tai lieu on tap toan lop 12 nam 2011

8/7/2019 Tai Lieu on Tap Toan Lop 12 Nam 2011

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Ti li u lu hnh n i b n t v rn lu n k nn mn Ton cho h c sinh l 2 n thi tt n hi THPT

TI LIU N TP KIN THC V RN LUYN K NNG CBNCHO HC SINH THI TT NGHIP THPT V B TC THPT

MN: TON

Nhm to iu kin v nh hng cho hc sinh n tp thi tt nghip THPT v b tc THPTt hiu qu, SGio dc v o to pht hnh ti liu lu hnh ni b v n tp v rn luyn knng cbn cho hc sinh thi tt nghip THPT v b tc THPT cc mn Ng vn, Ton v TingAnh. Ti liu c tnh cht to iu kin hc sinh n tp kin thc v rn luyn k nng theo chun

kin thc, k nng ca chng trnh Gio dc ph thng v l ti liu tham kho gio vin n tpcho hc sinh (Ti liu khng phi l cng n thi tt nghip THPT). Ban bin tp rt mong cs gp ca cn b, gio vin v cc em hc sinh ti liu ngy cng hon chnh hn.

Phn th nht:TM TT L THUYT V BI TP RN LUYN

I. KHO ST V V TH HM S

A. Cc bc kho st v v th hm s.1. Kho st v v th hm s bc ba 3 2 , 0y ax bx cx d a= + + + .

(1) Tp xc nh: D = R.(2) S bin thin:

* Chiu bin thin:- o hm y' = 3ax2 + 2bx + c.- Xt du y' t suy ra sng bin, nghch bin ca hm s.

* Cc tr:- Nu qua x0 m y' i du t (+) sang (-) th hm st cc i ti x0 ; yC = y(x0).- Nu qua x0 m y' i du t (-) sang (+) th hm st cc tiu ti x0 ;yCT = y(x0).

* Gii hn:

- 3 2, 0

lim ( ), 0xa

ax bx cx d a+

+ >+ + + = + + + = + + + =


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3. Kho st v v th hm phn thc : ( 0)ax b

y accx d

+=

+.

(1) Tp xc nh: D = \d

Rc

.

(2) S bin thin:* Chiu bin thin:

- o hm2

( )

ad cby

cx d

=

+

.

- Nu y' > 0 th hm sng bin trn mi khong ( ;d

c ) ,( ;

d

c + ).

- Nu y' < 0 th hm s nghch bin trn mi khong ( ;d

c ),( ;

d

c + ).

* Cc tr: Hm s khng c cc tr.* Gii hn v tim cn:

- Tm cc gii hn khi , ( )d

x xc

.

- th hm s nhn ng thng x =d

c lm tim cn ng v ng thng

y =a

clm tim cn ngang.

* Bng bin thin:(3) V th:

- V cc ng tim cn ln h trc to.- Tm giao im ca th vi cc trc to, cc im c bit v biu din chng ln h trcta .

B. Bi tp luyn tp.1. Cho hm s 3 2( ) 3 4y f x x x= = + .

a) Kho st v v th hm s .b) Bin lun s nghim phng trnh 3 23 0x x m+ + = tu theo gi tr ca tham s m.(S: m0 :1 nghim; m=-4 hoc m=0: 2 nghim; -4


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7. Cho hm s 4 21 3( )2 2

y f x x x= = + .

a) Kho st v v th hm s .b) Bin lun s nghim phng trnh 24 2 0x x m+ + = tu theo gi tr ca tham s m.

(S: m>0: v nghim; m=0: 1nghim; m1: khng c giao im; m=1: 1giao im; m 0)

10. Cho hm s 2 1( )1

xy f x

x

+= =

+(H).

a) Kho st v v th hm s .b) M l im bt k thuc (H). I l giao im hai tim cn . Tip tuyn ti M ct hai tim cn

ti A v B.i. Chng minh M l trung im AB.ii. Chng minh din tch tam gic IAB khng i.iii. Tm M IA+IB nh nht. (S: M(0;1) hoc M(-2;3).)

