tuyen tap de dh 2002-2012 theo chu de booklet

Tuyn tp cc thi i hc

2002-2012

theo ch

Chng 7.Tch phn v ng dng 60

p s

7.1pi2

32+

1

4

7.2 12

(2 ln 3 3 ln 2)

7.3 I=

2

3+23

ln 2 + ln 3

7.4 343

+ 10 ln 35

7.5

3 + 2pi3

+ 12

ln 23

2+3

7.6 pi4

+ ln(2pi8

+22

)

7.7 (1)

7.8 (3 ln 3 2)

7.9 (e+ pi4 1)

7.10 I = (53e24

)

7.11 (5e4132

)

7.12 (32 ln 216

)

7.13 (ln(e2 + e+ 1) 2)

7.14 ( e22 1)

7.15 (12

ln 2)

7.16 (116135

)

7.17 (2 ln 2 1)

7.18 (ln 32)

7.19 (432

4)

7.20 (14(3 + ln 27

16))

7.21 (13

+ ln 32)

7.22 (14ln5

3)

7.23 (113 4 ln 2)

7.24 (3427

)

7.25 (23)

7.26 (12ln(2+

3) 10

93)

7.27 ( 815 pi

4)

7.28 (13

+ 12ln1+2e

3)

7.29 (2pi + 43)

7.30 (1096

)

7.31 ( e2 1)

7.32 (pi(5e32)27

)

Mc lc

1 Phng trnh-Bt PT-H PT-H BPT 31.1 Phng trnh v bt phng trnh . . . . . . . . . . . . . . . . . . 3

1.1.1 Phng trnh, bt phng trnh hu t v v t . . . . . . . 31.1.2 Phng trnh lng gic . . . . . . . . . . . . . . . . . . 41.1.3 Phng trnh,bt phng trnh m v logarit . . . . . . . . 8

1.2 H Phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Phng php hm s, bi ton cha tham s . . . . . . . . . . . . 12p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Bt ng thc 172.1 Bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Gi tr nh nht- Gi tr ln nht . . . . . . . . . . . . . . . . . . 182.3 Nhn dng tam gic . . . . . . . . . . . . . . . . . . . . . . . . . 20p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Hnh hc gii tch trong mt phng 223.1 ng thng . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 ng trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Cnic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4 T hp v s phc 304.1 Bi ton m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2 Cng thc t hp . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3 ng thc t hp khi khai trin . . . . . . . . . . . . . . . . . . . 314.4 H s trong khai trin nh thc . . . . . . . . . . . . . . . . . . . 324.5 S phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Kho st hm s 365.1 Tip tuyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Cc tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.3 Tng giao th . . . . . . . . . . . . . . . . . . . . . . . . . . 405.4 Bi ton khc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6 Hnh hc gii tch trong khng gian 446.1 ng thng v mt phng . . . . . . . . . . . . . . . . . . . . . 446.2 Mt cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Phng php ta trong khng gian . . . . . . . . . . . . . . . 51p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

7 Tch phn v ng dng 577.1 Tnh cc tch phn sau: . . . . . . . . . . . . . . . . . . . . . . . 577.2 Tnh din tch hnh phng c gii hn bi cc ng sau: . . . . 597.3 Tnh th tch khi trn xoay c to bi hnh phng (H) khi quay

quanh Ox. Bit (H) c gii hn bi cc ng sau: . . . . . . . 59p S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60


Bi 7.22 (A-03).

I =

235

dx

xx2 + 4

.

Bi 7.23 (A-04).

I =

21

x

1 +x 1 dx.

Bi 7.24 (A-05).

I =

pi2

0

sin 2x+ sinx1 + 3 cosx

dx.

Bi 7.25 (A-06).

I =

pi2

0

sin 2xcos2 x+ 4 sin2 x

dx.

Bi 7.26 (A-08).

I =

pi6

0

tan4 x

cos 2xdx.

Bi 7.27 (A-09).

I =

pi2

0

(cos3 x 1) cos2 x dx.

Bi 7.28 (A-10).

I =

10

x2 + ex + 2x2ex

1 + 2exdx.

7.2 Tnh din tch hnh phng c gii hn bi ccng sau:

Bi 7.29 (B-02). y =

4 x24 v y = x2

4

2.

Bi 7.30 (A-02). y = |x2 4x+ 3| v y = x+ 3.Bi 7.31 (A-07). y = (e+ 1)x v y = (1 + ex)x.

7.3 Tnh th tch khi trn xoay c to bi hnhphng (H) khi quay quanh Ox. Bit (H) cgii hn bi cc ng sau:

Bi 7.32 (B-07). y = x lnx, y = 0 , x = e.


Bi 7.5 (B-11).

I =

pi3

0

1 + x sinx

cos2 xdx

Bi 7.6 (A-11).

I =

pi4

0

x sinx+ (x+ 1) cosx

x sinx+ cosxdx

Bi 7.7 (D-03).

I =

20

|x2 x| dx.

Bi 7.8 (D-04).

I =

32

ln(x2 x) dx.

Bi 7.9 (D-05).

I =

pi2

0

(esinx + cosx) cosx dx.

Bi 7.10 (D-06).

I =

10

(x 2)e2x dx.

Bi 7.11 (D-07).

I =

e1

x3 ln2 x dx.

Bi 7.12 (D-08).

I =

21

lnx

x3dx.

Bi 7.13 (D-09).

I =

31

dx

ex 1 .

Bi 7.14 (D-10).

I =

e1

(2x 3x

) lnx dx.

Bi 7.15 (B-03).

I =

pi4

0

1 2 sin2 x1 + sin 2x

dx.

Bi 7.16 (B-04).

I =

e1

1 + 3 lnx lnx

xdx.

Bi 7.17 (B-05).

I =

pi2

0

sin 2x cosx

1 + cos xdx.

Bi 7.18 (B-06).

I =

ln 5ln 3

dx

ex + 2ex 3 .

Bi 7.19 (B-08).

I =

pi4

0

sin(x pi4

)dx

sin 2x+ 2(1 + sin x+ cosx).

Bi 7.20 (B-09).

I =

31

3 + ln x

(x+ 1)2dx.

Bi 7.21 (B-10).

I =

e1

lnx

x(2 + lnx)2dx.

Chng 1

Phng trnh-Bt PT-H PT-HBPT

1.1 Phng trnh v bt phng trnh . . . . . . . . . . . . . . 31.1.1 Phng trnh, bt phng trnh hu t v v t . . . . . 31.1.2 Phng trnh lng gic . . . . . . . . . . . . . . . . 41.1.3 Phng trnh,bt phng trnh m v logarit . . . . . . 8

1.2 H Phng trnh . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Phng php hm s, bi ton cha tham s . . . . . . . . 12p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.1 Phng trnh v bt phng trnh

1.1.1 Phng trnh, bt phng trnh hu t v v t

Bi 1.1 (B-12). Gii bt phng trnh

x+ 1 +x2 4x+ 1 3x.

Bi 1.2 (B-11). Gii phng trnh sau:

3

2 + x 62 x+ 4

4 x2 = 10 3x (x R)

Chng 1.Phng trnh-Bt PT-H PT-H BPT 4

Bi 1.3 (D-02). Gii bt phng trnh sau:

(x2 3x)

2x2 3x 2 0.Bi 1.4 (D-05). Gii phng trnh sau:

2

x+ 2 + 2

x+ 1x+ 1 = 4.

Bi 1.5 (D-06). Gii phng trnh sau:

2x 1 + x2 3x+ 1 = 0. (x R)Bi 1.6 (B-10). Gii phng trnh sau:

3x+ 16 x+ 3x2 14x 8 = 0.

Bi 1.7 (A-04). Gii bt phng trnh sau:2(x2 16)x 3 +

x 3 > 7 x

x 3 .

Bi 1.8 (A-05). Gii bt phng trnh sau:

5x 1x 1 > 2x 4.Bi 1.9 (A-09). Gii phng trnh sau:

2 3

3x 2 + 36 5x 8 = 0.Bi 1.10 (A-10). Gii bt phng trnh sau:

xx12(x2 x+ 1) 1.

1.1.2 Phng trnh lng gic

Bi 1.11 (D-12). Gii phng trnh sin 3x+ cos 3x sinx+ cosx =

2 cos 2x

Bi 1.12 (B-12). Gii phng trnh

2(cosx+

3 sinx) cosx = cosx

3 sinx+ 1.

Chng 7

Tch phn v ng dng

7.1 Tnh cc tch phn sau: . . . . . . . . . . . . . . . . . . . . 577.2 Tnh din tch hnh phng c gii hn bi cc ng sau: 597.3 Tnh th tch khi trn xoay c to bi hnh phng (H)

khi quay quanh Ox. Bit (H) c gii hn bi cc ngsau: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

p S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

7.1 Tnh cc tch phn sau:Bi 7.1 (D-12).

I =

pi4

0

x(1 + sin 2x)dx

Bi 7.2 (B-12).

I =

10

x3

x4 + 3x2 + 2dx.

Bi 7.3 (A-12).

I =

31

1 + ln(x+ 1)

x2dx

Bi 7.4 (D-11).

I =

40

4x 12x+ 1 + 2

dx

Chng 6.Hnh hc gii tch trong khng gian 56

6.39 H(3; 0; 2); r = 4

6.40 x2 + y2 + (z + 2)2 = 25

6.41 V =a3

24

2

d =a6

6.42 V(SABH) = 7a311

96

6.43 V (S,ABC) = 13a27

4a = a

37

12

d [BC, SA] = 32HI = 3

2a42

12= a

428

6.44 V = 2a3

3;h = 6a7

6.45 V = 3a32

;h = a3

2

6.46 V(SMNCB) =

3a3;h = a1213

6.47 a3

2; a2

2

6.48 33a3

50

6.49 a3

6.50 a7

7

6.51 4a39

; 2a5

5

6.52 a314

48

6.53 a6; 90o

6.54 a

2

6.5526a3 tan

6.56 a32

36

6.57 a2

4

6.58 a33

3;55

6.59 9a3208

6.60 3a33

8; 7a12

6.61 a210

16

6.62 120o

6.633a3

12

6.643a3

96

6.65 a32

; 14

6.66 315a3

5

6.67 53a3

24; 23a19


Bi 1.13 (A-12). Gii phng trnh sau:

3 sin 2x+ cos 2x = 2 cos x 1Bi 1.14 (D-11). Gii phng trnh sau:

sin 2x+ 2 cosx sinx 1tanx+

3

= 0.


sin 2x cosx+ sinx cosx = cos 2x+ sinx+ cosx

Bi 1.16 (A-11). Gii phng trnh

1 + sin 2x+ cos 2x

1 + cot2 x=

2 sinx sin 2x.

Bi 1.17 (D-02). Tm x thuc on [0; 14] nghim ng ca phng trnh:

cos 3x 4 cos 2x+ 3 cosx 4 = 0.Bi 1.18 (D-03). Gii phng trnh sau:

sin2 (x

2 pi

4) tan2 x cos2 x

2= 0.


(2 cosx 1)(2 sinx+ cosx) = sin 2x sinx.Bi 1.20 (D-05). Gii phng trnh sau:

cos4 x+ sin4 x+ cos (x pi4

) sin (3x pi4

) 32

= 0.


cos 3x+ cos 2x cosx 1 = 0.Bi 1.22 (D-07). Gii phng trnh sau:

(sinx

2+ cos

x

2)2

+

3 cosx = 2.



2 sinx(1 + cos 2x) + sin 2x = 1 + 2 cosx.


3 cos 5x 2 sin 3x cos 2x sinx = 0.Bi 1.25 (D-10). Gii phng trnh sau:

sin 2x cos 2x+ 3 sinx cosx 1 = 0.Bi 1.26 (B-02). Gii phng trnh sau:

sin2 3x cos2 4x = sin2 5x cos2 6x.Bi 1.27 (B-03). Gii phng trnh sau:

cotx tanx+ 4 sin 2x = 2sin 2x

.


