; 0). Gi H l hnh chiu ca M0 trn (P). Gi I l giao im ca () vi (P). Suy ra phng trnh ng thng IH chnh l phng trnh hnh chiu ca () ln mt phng (P):

1 3

4 5 6x y z





60

Cu IV:

1. Tm 3

0

1 11x

x ximx

3 3

0 00

1 1 1 1 1 11 lim limx xx

x x x ximx x x

Xt tng s hng:

00

11 lim21 1 1 1xx

x ximx x x

3

00 23 3

1 11 11 lim[1 1 1 ]

xx

xximx x x x

20 3 3

1 1131 1x

imx x x

3

0

1 1 1 112 3x

x ximx

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 R1 = 3 (C2) c tm I2 (3; -4); bn knh R2 = 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 R1 = R2 = 3 nn d1 // d2 // I1I2

Phng trnh ng thng I1I2 = 0 2 2 2 2 0

3 6x y x y

Phng 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 5

3 5 2 2 3 5 2 0

c x yc

c x y





61

Hnh 17

Do tnh i xng, d3 v d4 ct nhai ti trung im I ca on I1I2 c to (3/2;1).

Phng trnh d3, d4 c dng y + 1 = k 3 31 02 2

kx kx y

Khong cch t I1 ti d3, d4 bng 2

32 12 3

1

k

k

Gii ra ta c : 1

2

0 144 333

k y

y xk

Cu V:

Cho , 0

54

x y

x y

Tm min S vi S = 4 14x y

Cch 1: S = 5

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

min S = 5

1 14

454

x yx y

x y

114

x

y

Cch 2: S = 4 1 ( )5 4

f xx x

0 < x < 5

4

I1 I2 I

d1

d2

d3





62

f(x) = 22

4 4 05 4x x

22 5 4

1504

x xx

x

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

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

x y x yx yx y

(3)

Du "=" (3) khi

54

2 1 4 1. 2 . 5 1

4 4

x y xx x y y

x y yx y

(3) 25 5 4 1 4 1 5

2 4 4 4x y x y

Vy min S = 5.





63

S 4

Cu I:

1. Gii bt phng trnh: 12312 xxx iu kin: x 3.

Bt phng trnh 12312 xxx

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

2

2 9 2 12 3 2 1

12 . 3 4

9 52 0 13 4

x x x x

x x

x x x

Do iu kin x 3, suy ra 3 4x 2. Gii phng trnh

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

iu kin:

1cos0cos

02

cos

0cos

xx

xx

Ta c: sin sin

21 tan tan 12 cos cos

2

xxxx xx

xxx

xx

xx

xxxx

cos1

2coscos

2cos

2coscos

2sinsin

2coscos

Phng trnh 2 sintan cos coscos

xx x xx

cosx(1 - cosx) = 0. Do iu kin cosx 0 nn phng 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]





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, hm t cc i ti x = 0 Vi m = -1 th y"(0) = 6> 0, hm t 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 - 3x c cho trn hnh 18. 3. Tm k h bt phng trnh c nghim

)2(1)1(log31log

21

)1(031

32

22

3

xx

kxx

iu kin: (x - 1)3 > 0 x > 1 Khi x > 1, bt phng trnh (1) (x - 3)3 - 3k < k (1') Bt phng trnh (2) log2x + log2(x - 1) 1 (x > 1). x(x - 1) 2

1022

xxx

1< x 2 Bi ton quy v xc nh k bt phng trnh (1') c nghim tha iu kin 1< x 2. Da vo th hm s v cu 2, xt ch trn khong 1< x 2, ta suy ra tp mi tr s k cn tm l k > -5 (k >

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

Cu III: 1. Gi H l trung im BC. Do ABC vung cn ti A nn AH BC; AB = AC SB = SC SH BC. Do AHS = 600.

Hnh 18





65

Ta c AH = .22aBC

Trong SAH vung ti A nn:

SA = 0 3tan 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 qua M2(0; 1; - 2) v vung gc cc vc t 2u

= (a; 3; 0), 2v

= ( 1; 0; - 3). Suy ra mt vc

t ch phng ca d2 l 2a

= [ 2u

, 2v

] = (3; - a; 1).

1 1M M

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

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

Ta c 1 2[ ; ]a a

. 1 1M 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 trnh tham s ca cc ng thng d1 v d2:

1 2

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

3z = t 12 '

3

x tay t

z t

Xt h phng trnh:

a + at ' (1)

1 1 ' (2)3

12 ' (3)3

tat 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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