11. Cho hm s2

1y

x=

; gi th hm s l (H)

a) Kho st s bin thin v v th hm s.b) Vit phng trnh tip tuyn vi th (H) bit tip tuyn song song vi ng thng d:

2 5 0x y+ = .

c) Tm gi tr ln nht ca hm s 4 6os siny c x x= + .12. Cho hm s 3 23 4y x x= + ; gi th hm s l (C)

a) Kho st s bin thin v v th hm s.b) Trn (C) ly im A c honh 2Ax = . Vit phng trnh ng thng d qua A v d tip xc vi (C).

c) Tm gi tr ln nht, gi tr nh nht ca hm s 3 2cosy x x= + trn 0;2

.

13 : Cho hm s 3 3 1y x x= ; gi th hm s l (C).a) Kho st s bin thin v v th hm s.

b) Dng th (C) bin lun theo m s nghim ca phng trnh 3 3 1 0x x m = .c) Tm gi tr ln nht, gi tr nh nht ca hm s 2cos 2 sin .cos 4y x x x= + .

14. Cho hm s4

4y

x=

; gi th hm s l (C).

a) Kho st s bin thin v v th hm s.b) Vit phng trnh tip tuyn (d) ca (C) ti im thuc (C) c honh l 3.

c)Tm gi tr ln nht, gi tr nh nht ca hm s 24y x x= + .

15. Cho hm s3

2y

x=

.

a) Kho st s bin thin v v th (C) ca hm s.b) Tm k ng thng (d) i qua gc ta , c h s gc k, ct (C) ti hai im phn bit.c) Tm gi tr ln nht, gi tr nh nht ca hm s 4 4sin cosy x x= +

16. Cho hm s2 3

1

xy

x

+=

+c th (C).


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a) Kho st s bin thin v v th (C).b) Tm cc im trn th ( )C ca hm s c ta l nhng s nguyn.

c)Tm gi tr ln nht v gi tr nh nht ca hm s 2(3 ) 1y x x= + trn on [0;2].17. Cho hm s y = x4 + 2x2 + 3 c th (C)

a) Kho st v v th (C) ca hm s.b) Da vo th (C), hy xc nh cc gi tr ca m phng trnh x4 2x2 + m = 0

c 4 nghim phn bit.

c) Tm gi tr ln nht, nh nht (nu c) ca hm s2

41

x xyx+ +=

+trn [0;3].

18. Cho hm s 3 23 4y x x= + ; gi th hm s l (C)a) Kho st s bin thin v v th ( )C ca hm s cho.

b) Tnh din tch hnh phng gii hn bi (C) v trc honh.c) Tm gi tr ln nht, gi tr nh nht ca hm s ( ) 3 23 2 5 1f x x x x= + trn [ ]0;3 .

19. Cho hm s 4 21

4 62

y x x= +

a) Kho st s bin thin v v th ca hm s cho.

b) Tm cc gi tr ca tham s m phng trnh 4 21 4 02

x x m + + = c 4 nghim phn bit.

c) Tm gi tr ln nht, gi tr nh nht ca hm s ( )4

1f x x

x= +

+trn on [ ]0;2 .

20. Cho hm s y = x4 - 2mx2 + 2m + m4 ; (l)a) Kho st s bin thin v v th ca hm sng vi m =1.b) Tm m th hm s (l) c 3 im cc tr.

c) Tm gi tr ln nht, nh nht (nu c) ca hm s:2

1

1

xy

x x

+=

+

II. TM GI TR LN NHT, GI TR NH NHTCA HM S ( )y f x= trn D

A. Hai cch thng dng.Cch 1: - Lp bng bin thin ca hm s ( )f x trn D.

- T bng bin thin suy ra GTLN, GTNN .Cch 2: Nu ( )f x lin tc trn D = [a;b]

- Tm cc im 1 2, , , nx x x trn khong (a;b) m ti , ( )f x bng 0 hoc , ( )f x khng

tn ti.- Tnh 1 2( ), ( ), ( ), , ( ), ( )nf a f x f x f x f b .- Tm s ln nht M v s nh nht m trong cc s trn.