5 sinx 2 = 3(1 sinx) tan2 x.Bi 1.29 (B-05). Gii phng trnh sau:

1 + sin x+ cosx+ sin 2x+ cos 2x = 0.


cotx+ sinx(1 + tan x tanx

2) = 4.


2 sin2 2x+ sin 7x 1 = sin x.Bi 1.32 (B-08). Gii phng trnh sau:

sin3 x

3 cos3 x = sinx cos2 x

3 sin2 x cosx.


sinx+ cosx sin 2x+

3 cos 3x = 2(cos 4x+ sin3 x).


6.6 M(0; 1; 3) hayM(67; 47; 127

)

6.7 m = 12

6.8 k = 1

6.9 aba2+b2

; a = b = 2

6.10 15x+11y17z10 = 0;SOAB =5

6.11 A(1;4; 1); x11

= y23 =z35

6.12 x2

= y21 =z21

;M(1; 0; 4)

6.13 D(52; 12;1); x+3

1= y12 =

z11

6.14 xz22 = 0;M(4; 1; 1),M(7; 4; 4)

6.15 5

6.16 x+43

= y+22

= z41

6.17 x+ 3y + 5z 13 = 0M(0; 1;1), N(0; 1; 1)

6.18 x+ 2y 4z + 6 = 0;M(2; 3;7)

6.19 4x+2y+7z15 = 0; 2x+3z5 =0x+326

= y11

= z12

6.20 b = c = 12;M(1; 0; 0),M(2; 0; 0)

6.21 2x z = 0;H(2; 3; 4)

6.22 a2b4

; ab

= 1

6.23 30o, 26

3;

2

6.24 I(3; 5; 7)orI(3;7; 1)x = t, y = 1, z = 4 + t

6.25 122

2x y+ z 1 = 0; x 2y z+ 1 = 0

6.26 x27

= y1

= z+14

6.27 H(3; 1; 4);x 4y + z 3 = 0

6.28 M(0; 1;3)orM(1835

; 5335

; 335

)

6.29 16

6.30 (S) : (x2)2 + (y1)2 + (z3)2 =25.

6.31 (x+ 1)2 + (y + 1)2 + (z 2)2 =17

6.32 x2 + y2 + (z 3)2 = 83

6.33 (x5)2+(y11)2+(z2)2 = 1(x+ 1)2 + (y + 1)2 + (z + 1)2 = 1

6.34 x y + z = 0;x y z = 0

6.35 (x 1)2 + y2 + (z 1)2 = 1

6.36 x2 + y2 + z2 3x 3y 3z = 0H(2; 2; 2)

6.37 x2 + (y + 2)2 + z2 = 57625

;172

6.38 y 2z = 0;M(1;1;3)


Bi 6.61 (A-02). Cho hnh chp tam gic u S.ABC nh S, c di cnh ybng a. Gi M, N ln lt l trung im ca cc cnh SB v SC. Tnh theo a dintch tam gic AMN, bit rng mt phng (AMN) vung gc vi mt phng (SBC).

Bi 6.62 (A-03). Cho hnh lp phng ABCD.ABCD. Tnh s o gc phngnh din [B,AC,D].

Bi 6.63 (A-06). Cho hnh tr c cc y l hai hnh trn tm O v O, bn knhy bng chiu cao v bng a. Trn ng trn tm O ly im A, trn ng trntm O ly im B sao cho AB=2a. Tnh th tch khi t din OOAB.

Bi 6.64 (A-07). Cho hnh chp S.ABCD c y l hnh vung cnh a, mt bnSAD l tam gic u v nm trong mt phng vung gc vi y. Gi M, N, P lnlt l trung im ca cc cnh SB, BC, CD. Chng minh rng AM vung gcvi BP v tnh th tch ca khi t din CMNP.

Bi 6.65 (A-08). Cho lng tr ABC.ABC c di cnh bn bng 2a, y ABCl tam gic vung ti A, AB=a, AC=a

3 v hnh chiu vung gc ca nh A

trn mt phng (ABC) l trung im ca cnh BC. Tnh theo a th tch khi chpAABC v tnh cosin ca gc gia hai ng thng AA v BC.

Bi 6.66 (A-09). Cho hnh chp S.ABCD c y ABCD l hnh thang vung tiA v D; AB=AD=2a, CD=a; gc gia hai mt phng (SBC) v (ABCD) bng 60o.Gi I l trung im ca cnh AD. Bit hai mt phng (SBI) v (SCI) cng vunggc vi mt phng (ABCD), tnh th tch khi chp S.ABCD theo a.

Bi 6.67 (A-10). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a.Gi M, N ln lt l trung im ca cc cnh AB v AD; H l giao im ca CNvi DM. Bit SH vung gc vi mt phng (ABCD) v SH=a

3. Tnh th tch

khi chp S.CDNM v tnh khong cch gia hai ng thng DM v SC theo a.

p s

6.1 M(1;1; 0) : M(73;5

3; 23).

6.2 6x+ 3y 4z + 12 = 0

6.3 x+12

= y+43

= z2

6.4 : x12

= y22

= z33

6.5 1. M = (4;7; 6); (5; 9;11)2. M(2; 1;5);M(14;35; 19)



(sin 2x+ cos 2x) cosx+ 2 cos 2x sinx = 0.Bi 1.35 (A-02). Tm ngim thuc khong (0; 2pi) ca phng trnh:

5

(sinx+

cos 3x+ sin 3x

1 + 2 sin 2x

)= cos 2x+ 3.


cotx 1 = cos 2x1 + tan x

+ sin2 x 12

sin 2x.


cos2 3x cos 2x cos2 x = 0.Bi 1.38 (A-06). Gii phng trnh sau:

2(cos6 x+ sin6 x) sinx cosx2 2 sinx = 0.


(1 + sin2 x) cosx+ (1 + cos2 x) sinx = 1 + sin 2x.


1

sinx+

1

sin (x 3pi2

)= 4 sin (

7pi

4 x).


(1 2 sinx) cosx(1 + 2 sinx)(1 sinx) =

3.


(1 + sin x+ cos 2x) sin (x+pi

4)

1 + tan x=

12

cosx.


1.1.3 Phng trnh,bt phng trnh m v logarit


log2(8 x2) + log 12(

1 + x+

1 x) 2 = 0 (x R)


2x2x 22+xx2 = 3.


2x2+x 4.2x2x 22x + 4 = 0.


log2 (4x + 15.2x + 27) + 2 log2 (

1

4.2x 3) = 0.

Bi 1.47 (D-08). Gii bt phng trnh sau:

log 12

x2 3x+ 2x

0.


42x+x+2 + 2x

3

= 42+x+2 + 2x

3+4x4 (x R)

Bi 1.49 (B-02). Gii bt phng trnh sau:

logx (log3 (9x 72)) 1.

Bi 1.50 (B-05). Chng minh rng vi mi x R, ta c:

(12

5)x

+ (15

4)x

+ (20

3)x

3x + 4x + 5x.

Khi no ng thc sy ra?


log5 (4x + 144) 4 log2 5 < 1 + log5 (2x2 + 1).


Bi 6.53 (B-02). Cho hnh lp phng ABCD.A1B1C1D1 c cnh bng a.a) Tnh theo a khong cch gia hai ng thng A1B v B1D.b) Gi M, N, P ln lt l trung im cc cnh BB1, CD, A1D1. Tnh gc gia haing thng MP v C1N.

Bi 6.54 (B-03). Cho hnh lng tr ng ABCD.ABCD c y ABCD l mthnh thoi cnh a, gc BAD = 60o. Gi M l trung im cnh AA v N l trungim cnh CC. Chng minh rng bn im B, M, D, N cng thuc mt phng.Hy tnh di cnh AA theo a t gic BMDN l hnh vung.

Bi 6.55 (B-04). Cho hnh chp t gic u S.ABCD c cnh y bng a, gcgia cnh bn v mt y bng (0o < < 90o). Tnh tan ca gc gia hai mtphng (SAB) v (ABCD) theo . Tnh th tch khi chp S.ABCD theo a v .

Bi 6.56 (B-06). Cho hnh chp S.ABCD c y ABCD l hnh ch nht viAB=a, AD=a

2, SA=a v SA vung gc vi mt phng (ABCD). Gi M, N ln

lt l trung im ca AD v SC, I l giao im ca BM v AC. Chng minh rngmt phng (SAC) vung gc vi mt phng (SMB). Tnh th tch khi t dinANIB.

Bi 6.57 (B-07). Cho hnh chp t gic u S.ABCD c y l hnh vung cnha. Gi E l im i xng ca D qua trung im ca SA, M l trung im caAE, N l trung im ca BC. Chng minh MN vung gc vi BD v tnh (theo a)khong cch gia hai ng thng MN v AC.

Bi 6.58 (B-08). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh 2a,SA=a, SB=a

3 v mt phng (SAB) vung gc vi mt phng y. Gi M, N ln

lt l trung im cc cnh AB, BC. Tnh theo a th tch ca khi chp S.BMDNv tnh cosin gc gia hai ng thng SM, DN.

Bi 6.59 (B-09). Cho hnh lng tr tam gic ABC.ABC c BB=a, gc giang thng BB v mt phng (ABC) bng 60o, tam gic ABC vung ti C vBAC = 60o. Hnh chiu vung gc ca im B ln mt phng (ABC) trng vitrng tm ca tam gic ABC. Tnh th tch khi t din AABC theo a.

Bi 6.60 (B-10). Cho hnh lng tr tam gic u ABC.ABC c AB=a, gc giahai mt phng (ABC) v (ABC) bng 60o. Gi G l trng tm tam gic ABC.Tnh th tch khi lng tr cho v tnh bn knh mt cu ngoi tip t dinGABC theo a.


Bi 6.45 (B-11). Cho lng trABCD.A1B1C1D1 c y ABCD l hnh ch nht.AB = a, AD = a

3. Hnh chiu vung gc ca im A1 trn mt phng (ABCD)

trng vi giao im AC v BD. Gc gia hai mt phng (ADD1A1) v (ABCD)bng 60o. Tnh th tch khi lng tr cho v khong cch t im B1 n mtphng (A1BD) theo a.

Bi 6.46 (A-11). Cho hnh chp S. ABC c y ABC l tam gic vung cn tiB, AB=BC=2a; hai mt phng (SAB) v (SAC) cng vung gc vi mt phng(ABC). Gi M l trung im ca AB; mt phng qua SM v song song vi BC,ct AC ti N. Bit gc gia hai mt phng (SBC) v (ABC) bng 60o. Tnh th tchkhi chp S. BCNM v khong cch gia hai ng thng AB v SN theo a.

Bi 6.47 (D-03). Cho hai mt phng (P) v (Q) vung gc vi nhau, c giao tuynl ng thng . Trn ly hai in A, B vi AB=a. Trong mt phng (P) lyin C, trong mt phng (Q) ly im D sao cho AC, BD cng vung gc vi v AC=BD=AB. Tnh bn knh mt cu ngoi tip t din ABCD v tnh khongcch t A n mt phng (BCD) theo a.

Bi 6.48 (D-06). Cho hnh chp tam gic S.ABC c y ABC l tam gic ucnh a, SA=2a v SA vung gc vi mt phng (ABC). Gi M, N ln lt l hnhchiu vung gc ca A trn cc ng thng SB v SC. Tnh th tch khi chpA.BCMN.

Bi 6.49 (D-07). Cho hnh chp S.ABCD c y l hnh thang, ABC = BAD =90o, BA=BC=a, AD=2a. Cnh bn SA vung gc vi dy v SA=a

2. Gi H l

hnh chiu vung gc ca A ln SB. Chng minh tam gic SCD vung v tnh(theo a) khong cch t H n mt phng (SCD).

Bi 6.50 (D-08). Cho lng tr ng ABC.ABC c y ABC l tam gic vung,AB=BC=a, cnh bn AA=a

2. Gi M l trung im cnh BC. Tnh theo a th

tch khi lng tr ABC.ABC v khong cch gia hai ng thng AM, BC.

Bi 6.51 (D-09). Cho lng tr ng ABC.ABC c y ABC l tam gic vungti B, AB=a, AA=2a, AC=3a. Gi M l trung im ca on thng AC, I l giaoim ca AM v AC. Tnh theo a th tch khi t din IABC v khong cch tim A n mt phng (IBC).

Bi 6.52 (D-10). Cho hnh chp S.ABCD c dy ABCD l hnh vung cnh a,cnh bn SA=a, hnh chiu vung gc ca nh S trn mt phng (ABCD) l im

H thuc on AC, AH=AC4. Gi CM l ng cao ca tam gic SAC. Chng

minh M l trung im ca SA v tnh th tch khi t din SMBC theo a.



(

2 1)x + (

2 + 1)x 2

2 = 0.


log0,7 (log6 (x2 + x

x+ 4)) < 0.


3.8x + 4.12x 18x 2.27x = 0.

Bi 1.55 (A-07). Gii bt phng trnh sau:

2 log3 (4x 3) + log 13

(2x+ 3) 2.


log2x1 (2x2 + x 1) + logx+1 (2x 1)2 = 4.

1.2 H Phng trnh

Bi 1.57 (D-12). Gii h phng trnh{xy + x 2 = 02x3 x2y + x2 + y2 2xy y = 0 ; (x; y R)

Bi 1.58 (A-12). Gii h phng trnh{x3 3x2 9x+ 22 = y3 + 3y2 9yx2 + y2 x+ y = 1

2

(x, y R).

Bi 1.59 (A-11). Gii h phng trnh:{5x2y 4xy2 + 3y3 2(x+ y) = 0xy(x2 + y2) + 2 = (x+ y)2

(x, y R)


Bi 1.60 (D-02). Gii h phng trnh sau:23x = 5y2 4y

4x + 2x+1

2x + 2= y.

Bi 1.61 (D-08). Gii h phng trnh sau:{xy + x+ y = x2 2y2x

2y yx 1 = 2x 2y (x, y R).

Bi 1.62 (D-09). Gii h phng trnh sau:{x(x+ y + 1) 3 = 0(x+ y)2 5

x2+ 1 = 0

(x, y R).

Bi 1.63 (D-10). Gii h phng trnh sau:{x2 4x+ y + 2 = 02 log2 (x 2) log2 y = 0 (x, y R).

Bi 1.64 (B-02). Gii h phng trnh sau:{3x y = x y

x+ y =x+ y + 2.

Bi 1.65 (B-03). Gii h phng trnh sau:3y =

y2 + 2

x2

3x =x2 + 2

y2.

Bi 1.66 (B-05). Gii h phng trnh sau:{ x 1 +2 y = 1

3 log9 (9x2) log3 y3 = 3.

Bi 1.67 (B-08). Gii h phng trnh sau:{x4 + 2x3y + x2y2 = 2x+ 9x2 + 2xy = 6x+ 6

(x, y R).


Bi 6.38 (B-07). Trong khng gian vi h ta Oxyz, cho mt cu(S): x2 +y2 +z22x+4y+2z3 = 0 v mt phng (P) : 2xy+2z14 = 0.1. Vit phng trnh mt phng (Q) cha trc Ox v ct (S) theo mt ng trnc bn knh bng 3.2. Tm ta im M thuc mt cu (S) sao cho khong cch t M n mt phng(P) ln nht.

Bi 6.39 (A-09). Trong khng gian vi h ta Oxyz, cho mt phng(P):2x 2y z 4 = 0 v mt cu (S):x2 + y2 + z2 2x 4y 6z 11 = 0.Chng minh rng mt phng (P) ct mt cu (S) theo mt ng trn. Xc ta nh tm v bn knh ca ng trn .

Bi 6.40 (A-10). Trong khng gian ta Oxyz, cho im A(0;0;2) v ngthng

:x+ 2

2=y 2

3=z + 3

2. Tnh khong cch t A n . Vit phng trnh

mt cu tm A, ct ti hai im B v C sao cho BC=8.

6.3 Phng php ta trong khng gian

Bi 6.41 (D-12). Cho hnh hp ng ABCD.ABCD c y l hnh vung, tamgic AAC vung cn, AC = a. Tnh th tch khi t din ABBC v khong ccht im A n mt phng (BCD) theo a.

Bi 6.42 (B-12). Cho hnh chp tam gic u S.ABC vi SA = 2a, AB = a. Gi Hl hnh chiu vung gc ca A trn cnh SC. Chng minh SC vung gc vi mtphng (ABH). Tnh th tch ca khi chp S.ABH theo a.

Bi 6.43 (A-12). Cho hnh chp S.ABC c y l tam gic u cnh a. Hnh chiuvung gc ca S trn mt phng (ABC) l im H thuc cnh AB sao cho HA=2HB. Gc gia ng thng SC v mt phng (ABC) bng 60o. Tnh th tchca khi chp S.ABC v tnh khong cch gia hai ng thng SA v BC theo a.

Bi 6.44 (D-11). Cho hnh chp S.ABC c y ABC l tam gic vung ti B,BA = 3a, BC = 4a; mt phng (SBC) vung gc vi mt phng (ABC). Bit SB=2a

3v SBC = 30o. Tnh th tch khi chp S.ABC v khong cch t im Bn mt phng (SAC) theo a.


6.2 Mt cu

Bi 6.30 (D-12). Trong khng gian vi h ta Oxyz, cho mt phng (P): 2x+y2z + 10 = 0 v im I (2; 1; 3). Vit phng trnh mt cu tm I ct (P) theomt ng trn c bn knh bng 4.

Bi 6.31 (B-12). Trong khng gian vi h ta Oxyz, cho ng thng d :x 1

2=

y

1=

z

2 v hai im A(2;1;0), B(-2;3;2). Vit phng trnh mt cu iqua A,B v c tm thuc ng thng d.

Bi 6.32 (A-12). Trong khng gian vi h ta Oxyz, cho ng thng d:x+ 1

1=y

2=z 2

1v im I (0; 0; 3). Vit phng trnh mt cu (S) c tm I

v ct d ti hai im A, B sao cho tam gic IAB vung ti I.

Bi 6.33 (D-11). Trong khng gian vi h ta Oxyz, cho ng thng :x 1

2=y 3

4=z

1v mt phng (P): 2x y + 2z = 0. Vit phng trnh mt

cu c tm thuc ng thng , bn knh bng 1 v tip xc vi mt phng (P).

Bi 6.34 (A-11). Trong khng gian vi h ta Oxyz, cho mt cu(S) : x2 + y2 + z2 4x 4y 4z = 0 v im A(4; 4; 0). Vit phng trnh mtphng (OAB), bit im B thuc (S) v tam gic OAB u.

Bi 6.35 (D-04). Trong khng gian vi h ta Oxyz cho ba im A(2;0;1),B(1;0;0), C(1;1;1) v mt phng (P) : x + y + z 2 = 0. Vit phng trnh mtcu i qua ba im A, B, C, v c tm thuc mt phng (P).

Bi 6.36 (D-08). Trong khng gian vi h ta Oxyz, cho bn im A(3;3;0),B(3;0;3), C(0;3;3), D(3;3;3).1. Vit phng trnh mt cu i qua bn im A, B, C, D.2. Tm ta tm ng trn ngoi tip tam gic ABC.

Bi 6.37 (B-05). Trong khng gian vi h ta Oxyz, cho hnh lng tr ngABC.A1B1C1 vi A(0;3;0), B(4;0;0), C(0;3;0), B1(4;0;4).1. Tm ta cc nh A1, C1. Vit phng trnh mt cu c tm l A v tip xcvi mt phng (BCC1B1).2. Gi M l trung im ca A1B1. Vit phng trnh mt phng (P) i qua haiim A, M v song song vi BC1. Mt phng (P) ct ng thng A1C1 ti imN. Tnh di on MN.


Bi 1.68 (B-09). Gii h phng trnh sau:{xy + x+ 1 = 7yx2y2 + xy + 1 = 13y2

(x, y R).

Bi 1.69 (B-10). Gii h phng trnh sau:{log2 (3y 1) = x4x + 2x = 3y2.

Bi 1.70 (A-03). Gii h phng trnh sau: x1

x= y 1

y2y = x3 + 1.

Bi 1.71 (A-04). Gii h phng trnh sau: log 14 (y x) log41

y= 1

x2 + y2 = 25.

Bi 1.72 (A-06). Gii h phng trnh sau:{x+ y xy = 3x+ 1 +

y + 1 = 4.

Bi 1.73 (A-08). Gii h phng trnh sau:x2 + y + x3y + xy2 + xy = 5

4

x4 + y2 + xy(1 + 2x) = 54.

Bi 1.74 (A-09). Gii h phng trnh sau:{log2 (x

2 + y2) = 1 + log2 (xy)

3x2xy+y2 = 81.

Bi 1.75 (A-10). Gii h phng trnh sau:{(4x2 + 1)x+ (y 3)5 2y = 04x2 + y2 + 2

3 4x = 7.


1.3 Phng php hm s, bi ton cha tham s

Bi 1.76 (D-11). Tm m h phng trnh sau c nghim{2x3 (y + 2)x2 + xy = mx2 + x y = 1 2m (x, y R)

Bi 1.77 (D-04). Tmm h phng trnh sau c nghim:{ x+y = 1

xx+ y

y = 1 3m.

Bi 1.78 (D-04). Chng minh rng phng trnh sau c ng mt nghim:

x5 x2 2x 1 = 0.

Bi 1.79 (D-06). Chng minh rng vi mi a > 0, h phng trnh sau c nghimduy nht: {

ex ey = ln (1 + x) ln (1 + y)y x = a.

Bi 1.80 (D-07). Tm gi tr ca tham sm phng trnh sau c nghim thc:x+

1

x+ y +

1

y= 5

x3 +1

x3+ y3 +

1

y3= 15m 10.

Bi 1.81 (B-04). Xc nhm phng trnh sau c nghim

m(

1 + x2

1 x2)

= 2

1 x4 +

1 + x2

1 x2.

Bi 1.82 (B-06). Tmm phng trnh sau c hai nghim thc phn bit:x2 +mx+ 2 = 2x+ 1.

Bi 1.83 (B-07). Chng minh rng vi mi gi tr dng ca tham s m, phngtrnh sau c hai nghim thc phn bit:

x2 + 2x 8 =m(x 2).


Bi 6.25 (A-06). Trong khng gian vi h ta Oxyz, cho hnh lp phngABCD.ABCD vi A(0;0;0), B(1;0;0), D(0;1;0), A(0;0;1). Gi M v N ln ltl trung im ca AB v CD.1. Tnh khong cch gia hai ng thng AC v MN.2. Vit phng trnh mt phng cha AC v to vi mt phng Oxy mt gc

bit cos =16.

Bi 6.26 (A-07). Trong khng gian vi h ta Oxyz, cho hai ng thng

d1 :x

2=y 11 =

z + 2

1v d2 :

x = 1 + 2ty = 1 + tz = 3.

1. Chng minh rng d1 v d2 cho nhau.2. Vit phng trnh ng thng d vung gc vi mt phng (P): 7x+y4z = 0v ct hai ng thng d1, d2.

Bi 6.27 (A-08). Trong khng gian vi h ta Oxyz, cho im A(2;5;3) vng thng d:

x 12

=y

1=z 2

2.