- Ta c [ ; ] [ ; ]min ( ) ,max ( )a b a bf x m f x M = = .B. Bi tp.

1. Tm GTLN, GTNN ca hm s 3 2( ) 9f x x x x= + + trn on [-3;5].(S:

[ 3;5] [ 3;5]min ( ) ( 3) 45,max ( ) (5) 195f x f f x f

= = = = )

2. Tm GTLN, GTNN ca hm s 4( )2

f x xx

= +

trn on [3;5].

(S:[3;5] [3;5]

min ( ) (4) 6,max ( ) (3) 7f x f f x f = = = = )

3.Tm GTLN, GTNN ca hm s ( ) 2 1

2

2f x x x= + + trn khong

5

( ; )2 .(S:

5( ; )

2

max ( ) ( 3) 9f x f

= = , ( )f x khng c GTNN )

4. Tm GTLN, GTNN ca hm s 2( ) 8f x x x= + .


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(S: max ( ) (2) 4,min ( ) ( 8) 8f x f f x f = = = = )

5. Tm GTLN, GTNN ca hm s ( ) 9 3f x x= trn on [-2;2].(S:

[ 2;2] [ 2;2]min ( ) (2) 3,max ( ) ( 2) 15f x f f x f

= = = = )

6. Tm GTLN, GTNN ca hm s 2 1( ) sin cos2

f x x x= + .

(S:23 6min ( )

742

6

x kf x

x k

= +=

= +

,2

3max ( ) 2

2f x x k

= = + )

7. Tm GTLN, GTNN ca hm s 3 2( ) cos cos 1f x x x= + trn on 3[0; ]2

.

(S: min ( ) 1

3

2

2

x

f x x

x

== =

=

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

8. Tm GTLN, GTNN ca hm s 4 4( ) cos sinf x x x= + .(S:

1min ( )

2 4 2f x x k

= = + , max ( ) 1

2f x x k

= = )

9. Tm GTLN, GTNN ca hm s ( ) xx

ef x

e e=

+trn on [ ln 2 ; ln4] .

(S:[ln 2;ln 4] [ln 2;ln 4]

2min ( ) (ln 2) , max ( ) (ln 4)

2 4

4f x f f x f

e e= = = =

+ +)

10. Tm GTLN, GTNN ca hm s 2( ) ln( 5 )f x x x= + + trn on [-2;2].(S:

[ 2;2] [ 2;2]min ( ) ( 2) 0, max ( ) (2) ln 5f x f f x f

= = = = )

III. NGUYN HM TCH PHN

1. ( )F x l mt nguyn hm ca ( ) tan .sin 2f x x x=

a) Tnh ( )6

''F

b) Bit th hm s y = ( )F x ct trc tung Oy ti im c tung bng 5. Hy xc nh ( )Fx .2. 3( ) ( .sin .cos )xF x a x b xe += l mt nguyn hm ca

3( ) (3.sin 4.cos )xf x x xe = . Hy xc nh cc gi tr ca a v b.3. ( )F x , ( )G x ln lt l cc nguyn hm ca

22

1( )

sin . osf x

x c x= v

3 1( )

x

xe

g xe

+= . Bit

1( ) ( ln 2 )

4 2F G

= = .

Hy tnh :3

( )4

F

v ( )0G .