1. Tm ta hnh chiu vung gc ca im A trn ng thng d.2. Vit phng trnh mt phng () cha d sao cho khong cch t A n () lnnht.

Bi 6.28 (A-09). Trong khng gian vi h ta Oxyz, chomt phng (P):x2y+2z1 = 0 v hai ng thng 1 : x+ 1

1=y

1=z + 9

6,

2 :x 1

2=y 3

1=z + 1

2 . Xc nh ta im M thuc ng thng 1sao cho khong cch t M n ng thng 2 v khong cch t M n mtphng (P) bng nhau.

Bi 6.29 (A-10). Trong khng gian ta Oxyz, cho ng thng :x 1

2=

y

1=z + 2

1 v mt phng (P):x 2y + z = 0. Gi C l giao im ca vi (P),M l im thuc . Tnh khong cch t M n (P), bit MC=

6.


Bi 6.20 (B-10).1. Trong khng gian vi h ta Oxyz, cho cc im A(1;0;0), B(0;b;0),C(0;0;c), trong b, c dng v mt phng (P):y z + 1 = 0. Xc nh b vc, bit mt phng (ABC) vung gc vi mt phng (P) v khong cch t im O

n mt phng (ABC) bng1

3.

2. Trong khng gian vi h ta Oxyz, cho ng thng :x

2=y 1

1=z

2.

Xc nh ta im M trn trc honh sao cho khong cch t M n bngOM.

Bi 6.21 (A-02). Trong khng gian vi h ta Oxyz cho hai ng thng:

1 :

{x 2y + z 4 = 0x+ 2y 2z + 4 = 0 v 2 :

x = 1 + ty = 2 + tz = 1 + 2t

.

1. Vit phng trnh mt phng (P) cha ng thng 1 v song song vi ngthng 2.2. Cho im M(2;1;4). Tm ta im H thuc 2 sao cho on MH c dinh nht.

Bi 6.22 (A-03). Trong khng gian vi h ta Oxyz, cho hnh hp ch nhtABCD.ABCD c A trng vi gc ca h trc ta , B(a;0;0), D(0;a;0),A(0;0;b) (a> 0,b> 0). Gi M l trung im cnh CC.1. Tnh th tch khi t din BDAM theo a v b.2. Xc nh t s

a

b hai mt phng (ABD) v (MBD) vung gc vi nhau.

Bi 6.23 (A-04). Trong khng gian vi h ta Oxyz cho hnh chp S.ABCDc y ABCD l hnh thoi, AC ct BD ti gc ta O. Bit A(2;0;0), B(0;1;0),S(0;0;2

2). Gi M l trung im ca cnh SC.

1. Tnh gc v khong cch gia hai ng thng SA, BM.2. Gi s mt phng (ABM) ct ng thng SD ti im N. Tnh th tch khichp S.ABMN.

Bi 6.24 (A-05). Trong khng gian vi h trc ta Oxyz chong thng d:

x 11 =

y + 3

2=z 3

1v mt phng (P):2x+ y 2z + 9 = 0.

1. Tm ta im I thuc d sao cho khong cch t I n mt phng (P) bng 2.2. Tm ta giao im A ca ng thng d v mt phng (P). Vit phng trnhtham s ca ng thng nm trong mt phng (P), bit i qua A v vunggc vi d.


Bi 1.84 (A-02). Cho phng trnh:

log23 x+

log23 x+ 1 2m 1 = 0 (m l tham s).

1. Gii phng trnh khim = 2.2. Tmm phng trnh c t nht mt nghim thuc on [1; 3

3].

Bi 1.85 (A-07). Tmm phng trnh sau c nghim thc:

3x 1 +mx+ 1 = 2 4

x2 1.

Bi 1.86 (A-08). Tm cc gi tr ca tham s m phng trnh sau c ng hainghim thc phn bit:

4

2x+

2x+ 2 4

6 x+ 26 x = m (m R).

p s

1.1[

0 x 14

x 4

1.2 x = 65

1.3

x 12x = 2x 3

1.4 x = 3

1.5 x = 1 x = 22

1.6 x = 5

1.7 x > 1034

1.8 2 x < 10

1.9 x = 2

1.10 x = 35

2

1.11[x = pi

12+ k2pi

x = 7pi12

+ k2pi

1.12[x = 2pi

3+ k2pi

x = k2pi

1.13

x = pi2 + kpix = k2pix = 2pi

3+ k2pi

1.14 x = pi3

+ k2pi

1.15 cosx = 1; cosx = 12

1.16[x = pi

2+ kpi

x = pi4

+ k2pi


1.17 x = pi2; x = 3pi

2; x = 5pi

2; x = 7pi

2

1.18

[x = pi + k2pi

x = pi4

+ kpi(k Z)

1.19[x = pi

3+ k2pi

x = pi4

+ kpi(k Z)

1.20 x =pi

4+ kpi (k Z)

1.21

[x = kpi

x = 2pi3

+ k2pi(k Z)

1.22[x = pi

2+ k2pi

x = pi6

+ k2pi(k Z)

1.23[x = 2pi

3+ k2pi

x = pi4

+ kpi(k Z)

1.24[x = pi

18+ k pi

3

x = pi6

+ k pi2

(k Z)

1.25[x = pi

6+ k2pi

x = 5pi6

+ k2pi(k Z)

1.26[x = kpi

9

x = kpi2

(k Z)

1.27 x = pi3

+ kpi (k Z)

1.28[x = pi

6+ k2pi

x = 5pi6

+ k2pi(k Z)

1.29[x = pi

4+ kpi

x = 2pi3

+ k2pi(k Z)

1.30[x = pi

12+ kpi

x = 5pi12

+ kpi(k Z)

1.31 x = pi8

+ k pi4

x = pi18

+ k 2pi3

x = 5pi18

+ k 2pi3

1.32[x = pi

4+ k pi

2

x = pi3

+ kpi(k Z)

1.33[x = pi

6+ k2pi

x = pi42

+ k 2pi7

(k Z)

1.34 x = pi4

+ k pi2

(k Z)

1.35[x = pi

3

x = 5pi3

1.36 x = pi4

+ kpi (k Z)

1.37 x = k pi2

(k Z)

1.38 x = 5pi4

+ k2pi (k Z)

1.39 x = pi4

+ kpix = pi

2+ k2pi

x = k2pi

1.40 x = pi4

+ kpix = pi

8+ kpi

x = 5pi8

+ kpi

1.41 x = pi18

+ k 2pi3

(k Z)

1.42[x = pi

6+ k2pi

x = 7pi6

+ k2pi(k Z)

1.43 x = 0


2. Trong khng gian vi h ta Oxyz, cho hai ng thng 1 :

x = 3 + ty = tz = t

v 2 :x 2

2=y 1

1=z

2. Xc nh ta im M thuc 1 sao cho khong

cch t M n 2 bng 1.

Bi 6.15 (B-03). Trong khng gian vi h ta Oxyz, cho hai im A(2;0;0),B(0;0;8) v im C sao cho

AC =(0;6;0). Tnh khong cch t trung im I ca

BC n ng thng OA.

Bi 6.16 (B-04). Trong khng gian vi h ta Oxyz, cho im A(4;2;4) v

ng thng d:

x = 3 + 2ty = 1 tz = 1 + 4t

. Vit phng trnh ng thng i qua im

A, ct v vung gc vi ng thng d.

Bi 6.17 (B-06). Trong khng gian vi h ta Oxyz, cho im A(0;1;2) v haing thng:

d1 :x

2=y 1

1=z + 1

1 , d2 :

x = 1 + ty = 1 2tz = 2 + t.

1. Vit phng trnh mt phng (P) qua A, ng thi song song vi d1 v d2.2. Tm ta cc im M thuc d1, N thuc d2 sao cho ba im A, M, N thnghng.

Bi 6.18 (B-08). Trong khng gian vi h ta Oxyz, cho ba im A(0;1;2),B(2;2:1), C(2;0;1).1. Vit phng trnh mt phng i qua ba im A, B, C.2. Tm ta imM thuc mt phng 2x+2y+z3 = 0 sao choMA=MB=MC.Bi 6.19 (B-09).1. Trong khng gian vi h ta Oxyz, cho t din ABCD c cc nh A(1;2;1),B(2;1;3), C(2;1;1) v D(0;3;1). Vit phng trnh mt phng (P) i qua A, Bsao cho khong cch t C n (P) bng khong cch t D n (P).2. Trong khng gian vi h ta Oxyz, cho mt phng (P): x 2y+ 2z 5 = 0v hai im A(3;0;1), B(1;1;3). Trong cc ng thng i qua A v song songvi (P), hy vit phng trnh ng thng m khong cch t B n ng thng nh nht.


Bi 6.10 (D-05). Trong khng gian vi h ta Oxyz cho hai ng thng

d1 :x 1

3=y + 2

1 =z + 1

2v d2 :

{x+ y z 2 = 0x+ 3y 12 = 0.

a) Chng minh rng d1 v d2 song song vi nhau. Vit phng trnh mt phng(P) cha c hai ng thng d1 v d2.b) Mt phng ta Oxy ct c hai ng thng d1, d2 ln lt ti cc im A, B.Tnh din tch tam gic OAB (O l gc ta ).

Bi 6.11 (D-06). Trong khng gian vi h ta Oxyz, cho im A(1;2;3) v haing thng:

d1 :x 2

2=y + 2

1 =z 3

1, d2 :

x 11 =

y 12

=z + 1

1.

1. Tm ta im A i xng vi im A qua ng thng d1.2. Vit phng trnh ng thng i qua A, vung gc vi d1 v ct d2.

Bi 6.12 (D-07). Trong khng gian vi h ta Oxyz, cho hai im A(1;4;2),B(1;2;4) v ng thng : x 11 =

y + 2

1=z

2.

1. Vit phng trnh ng thng d i qua trng tm G ca tam gic OAB vvung gc mt phng (OAB).2. Tm ta im M thuc ng thng sao cho MA2+MB2 nh nht.

Bi 6.13 (D-09).1. Trong khng gian vi h ta Oxyz,cho cc im A(2;1;0), B(1;2;2), C(1;1;0)v mt phng (P): x+ y+ z 20 = 0. Xc nh ta im D thuc ng thngAB sao cho ng thng CD song song vi mt phng (P).

2. Trong khng gian vi h ta Oxyz, cho ng thng :x+ 2

1=y 2

1=

z

1 v mt phng (P):x+ 2y 3z+ 4 = 0. Vit phng trnh ng thng d nmtrong (P) sao cho d ct v vung gc vi ng thng .

Bi 6.14 (D-10).1. Trong khng gian vi h ta Oxyz, cho hai mt phng (P): x+y+ z3 = 0v (Q):x y + z 1 = 0. Vit phng trnh mt phng (R) vung gc vi (P) v(Q) sao cho khong cch t O n (R) bng 2.


1.44[x = 1x = 2

1.45 x = 0 x = 1

1.46 x = log2 3

1.47 S = [22; 1) (2; 2 +2]

1.48 x = 1 x = 2

1.49 log9 73 < x 2

1.50 x = 0

1.51 2 < x < 4

1.52 x = 1 x = 1

1.53 S = (4;3) (8; +)

1.54 x = 1

1.55 34< x 3

1.56 x = 2 x = 54

1.57

(x; y) = (1; 1)(1+52

;

5)

(15

2;5)

1.58 (x; y) =(32;1

2

);(12; 3

2

)1.59 (1; 1); (1;1); (2

25

;25);

(225

;25)

1.60

{x = 0

y = 1{x = 2

y = 4

1.61 (x; y) = (5; 2)

1.62 (x; y) = (1; 1); (2;32

)

1.63 (x; y) = (3; 1)

1.64 (x; y) = (1; 1); (32; 12)

1.65 x = y = 1

1.66 (x; y) = (1; 1); (2; 2)

1.67 (x; y) = (4; 174

)

1.68 (x; y) = (1; 13); (3; 1)

1.69 (x; y) = (1; 12)

1.70 (x; y) = (1; 1); (1+5

2; 1+

5

2)

(15

2; 1

5

2)

1.71 (x; y) = (3; 4)

1.72 (x; y) = (3; 3)

1.73 (x; y) = ( 3

54; 3

2516

) = (1;32)

1.74 x = y = 2x = y = 2

1.75 (x; y) = (12; 2)

1.76 m 23

2

1.77 0 m 14


1.78 f(x) = vt b trn[1; +)

1.80

[ 74 m 2

m 22

1.81

2 1 m 1

1.82 m 92

1.83

1.84 1.x = 33

2.0 m 2

1.85 1 < m 13

1.86 2

6 + 2 4

6 m < 32 + 6


Bi 6.4 (D-11). Trong khng gian vi h ta Oxyz, cho im A(1; 2; 3) v

ng thng d :x+ 1

2=y

1=z 32 . Vit phng trnh ng thng i qua

im A, vung gc vi ng thng d v ct trc Ox.

Bi 6.5 (B-11).

1. Trong khng gian h to Oxyz, cho ng thng

:x 2

1=y + 1

2 =z

1 v mt phng (P): x + y + z 3 = 0. Gi Il giao im ca v (P). Tm ta im M thuc (P) sao cho MI vunggc vi v MI =4

14.

2. Trong khng gian vi h to Oxyz, cho ng thng

:x+ 2

1=y 1

3=z + 5

2 v hai im A(2; 1; 1);B(3;1; 2). Tmta im M thuc ng thng sao cho tam gic MAB c din tchbng 3

5.

Bi 6.6 (A-11). Trong khng gian vi h ta Oxyz, cho hai imA(2; 0; 1), B(0;2; 3) v mt phng (P ) : 2x y z + 4 = 0. Tm ta imM thuc (P ) sao choMA = MB = 3.

Bi 6.7 (D-02). Trong khng gian vi h ta Oxyz, cho mt phng(P) : 2x y + 2 = 0 v ng thngdm :

{(2m+ 1)x+ (1m)y +m 1 = 0mx+ (2m+ 1)z + 4m+ 2 = 0

(m l tham s).Xc nhm ng thng dm song song vi mt phng (P).

Bi 6.8 (D-03). Trong khng gian vi h ta Oxyz, cho mt phng

(P) : x y2z+ 5 = 0 v ng thng dk :{x+ 3ky z + 2 = 0kx y + z + 1 = 0 (k l tham

s).Xc nh k ng thng dk vung gc vi mt phng (P).

Bi 6.9 (D-04). Trong khng gian vi h ta Oxyz cho hnh lng tr ngABC.A1B1C1. Bit A(a;0;0), B(a;0;0), C(0;1;0), B1(a;0;b), a> 0, b> 0.a) Tnh khong cch gia hai ng thng B1C v AC1 theo a, b.b) Cho a, b thay i, nhng lun tha mn a+b = 4. Tm a, b khong cch giahai ng thng B1C v AC1 ln nht.

Chng 6

Hnh hc gii tch trong khng gian

6.1 ng thng v mt phng . . . . . . . . . . . . . . . . . . 446.2 Mt cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Phng php ta trong khng gian . . . . . . . . . . . . 51p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

6.1 ng thng v mt phng

Bi 6.1 (D-12). Trong khng gian vi h ta Oxyz, cho ng thng d :x 1

2=

y + 1

1 =z

1v hai im A (1; -1; 2), B (2; -1; 0). Xc nh ta

im M thuc d sao cho tam gic AMB vung ti M.

Bi 6.2 (B-12). Trong khng gian vi h ta Oxyz, cho A(0;0;3), M(1;2;0).Vit phng trnh mt phng (P) qua A v ct cc trc Ox, Oy ln lt ti B, Csao cho tam gic ABC c trng tm thuc ng thng AM.

Bi 6.3 (A-12). Trong khng gian vi h ta Oxyz, cho ng thngd:x+ 1

2=y

1=z 2

1, mt phng (P) :x + y 2z + 5 = 0 v im A (1; -1;

2). Vit phng trnh ng thng ct d v (P) ln lt ti M v N sao cho A ltrung im ca on thng MN.

Chng 2

Bt ng thc

2.1 Bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Gi tr nh nht- Gi tr ln nht . . . . . . . . . . . . . . . 182.3 Nhn dng tam gic . . . . . . . . . . . . . . . . . . . . . . 20p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1 Bt ng thc

Bi 2.1 (A-09). Chng minh rng vi mi s thc dng x, y, ztha mn x(x+ y + z) = 3yz, ta c:

(x+ y)3 + (x+ z)3 + 3(x+ y)(x+ z)(y + z) 5(y + z)3.Bi 2.2 (A-05). Cho x, y, z l cc s dng tha mn

1

x+

1

y+

1

z= 4. Chng

minh rng1

2x+ y + z+

1

x+ 2y + z+

1

x+ y + 2z 1.

Bi 2.3 (A-03). Cho x, y, z l ba s dng v x+ y + z 1. Chng minh rngx2 +

1

x2+

y2 +

1

y2+

z2 +

1

z2

82.

Chng 2.Bt ng thc 18

Bi 2.4 (D-07). Cho a b > 0. Chng minh rng :(

2a +1

2a

)b(

2b +1

2b

)a.

Bi 2.5 (D-05). Cho cc s dng x, y, z tha mn xyz = 1. Chng minh rng1 + x3 + y3

xy+

1 + y3 + z3

yz+

1 + z3 + x3

zx 3

3.

Khi no ng thc xy ra?

2.2 Gi tr nh nht- Gi tr ln nht

Bi 2.6 (D-12). Cho cc s thc x, y tha mn (x4)2 + (y4)2 + 2xy 32. Tmgi tr nh nht ca biu thc A = x3 + y3 + 3(xy1)(x+ y2).

Bi 2.7 (B-12). Cho cc s thc x, y, z tha mn cc iu kin x+ y + z = 0 v

x2 + y2 + z2 = 1.

Tm gi tr ln nht ca biu thc

P = x5 + y5 + z5.

Bi 2.8 (A-12). Cho cc s thc x, y, z tha mn iu kin x + y + z = 0. Tmgi tr nh nht ca biu thc

P = 3|xy| + 3|yz| + 3|zx|

6x2 + 6y2 + 6z2

Bi 2.9 (B-11). Cho a v b l cc s thc dng tha mn2(a2 + b2) + ab = (a + b)(ab + 2). Tm gi tr nh nht ca biu thc

P= 4(a3

b3+b3

a3

) 9

(a2

b2+b2

a2

).

Bi 2.10 (A-11). Cho x, y, z l ba s thc thuc on [1; 4] v x y, x z. Tmgi tr nh nht ca biu thc

P =x

2x+ 3y+

y

y + z+

z

z + x

.

Chng 5.Kho st hm s 43

5.24 I(13< m < 1,m 6= 0); II(m =

1)

5.25 I(0 < m < 1); II(m = 26)

5.26 m = 2

5.27 12< m < 0

5.28 m = 15

2

5.29 4 < m < 5

5.30 14< m < 1,m 6= 0

5.31 m = 0 orm = 2

5.32 m > 0

5.33 m = 1


Bi 5.32 (B-03). Cho hm s : y = x3 3x2 +m (1) (m l tham s).1. Tm m th hm s (1) c hai im phn bit i xng vi nhau qua gcta .2. Kho st s bin thin v v th ca hm s (1) ng vi m= 2.

Bi 5.33 (A-08). Cho hm s y =mx2 + (3m2 2)x 2

x+ 3m(1), vi m l tham

s thc.1. Kho st s bin thin v v th ca hm s (1) khim = 1.2. Tm cc gi tr ca tham s m gc gia hai ng tim cn ca th hms (1) bng 45o.

p s

5.1 m = 1

5.2 1 + 4 ln 43;m 6= 1

5.3 m = 4

5.4 M(12;2);M(1; 1)

5.5 y = 6x+ 10

5.6 y = x+ 83

5.7 y = x+225; y = x225

5.8 y = 24x+ 15; y = 154x 21

4

5.9 y = x 2

5.10 m = 23

5.11 m = 2

5.12 m = 0

5.13 m = 2 22

5.14 m < 3 or 0 < m < 3

5.15 M(2;m 3);N(0;m+ 1)

5.16 m = 12

5.17 1 < k < 3, k 6= 0, k 6= 2y = 2xm2 +m

5.18 m = 1

5.19 m = 4 26

5.20 k = 3

5.21 m > 1

5.22 m > 154,m 6= 24

5.23

Chng 2.Bt ng thc 19

Bi 2.11 (D-11). Tm gi tr nh nht v gi tr ln nht ca hm s y =2x2 + 3x+ 3

x+ 1trn on [0; 2].

Bi 2.12 (A-07). Cho x, y, z l cc s thc dng thay i v tha mn iu kinxyz = 1. Tm gi tr nh nht ca biu thc:

P =x2(y + z)

yy + 2z

z

+y2(z + x)

zz + 2x

x

+z2(x+ y)

xx+ 2y

y.

Bi 2.13 (A-06). Cho hai s thc x 6= 0, y 6= 0 thay i v tha mn iu kin:

(x+ y)xy = x2 + y2 xy.

Tm gi tr ln nht ca biu thc

A =1

x3+

1

y3.

Bi 2.14 (B-10). Cho cc s thc khng m a, b, c tha mn a + b + c = 1. Tmgi tr nh nht ca biu thc

M = 3(a2b2 + b2c2 + c2a2) + 3(ab+ bc+ ca) + 2a2 + b2 + c2.

Bi 2.15 (B-09). Cho cc s thc x, y thay i v tha mm (x+ y)3 + 4xy 2.Tm gi tr nh nht ca biu thc

A = 3(x4 + y4 + x2y2) 2(x2 + y2) + 1.

Bi 2.16 (B-08). Cho hai s thc x, y thay i v tha mn h thc x2 + y2 = 1.Tm gi tr ln nht v gi tr nh nht ca biu thc

P =2(x2 + 6xy)

1 + 2xy + 2y2.

Bi 2.17 (B-07). Cho x, y, z l ba s thc dng thay i. Tm gi tr nh nhtca biu thc:

P = x

(x

2+

1

yz

)+ y

(y

2+

1

zx

)+ z

(z

2+

1

xy

).

Chng 2.Bt ng thc 20

Bi 2.18 (B-06). Cho x, y l cc s thc thay i. Tm gi tr nh nht ca biuthc:

A =

(x 1)2 + y2 +

(x+ 1)2 + y2 + |y 2|.Bi 2.19 (B-03). Tm gi tr ln nht v gi tr nh nht ca hm sy = x+

4 x2.

Bi 2.20 (D-10). Tm gi tr nh nht ca hm s

y =x2 + 4x+ 21

x2 + 3x+ 10.

Bi 2.21 (D-09). Cho cc s thc khng m x, y thay i v tha mn x+ y = 1.Tm gi tr ln nht v gi tr nh nht ca biu thc

S = (4x2 + 3y)(4y2 + 3x) + 25xy.

Bi 2.22 (D-08). Cho x, y l hai s thc khng m thay i. Tm gi tr ln nhtv gi tr nh nht ca biu thc

P =(x y)(1 xy)(1 + x)2(1 + y)2

.

Bi 2.23 (D-03). Tm gi tr ln nht v gi tr nh nht ca hm s y =x+ 1x2 + 1

trn on [1; 2].

2.3 Nhn dng tam gic

Bi 2.24 (A-04). Cho tam gic ABC khng t tha mn iu kin

cos 2A+ 2

2 cosB + 2

2 cosC = 3.

Tnh ba gc ca tam gic ABC.

p s


2. Vi gi tr no ca m, phng trnh x2|x2 2| = m c ng 6 nghim thcphn bit?II. Tm cc gi tr ca tham s m ng thng y = x + m ct th hm sy =

x2 1x

ti hai im phn bit A, B sao cho AB= 4.

Bi 5.26 (B-10). Cho hm s y =2x+ 1

x+ 1.

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm m ng thng y = 2x+m ct th (C) ti hai im phn bit A, Bsao cho tam gic OAB c din tch bng

3 (O l gc ta ).

Bi 5.27 (A-03). Cho hm s y =mx2 + x+m

x 1 (1) (ml tham s).1. Kho st s bin thin v v th hm s (1) khim = 1.2. Tm m th hm s (1) ct trc honh ti hai im phn bit v hai im c honh dng.

Bi 5.28 (A-04). Cho hm s y =x2 + 3x 3

2(x 1) (1).1. Kho st hm s (1).2. Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao choAB= 1.

Bi 5.29 (A-06).1. Kho st s bin thin v v th hm s y = 2x3 9x2 + 12x 4.2. Tmm phng trnh sau c 6 nghim phn bit: 2|x3| 9x2 + 12|x| = m.Bi 5.30 (A-10). Cho hm s y = x3 2x2 + (1 m)x + m (1), m l tham sthc.1. Kho st s bin thin v v th ca hm s (1) khim = 1.2. Tm m th ca hm s (1) ct trc honh ti 3 im phn bit c honh x1, x2, x3 tha mn iu kin x21 + x

22 + x

23 < 4.

5.4 Bi ton khc

Bi 5.31 (D-04). Cho hm s : y = x3 3mx2 + 9x + 1 (1) (m l thams).1. Kho st hm s (1) ng vim = 2.2. Tm m im un ca th hm s (1) thuc ng thng y = x+ 1.


5.3 Tng giao th

Bi 5.20 (D-11). Cho hm s y =2x+ 1

x+ 1

1. Kho st s bin thin v v th (C) ca hm s cho

2. Tm k ng thng y = kx+ 2k+ 1 ct th (C) ti hai im phn bitA, B sao cho khong cch t A v B n trc honh bng nhau.

Bi 5.21 (D-03).

1. Kho st s bin thin v v th ca hm s y =x2 2x+ 4

x 2 (1).2. Tm m ng thng dm: y = mx + 2 2m ct th hm s (1) ti haiim phn bit.

Bi 5.22 (D-06). Cho hm s : y = x3 3x+ 2.1. Kho st s bin thin v v th (C) ca hm s cho.2. Gi d l ng thng i qua im A(3; 20) v c h s gc l m. Tm m ng thng d ct th (C) ti 3 im phn bit.

Bi 5.23 (D-08). Cho hm s : y = x3 3x2 + 4 (1).1. Kho st s bin thin v v th ca hm s (1).2. Chng minh rng mi ng thng i qua im I(1; 2) vi h s gc k (k> 3)u ct th ca hm s (1) ti ba im phn bit I, A, B ng thi I l trungim ca on thng AB.

Bi 5.24 (D-09).I. Cho hm s y = x4 (3m+ 2)x2 + 3m c th l (Cm), m l tham s.1. Kho st s bin thin v v th ca hm s cho khi m= 0.2. Tm m ng thng y = 1 ct th (Cm) ti 4 im phn bit u chonh nh hn 2.II. Tm cc gi tr ca tham sm ng thng y = 2x+m ct th hm sy =

x2 + x 1x

ti hai im phn bit A, B sao cho trung im ca on thngAB thuc trc tung.

Bi 5.25 (B-09).I. Cho hm s y = 2x4 4x2 (1).1. Kho st s bin thin v v th ca hm s (1).

Chng 2.Bt ng thc 21

2.6 Amin = 1755

4

2.7 P =5

6

36

2.8 Pmin = 3

2.9 minP = 234

2.10 Pmin = 3433

2.11 GTLN l 173;GTNN

l 3

2.12 Pmin = 2

2.13 Amax = 16

2.14 Mmin = 2

2.15 Amin =9

16

2.16 Pmax = 3;Pmin =6

2.17 Pmin =9

2

2.18 Amin = 2 +

3

2.19 max[2;2]

y = 2

2

min[2;2]

y = 2

2.20 ymin =

2

2.21 Smax = 252 ;Smin =19116

2.22 Pmin =1

4;Pmax =

14

2.23 ymax =

2; ymin =0

2.24 A = 90o;B = C =45o

Chng 3

Hnh hc gii tch trong mt phng

3.1 ng thng . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 ng trn . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Cnic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.1 ng thng

Bi 3.1 (D-12). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD.Cc ng thng AC v AD ln lt c phng trnh l x+3y = 0 v xy+4 = 0;ng thng BD i qua im M (1

3; 1). Tm ta cc nh ca hnh ch nht

ABCD.

Bi 3.2 (A-12). Trong mt phng vi h ta Oxy, cho hnh vung ABCD. GiM l trung im ca cnh BC, N l im trn cnh CD sao cho CN = 2ND. Gi sM(112

; 12

)v ng thng AN c phng trnh 2x y 3 = 0. Tm ta im

A.

Bi 3.3 (D-11). Trongmt phng ta Oxy, cho tam gic ABC c nhB(4; 1),trng tm G(1; 1) v ng thng cha phn gic trong ca gc A c phng trnhx y 1 = 0. Tm ta cc nh A v C.


Bi 5.14 (B-02). Cho hm s : y = mx4 + (m2 9)x2 + 10 (1) (m ltham s).1. Kho st s bin thin v v th ca hm s (1) ng vi m= 1.2. Tm m hm s (1) c ba im cc tr.

Bi 5.15 (B-05). Gi (Cm) l th ca hm s y =x2 + (m+ 1)x+m+ 1

x+ 1(*)

(m l tham s).1. Kho st v v th hm s (*) khi m= 1.2. Chng minh rng vi m bt k, th (Cm) lun lun c im cc i, imcc tiu v khong cch gia hai im bng

20.

Bi 5.16 (B-07). Cho hm s: y = x3 + 3x2 + 3(m2 1)x 3m2 1 (1), m ltham s.1. Kho st s bin thin v v th hm s (1) khi m= 1.2. Tm m hm s (1) c cc i, cc tiu v cc im cc tr ca th hm s(1) cch u gc ta O.

Bi 5.17 (A-02). Cho hm s: y = x3 + 3mx2 + 3(1m2)x+m3 m2 (1)(m l tham s).1. Kho st s bin thin v v th ca hn s (1) khim = 1.2. Tm k phng trnh: x3 + 3x2 + k3 3k2 = 0 c ba nghim phnbit.3. Vit phng trnh ng thng i qua hai im cc tr ca th hm s (1).

Bi 5.18 (A-05). Gi(Cm) l th ca hm s y = mx+1

x(*) (m l tham

s).

1. Kho st s bin thin v v th ca hm s (*) khim =1

4.

2. Tm m hm s (*) c cc tr v khong cch t im cc tiu ca (Cm) n

tim cn xin ca (Cm) bng12.

Bi 5.19 (A-07). Cho hm s y =x2 + 2(m+ 1)x+m2 + 4m

x+ 2(1),m l tham

s.1. Kho st s bin thin v v th ca hm s (1) khim = 1.2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca th cng vi gc ta O to thnh mt tam gic vung ti O.


Bi 5.9 (A-09). Cho hm s y =x+ 2

2x+ 3(1).

1. Kho st s bin thin v v th ca hm s (1).2. Vit phng trnh tip tuyn ca th hm s (1), bit tip tuyn ct trchonh, trc tung ln lt ti hai im phn bit A, B v tam gic OAB cn ti gcta O.

5.2 Cc tr

Bi 5.10 (D-12). Cho hm s y = 23x3mx22(3m21)x + 2

3(1), m l tham s

thc.

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

2. Tmm hm s (1) c hai im cc tr x1 vx2 sao cho x1.x2+2(x1+x2) =1

Bi 5.11 (B-12). Cho hm s y = x3 3mx2 + 3m2, (1),m l tham s thc.

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

2. Tm m th hm s (1) c hai im cc tr A v B sao cho tam gicOAB c din tch bng 48.

Bi 5.12 (A-12). Cho hm s y = x4 2(m + 1)x2 + m2, (1) ,vi m l thams.

1. Kho st s bin thin v v th hm s (1) khim = 0.

2. Tm m th hm s (1) c ba im cc tr to thnh ba nh ca mttam gic vung.

Bi 5.13 (B-11). Cho hm s y = x4 2(m+ 1)x2 +m, (1), m l tham s.

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

2. Tm m th hm s (1) c ba im cc tr A, B, C sao cho OA = BC, Ol gc ta , A l cc tr thuc trc tung, B v C l hai im cc tr cn li.

Chng 3.Hnh hc gii tch trong mt phng 23

Bi 3.4 (B-11). Trong mt phng to Oxy, cho hai ng thng : xy4 =0 v d : 2x y 2 = 0. Tm ta im N thuc ng thng d sao cho ngthng ON ct ng thng ti im M tha mn OM.ON = 8.

Bi 3.5 (A-10). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC cn ti A c nh A(6;6), ng thng i qua trung im ca cc cnhAB v AC c phng trnh x+ y 4 = 0. Tm ta cc nh B v C, bit imE(1;-3) nm trn ng cao i qua nh C ca tam gic cho.

Bi 3.6 (A-09). Trong mt phng vi h ta cac vung gc Oxy, cho hnhch nht ABCD c im I(6;2) l giao im ca hai ng cho AC v BD. imM(1;5) thuc ng thng AB v trung im E ca cnh CD thuc ng thng : x+ y 5 = 0. Vit phng trnh ng thng AB.Bi 3.7 (A-06). Trong mt phng vi h ta cac vung gc Oxy, cho ccng thng :

d1 : x+ y + 3 = 0, d2 : x y 4 = 0, d3 : x 2y = 0.Tm ta imM nm trn ng thng d3 sao cho khong cch t M n ngthng d1 bng hai ln khong cch t M n ng thng d2.

Bi 3.8 (A-05). Trong mt phng vi h ta cac vung gc Oxy, cho haing thng :

d1 : x y = 0 v d2 : 2x+ y 1 = 0.Tm ta cc nh ca hnh vung ABCD bit rng nh A thuc d1, nh Cthuc d2 v cc nh B, D thuc trc honh.

Bi 3.9 (A-04). Trong mt phng vi h ta cac vung gc Oxy, cho haiim A(0;2) v B(3;1). Tm ta trc tm v ta tm ng trn ngoitip ca tam gic OAB.

Bi 3.10 (A-02). Trong mt phng vi h ta cac vung gc Oxy, xt tamgic ABC vung ti A, phng trnh ng thng BC l

3x y 3 = 0, cc

nh A v B thuc trc honh v bn knh ng trn ni tip bng 2. Tm ta trng tm G ca tam gic ABC.

Bi 3.11 (B-10). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC vung ti A, c nh C(-4;1), phn gic trong gc A c phng trnhx + y 5 = 0. Vit phng trnh ng thng BC, bit din tch tam gic ABCbng 24 v nh A c honh dng.


Bi 3.12 (B-09). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC cn ti A c nh A(-1;4) v cc nh B, C thuc ng thng :xy4 = 0. Xc nh ta cc im B v C, bit rng din tch tam gic ABCbng 18.

Bi 3.13 (B-08). Trong mt phng vi h ta cac vung gc Oxy, hy xcnh ta nh C ca tam gic ABC bit rng hnh chiu vung gc ca C trnng thng AB l im H(-1;-1), ng phn gic trong ca gc A c phngtrnh x y + 2 = 0 v ng cao k t B c phng trnh 4x+ 3y 1 = 0.Bi 3.14 (B-07). Trong mt phng vi h ta cac vung gc Oxy, cho imA(2;2) v cc ng thng :

d1 : x+ y 2 = 0, d2 : x+ y 8 = 0.Tm ta im B v C ln lt thuc d1 v d2 sao cho tam gic ABC vung cnti A.

Bi 3.15 (B-04). Trong mt phng vi h ta cac vung gc Oxy, cho haiim A(1;1), B(4;-3). Tm im C thuc ng thng x 2y 1 = 0 sao chokhong cch t C n ng thng AB bng 6.

Bi 3.16 (B-03). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c AB=AC, BAC = 90o. Bit M(1;-1) l trung im cnh BC v

G(2

3; 0) l trng tm tam gic ABC. Tm ta cc nh A, B, C.

Bi 3.17 (B-02). Trong mt phng vi h ta cac vung gc Oxy, cho hnhch nht ABCD c tm I(

1

2; 0), phng trnh ng thng AB l x 2y + 2 = 0

v AB=2AD. Tm ta cc nh A, B, C, D bit rng nh A c honh m.

Bi 3.18 (D-10). Trong mt phng vi h ta cac vung gc Oxy, cho imA(0;2) v l ng thng i qua O. Gi H l hnh chiu vung gc ca A trn. Vit phng trnh ng thng , bit rng khong cch t H n trc honhbng AH.

Bi 3.19 (D-09). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c M(2;0) l trung im ca cnh AB. ng trung tuyn v ng caoi qua nh A ln lt c phng trnh l 7x 2y 3 = 0 v 6x y 4 = 0. Vitphng trnh ng thng AC.

Bi 3.20 (D-04). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c cc nh A(-1;0); B(4;0); C(0;m) vi m 6= 0. Tm ta trng tmG ca tam gic ABC theo m. Xc nh m tam gic GAB vung ti G.


2. Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc ta .3. Tm m th hm s (1) tip xc vi ng thng y = x.

Bi 5.3 (D-05). Gi (Cm) l th hm s y =1

3x3 m

2x2 +

1

3(*) (m l

tham s).1. Kho st s bin thin v v th ca hm s (*) ng vim = 2.2. Gi M l im thuc (Cm) c honh bng 1 . Tm m tip tuyn ca(Cm) ti im M song song vi ng thng 5x y = 0.

Bi 5.4 (D-07). Cho hm s y =2x

x+ 1.

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm ta im M thuc (C), bit tip tuyn ca (C) ti M ct hai trc Ox, Oyti

A, B v tam gic OAB c din tch bng1

4.

Bi 5.5 (D-10). Cho hm s y = x4 x2 + 6.1. Kho st s bin thin v v th (C) ca hm s cho.2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi ng

thng y =1

6x 1.

Bi 5.6 (B-04). Cho hm s y =1

3x3 2x2 + 3x (1) c th (C).

1. Kho st hm s (1).2. Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng ltip tuyn ca (C) c h s gc nh nht.

Bi 5.7 (B-06). Cho hm s y =x2 + x 1x+ 2

1. Kho st s bin thin v v th (C) ca hm s cho2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vitim cn xin ca (C).

Bi 5.8 (B-08). Cho hm s y = 4x3 6x2 + 1 (1).1. Kho st s bin thin v v th ca hm s (1).2. Vit phng trnh tip tuyn ca th hm s (1), bit rng tip tuyn iqua im M(1;9).

Chng 5

Kho st hm s

5.1 Tip tuyn . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Cc tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.3 Tng giao th . . . . . . . . . . . . . . . . . . . . . . . 405.4 Bi ton khc . . . . . . . . . . . . . . . . . . . . . . . . . . 41p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

5.1 Tip tuyn

Bi 5.1 (A-11). Cho hm s y =x+ 12x 1

1. Kho st s bin thin v v th (C) ca hm s cho.

2. Chng minh rng vi mi m ng thng y = x + m lun ct th (C)ti hai im phn bit A v B. Gi k1, k2 ln lt l h s gc ca cc tiptuyn vi (C) ti A v B. Tmm tng k1 + k2 t gi tr ln nht.

Bi 5.2 (D-02). Cho hm s : y =(2m 1)xm2

x 1 (1) (m l tham s).1. Kho st s bin thin v v th (C) ca hm s (1) ng vi m= 1.


3.2 ng trn

Bi 3.21 (D-12). Trong mt phng vi h ta Oxy, cho ng thng d : 2xy+3 = 0. Vit phng trnh ng trn c tm thuc d, ct trc Ox ti A v B, cttrc Oy ti C v D sao cho AB = CD = 2.

Bi 3.22 (B-12). Trong mt phng c h ta Oxy, cho cc ng trn (C1) :x2 + y2 =, (C2) : x2 + y2 12x + 18 = 0 v ng thng d : x y 4 = 0.Vit phng trnh ng trn c tm thuc (C2), tip xc vi d v ct (C1) ti haiim phn bit A v B sao cho AB vung gc vi d.

Bi 3.23 (D-11). Trong mt phng ta Oxy, cho im A(1; 0) v ng trn(C) : x2 + y2 2x + 4y 5 = 0. Vit phng trnh ng thng ct (C) tiim M v N sao cho tam gic AMN vung cn ti A.

Bi 3.24 (B-11). Trong mt phng ta Oxy, cho tam gic ABC c nhB(12; 1).

ng trn ni tip tam gic ABC tip xc vi cc cnh BC, CA, AB tng ngti cc im D, E, F. Cho D(3; 1) v ng thng EF c phng trnh y 3 = 0.Tm ta nh A, bit A c tung dng.

Bi 3.25 (A-11). Trong mt phng ta Oxy, cho ng thng : x+y+2 = 0v ng trn (C) : x2 +y24x2y = 0. Gi I l tm ca (C), M l im thuc. Qua M k cc tip tuyn MA v MB n (C) (A v B l cc tip im). Tmta im M, bit t gic MAIB c din tch bng 10.

Bi 3.26 (A-10). Trong mt phng vi h ta cac vung gc Oxy, cho haing thng d1 :

3x + y = 0 v d2 :

3x y = 0. Gi (T) l ng trn tip

xc vi d1 ti A, ct d2 ti hai im B v C sao cho tam gic ABC vung ti B.

Vit phng trnh ca (T), bit rng tam gic ABC c din tch bng

3

2v im

A c honh dng.

Bi 3.27 (A-09). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C) : x2 + y2 + 4x+ 4y + 6 = 0 v ng thng : x+my 2m+ 3 = 0,vi m l tham s thc. Gi I l tm ca ng trn (C). Tm m ct (C) tihai im phn bit A v B sao cho din tch tam gic IAB ln nht.

Bi 3.28 (A-07). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c A(0;2), B(-2;-2), v C(4;-2). Gi H l chn ng cao k t B; M vN ln lt l trung im ca cc cnh AB v BC. Vit phng trnh ng trn iqua cc im H, M, N.


Bi 3.29 (B-09). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 2)2 + y2 = 4

5v hai ng thng 1 : x y = 0,2 : x 7y = 0.

Xc nh ta tm K v bn knh ca ng trn (C1); bit ng trn (C1) tipxc vi cc ng thng 1, 2 v tm K thuc ng trn (C).

Bi 3.30 (B-06). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): x2 + y2 2x 6y + 6 = 0 v im M(-3;1). Gi T1 v T2 l cc tipim ca cc tip tuyn k t M n (C). Vit phng trnh ng thng T1T2.

Bi 3.31 (B-05). Trong mt phng vi h ta cac vung gc Oxy, cho haiim A(2;0) v B(6;4). Vit phng trnh ng trn (C) tip xc vi trc honhti im A v khong cch t tm ca (C) n im B bng 5.

Bi 3.32 (D-10). Trong mt phng vi h ta cac vung gc Oxy, cho tamgic ABC c nh A(3;-7), trc tm l H(3;-1), tm ng trn ngoi tip l I(-2;0). Xc nh ta nh C, bit C c honh dng.

Bi 3.33 (D-09). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 1)2 + y2 = 1. Gi I l tm ca (C). Xc nh ta im M thuc(C) sao cho IMO = 30o.

Bi 3.34 (D-07). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 1)2 + (y + 2)2 = 9 v ng thng d: 3x 4y +m = 0.Tm m trn d c duy nht mt im P m t c th k c hai tip tuynPA, PB ti (C) (A, B l cc tip im) sao cho tam gic PAB u.

Bi 3.35 (D-06). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): x2 + y2 2x 2y + 1 = 0 v ng thng d: x y + 3 = 0. Tm ta im M nm trn d sao cho ng trn tm M, c bn knh gp i bn knhng trn (C), tip xc ngoi vi ng trn (C).

Bi 3.36 (D-03). Trong mt phng vi h ta cac vung gc Oxy, cho ngtrn (C): (x 1)2 + (y 2)2 = 4 v ng thng d: x y 1 = 0.1. Vit phng trnh ng trn (C) i xng vi ng trn (C) qua ng thngd.2. Tm ta cc giao im ca (C) v (C).

3.3 Cnic

Bi 3.37 (B-12). Trong mt phng vi h ta Oxy, cho hnh thoi ABCD cAC = 2BD v ng trn tip xc vi cc cnh ca hnh thoi c phng trnh

Chng 4.T hp v s phc 35

4.1 P =443

506

4.2 C13 .C412.C21 .C48 .C11 .C44 = 207900

4.3 C215.C210.C15 + C215.C110.C25 +C315.C

110.C

15 = 56875

4.4 n = 8

4.5 C412 (C25 .C14 .C13 + C15 .C24 .C13 +C15 .C

14 .C

23) = 225

4.7 k = 9

4.8 M = 34

4.10 n = 1002

4.113n+1 2n+1

n+ 1

4.12 n = 6

4.13 3516.x5

4.14 a8 = 28C812 = 126720

4.15 C610 = 210

4.16 C38 .C23 + C48 .C04 = 238

4.17 C412 = 495

4.18 n = 7, x = 4

4.19 C1011 .21 = 22

4.20 (2)4C45 + 33.C310 = 3320

4.21 C47 = 35

4.22 n = 5

4.23 z = 12i; z = 2i

4.24 |w| = 5

4.25 z1 = 2(cos 2pi3 + i sin2pi3

)z2 = 2(cos

pi3

+ i sin pi3)

4.26 |w| = 4 + 9 = 13

4.27 z = 2 i

4.28

1. z = 13i; z = 23i2. z = 2 + 2i

4.29 z = 0, z = 12 1

2i

|z| =23

4.30 Phn o z l: 2|z + iz| = 82

4.31 A = 20

4.32 x2 + (y + 1)2 = 2

4.33 z = 3 + 4i hoc z = 5

4.34 (x 3)2 + (y + 4)2 = 4

4.35 1 + i; 1 i;1 + i;1 i


Bi 4.27 (D-11). Tm s phc z, bit :z (2 + 3i)z = 1 9i.Bi 4.28 (B-11).

1. Tm s phc z, bit: z 5 + i

3

z 1 = 0.

2. Tm phn thc v phn o ca s phc z =

(1 + i

3

1 + i

)3.

Bi 4.29 (A-11).

1. Tm tt c cc s phc z, bit z2 = |z|2 + z.2. Tnh mun ca s phc z, bit:

(2z 1)(1 + i) + (z + 1)(1 i) = 2 2i.

Bi 4.30 (A-10).1. Tm phn o ca s phc z, bit

z = (

2 + i)2(12i).

2. Cho s phc z tha mnz =

(13i)31 i . Tm mun ca s phc

z + iz.

Bi 4.31 (A-09). Gi z1 v z2 l hai nghim phc ca phng trnh z2+2z+10 =0. Tnh gi tr ca biu thc A = |z1|2 + |z2|2.Bi 4.32 (B-10). Trong mt phng vi h ta cac vung gc Oxy, tm tphp im biu din cc s phc z tha mn:

|z i| = |(1 + i)z|.

Bi 4.33 (B-09). Tm s phc z tha mn: |z (2 + i)| = 10 v zz = 25.Bi 4.34 (D-09). Trong mt phng vi h ta cac vung gc Oxy, tm tphp im biu din cc s phc z tha mn iu kin |z (3 4i)| = 2.Bi 4.35 (D-10). Tm s phc z tha mn: |z| = 2 v z2 l s thun o.

p s


x2 + y2 = 4. Vit phng trnh chnh tc ca elip (E) i qua cc nh A, B, C, Dca hnh thoi. Bit A thuc Ox.

Bi 3.38 (A-12). Trong mt phng vi h ta Oxy, cho ng trn(C) : x2 +y2 = 8. Vit phng trnh chnh tc elip (E), bit rng (E) c di trcln bng 8 v (E) ct (C) ti bn im to thnh bn nh ca mt hnh vung.

Bi 3.39 (A-11). Trong mt phng ta Oxy, cho elip (E) :x2

4+y2

1= 1. Tm

ta cc im A v B thuc (E), c honh dng sao cho tam gic OAB cnti O v c din tch ln nht.

Bi 3.40 (A-08). Trong mt phng vi h ta cac vung gc Oxy, hy vit

phng trnh chnh tc ca elip (E) bit rng (E) c tm sai bng

5

3v hnh ch

nht c s ca (E) c chu vi bng 20.

Bi 3.41 (B-10). Trong mt phng vi h ta cac vung gc Oxy, cho im

A(2;

3) v elip (E):x2

3+y2

2= 1. Gi F1 v F2 l cc tiu im ca (E) (F1 c

honh m), M l giao im c tung dng ca ng thng AF1 vi (E), Nl im i xng ca F2 qua M. Vit phng trnh ng trn ngoi tip tam gicANF2 .

Bi 3.42 (D-08). Trong mt phng vi h ta cac vung gc Oxy, choparabol (P): y2 = 16x v im A(1;4). Hai im phn bit B,C (B v C khcA) di ng trn (P) sao cho gc BAC = 90o. Chng minh rng ng thng BClun i qua mt im c nh.

Bi 3.43 (D-02). Trong mt phng vi h ta cac vung gc Oxy, cho elip

(E) c phng trnhx2

16+y2

9= 1. Xt im M chuyn ng trn tia Ox v im N

chuyn ng trn tia Oy sao cho ng thng MN lun tip xc vi (E). Xc nhta ca M, N on MN c di nh nht. Tnh gi tr nh nht .

Bi 3.44 (D-05). Trong mt phng vi h ta cac vung gc Oxy, cho im

C(2;0) v elp (E):x2

4+y2

1= 1. Tm ta cc im A, B thuc (E), bit rng

A, B i xng vi nhau qua trc honh v tam gic ABC l tam gic u.

p s


3.1 A(3; 1);D(1; 3).B(1;3).C(3;1)

3.2 A(1;1);A(4; 5).

3.3 A(4; 3);C(3;1)

3.4 N(0;4),M(0;2)N(6; 2);M(6

5; 25)

3.5 B(0;4), C(4; 0)hocB(6; 2), C(2;6)

3.6 y 5 = 0; x 4y + 19 = 0

3.7 M(22;11),M(2; 1)

3.8 A(1; 1), B(0; 0), C(1;1), D(2; 0)A(1; 1), B(2; 0), C(1;1), D(0; 0)

3.9 H(

3;1), I(3; 1)

3.10 G1(7+43

3; 6+2

3

3)

G2(431

3; 62

3

3)

3.11 3x 4y + 16 = 0

3.12 B(112

; 32);C(3

2;5

2)

B(32;5

2);C(11

2; 32)

3.13 C(103

; 34)

3.14 B(1; 3), C(3; 5)B(3;1), C(5; 3)

3.15 C = (7; 3); (4311

;2711

)

3.16 B,C = (4; 0); (2;2)

3.17 A(2; 0), B(2; 2), C(3; 0), D(1;2)

3.18 (

5 1)x 2

5 2y = 0

3.19 3x 4y + 5 = 0

3.20 m = 36

3.21 (x+ 3)2 + (y + 3)2 = 10

3.22 (C) : x2 + y2 6x 6y+ 10 = 0

3.23 : y = 1; y = 3

3.24 A(3; 133

)

3.25 M(2;4) vM(3; 1)

3.26 (x+ 123)2 + (y + 2

3)2 = 1

3.27 m = 0 m = 815

3.28 x2 + y2 x+ y 2 = 0

3.29 K(85; 45);R = 2

2

5

3.30 2x+ y 3 = 0

3.31 (x 2)2 + (y 1)2 = 1(x 2)2 + (y 7)2 = 49

3.32 C(2 +65; 3)

3.33 M(32;32

)

3.34 m = 19 m = 41

3.35 M = (1; 4); (2; 1)


Bi 4.18 (A-02). Cho khai trin nh thc:(2x12 + 2

x3

)n= C0n

(2x12

)n+C1n

(2x12

)n1 (2x3

)+ +Cn1n

(2x12

)(2x3

)n1+Cnn

(2x3

)n.

(n nguyn dng). Bit rng trong khai trin C3n = 5C1n v s hng th t

bng 20n, tm n v x.

Bi 4.19 (B-07). Tm h s ca s hng cha x10 trong khai trin nh thc Niutonca (2 + x)n, bit:

3nC0n 3n1C1n + 3n2C2n 3n3C3n + + (1)nCnn = 2048(n l s nguyn dng).

Bi 4.20 (D-07). Tm h s ca x5 trong khai trin thnh a thc ca:

x(1 2x)5 + x2(1 + 3x)10.Bi 4.21 (D-04). Tm s hng khng cha x trong khai trin nh thc Niuton ca(

3x+

14x

)7vi x > 0.

Bi 4.22 (D-03). Vi n l s nguyn dng, gi a3n3 l h s ca x3n3 trongkhai trin thnh a thc ca (x2 + 1)n(x+ 2)n. Tm n a3n3 = 26n.

4.5 S phc

Bi 4.23 (D-12). Gii phng trnh z2 + 3(1 + i)z + 5i = 0 trn tp hp cc sphc.

Bi 4.24 (D-12). Cho s phc z tha mn (2 + i)z +2(1 + 2i)

1 + i= 7 + 8i. Tm

mun ca s phc w = z + 1 + i.

Bi 4.25 (B-12). Gi z1 v z2 l hai nghim phc ca phng trnh z2 2

3iz4 = 0. Vit dng lng gic ca z1 v z2

Bi 4.26 (A-12). Cho s phc z tha5(z + i)

z + 1= 2 i. Tnh mun ca s phc

w = 1 + z + z2.


Bi 4.10 (A-05). Tm s nguyn dng n sao cho

C12n+1 2.2C22n+1 + 3.22C32n+1 4.23C42n+1 + + (2n+ 1).22nC2n+12n+1 = 2005.Bi 4.11 (B-03). Cho n nguyn dng. Tnh tng

C0n +22 1

2C1n +

23 13

C2n + +2n+1 1n+ 1

Cnn .

Bi 4.12 (D-08). Tm s nguyn dng n tha mn h thc

C12n + C32n + + C2n12n = 2048.

4.4 H s trong khai trin nh thc

Bi 4.13 (A-12). Cho n l s nguyn dng tha mn 5Cn1n = C3n. Tm s hng

cha x5 trong khai trin nh thc Niu-tn(nx2

14 1x

)n, x 6= 0.

Bi 4.14 (A-08). Cho khai trin (1 + 2x)n = a0 + a1x + + anxn, trong n N v cc h s a0, a1, , an tha mn h thc a0 + a1

2+ + an

2n= 4096.

Tm h s ln nht trong cc s a0, a1, , an.Bi 4.15 (A-06). Tm h s ca s hng cha x26 trong khai trin nh thc Niuton

ca(

1

x4+ x7

)n, bit rng

C12n+1 + C22n+1 + + Cn2n+1 = 220 1

(n l s nguyn dng).

Bi 4.16 (A-04). Tm h s ca x8 trong khai trin thnh a thc ca [1 + x2(1x)]8.

Bi 4.17 (A-03). Tm h s ca s hng cha x8 trong khai trin nh thc Niuton

ca(

1

x3+x5)n

, bit rng

Cn+1n+4 Cnn+3 = 7(n+ 3)(n l s nguyn dng, x > 0).


3.36 (x 3)2 + y2 = 4A(1; 0), B(3; 2)

3.37 x220

+ y2

5= 1

3.38 x216

+ y2

163

= 1

3.39 A,B = (

2;22

); (

2;22

)

3.40 x29

+ y2

4= 1

3.41 (x 1)2 + (y 23

3)2 = 4

3

3.42 I(17;4)

3.43 M(2

7; 0);N(0;

21)gtnn(MN) = 7

3.44 A,B = (27; 43

7); (2

7;4

3

7)

Chng 4

T hp v s phc

4.1 Bi ton m . . . . . . . . . . . . . . . . . . . . . . . . . . 304.2 Cng thc t hp . . . . . . . . . . . . . . . . . . . . . . . . 314.3 ng thc t hp khi khai trin . . . . . . . . . . . . . . . 314.4 H s trong khai trin nh thc . . . . . . . . . . . . . . . . 324.5 S phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33p s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.1 Bi ton m

Bi 4.1 (B-12). Trong mt lp hc gm c 15 hc sinh nam v 10 hc sinh n.Gio vin gi ngu nhin 4 hc sinh ln bng gii bi tp. Tnh xc sut 4 hcsinh c gi c c nam v n.

Bi 4.2 (B-05). Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3n. Hi c bao nhiu cch phn cng i thanh nin tnh nguyn v gip 3tnh mim ni, sao cho mi tnh c 4 nam v 1 n?

Bi 4.3 (B-04). Trong mt mn hc, thy gio c 30 cu hi khc nhau gm 5 cuhi kh, 10 cu hi trung bnh, 15 cu hi d. T 30 cu hi c th lp c


bao nhiu kim tra, mi gm 5 cu hi khc nhau, sao cho trong mi nhtthit phi 3 loi cu hi (kh, trung bnh, d) v s cu hi d khng t hn 2?

Bi 4.4 (B-02). Cho a gic u A1A2 A2n (n 2, n nguyn) ni tip ngtrn (O). Bit rng s tam gic c cc nh l 3 trong 2n im A1, A2, , A2nnhiu gp 20 ln s hnh ch nht c cc nh l 4 trong 2n imA1, A2, , A2n,tm n.

Bi 4.5 (D-06). i thanh nhin xung kch ca mt trng ph thng c 12 hcsinh, gm 5 hc sinh lp A, 4 hc sinh lp B v 3 hc sinh lp C. Cn chn 4 hcsinh i lm nhim v, sao cho 4 hc sinh ny thuc khng qu 2 trong 3 lp trn.Hi c bao nhiu cch chn nh vy?

4.2 Cng thc t hp

Bi 4.6 (B-08). Cho n, k nguyn dng, k n. Chng minh rngn+ 1

n+ 2

(1

Ckn+1+

1

Ck+1n+1

)=

1

Ckn.

Bi 4.7 (B-06). Cho tp hp A gm n phn t (n 4). Bit rng, s tp con gm 4phn t ca A bng 20 ln s tp con gm 2 phn t ca A. Tm k {1, 2, , n}sao cho tp con gm k phn t ca A l ln nht.

Bi 4.8 (D-05). Tnh gi tr ca biu thcM =A4n+1 + 3A

3n

(n+ 1)!Bit rng C2n+1 + 2C

2n+2 + 2C

2n+3 + C

2n+4 = 149 (n l s nguyn dng).

4.3 ng thc t hp khi khai trin

Bi 4.9 (A-07). Chng minh rng :

1

2C12n +

1

4C32n +

1

6C52n + +

1

2nC2n12n =

22n 12n+ 1

(n l s nguyn dng).

tuyen tap de dh 2002-2012 theo chu de booklet

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