4. Tm nguyn hm ca hm s21

: ( )1 2

xf f x

x

=

5. Tm nguyn hm ca hm s 21 1 1: ( ) sin osf f x cx x x=

6. Tm nguyn hm ca hm s1

: ( )2

x

xe

f f xe

+=

+

7. Tm nguyn hm ca hm s 3: ( ) 3 logxf f x x= +


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8. Tm h nguyn hm : I = 3 342011( ) . .x xe e dx+ ; J = 2011(1 ) .dxx x 9. Tm h nguyn hm : I =

1( .ln ).

.lnx x dx

x x

10. Tnh tch phn : I =1

4

0

(1 2 ) .x dx ; J =2

3

1

20

.

( 1)

x dx

x +

11. Chng t : 2 21

0

(1 ) .x x dx

= 3 3ln2 ln15

log log 26

e+

12. Tnh tch phn : J = ( )4

2

0

cos3 .s inx tan 3x x dx

+


2 2

1

2x xe e d x

+ ; J =2

4 2

2

2 1x x d x

+


0

os( )3 3 3

x xc dx

; J = 2

1

( 1) lne

x x x dx+ +

15. Tnh tch phn : I =ln2

0

3x

xdx

e

; I =2

0

cosxe x dx


lnln3

xx dxe + ; J =21

0

..

2

xxdx

e


2

0.tanx x dx

; I =2

2

0

.cos .x x dx

; J = 2 32

0

sin . osx c x d x


30

cos .(2 sin )x dxx

+ ; J =6

0os ( ).cos 2 .6c x x dx


0

sin 5 2cos .x x dx

; J = 21

3 ln.

(1 )

xdx

x+

20. Tnh tch phn : I = 21

22 4. ln ( 1) .x dx

xx + + ; I = 0

cos( ).sin .x x dxxe

+


2

0

tansin2.

os

xxdx

c x

e

+


3

0

. 1 . dxx x+ ; J = 31

2

0

. 1 .x dxx +


4 2 1.

1 2 1

xdx

x

++ + ; I =

32

0

3 x d x

23. Tnh tch phn : K =2

32

0

8 1

4 3

xd x

x

++ ; 2

0

1

9

6 5

xI d x

x x

=

+

24. Tnh tch phn : 2

1

0

10 3

6 9

xI d xx x= + ; 2

4

3

4 11

6 10

xI d xx x= +


0 1 sin cos

dx

x x

+ +


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26. a) Cho hm s y = f (x) l hm s l v lin tc trn on [ ];a a .

Chng minh rng : ( ) 0a

a

f x dx

= .

b) Vn dng kt qu trn, hy tnh tch phn: G =34

2

4

3 sin 6

os

x x xdx

c x

+

27. a) Cho hm s y = f (x) lin tc trn on [ ];a b .

Chng minh rng : ( ) ( )b b

a a

f x dx f a b x dx= + .

b) Vn dng kt qu trn, hy tnh tch phn : K =4

0(1 tan )ln x dx

+

28. a) Tnh din tch hnh phng (H) gii hn bi (C):2

1

xy

x

=

+v hai trc ta .

b) Tnh th tch khi trn xoay c to thnh khi quay hnh (H) quanh trc Ox.

29. Tnh din tch hnh phng ( H ) gii hn bi cc ng sau :4 21 ; 3 ; 0 ; 2 3x x y y x x= = = = + + 30. Tnh th tch khi trn xoay do hnh phng gii hn bi cc ng sau quay quanh trc Ox :

a) sin , 0 , 0 ,2 4

xy y x x

= = = = ; b) ln , 0 ,y x y x e= = = .

IV. LGART

1. a) Rt gn biu thc sau: E = 3 4 5 6 8log 2.log 3.log 4. log 5. log 7

b) Cho bit lg2 = a, lg3 = b. Tnh lg24

25theo a v b

2. Tm tp xc nh ca hm s: y = 13

1 log ( 2)x .

3. Cho hm s y = 2ln( 1)x + .a) Tnh y.b) Gii phng trnh . 1 0y + = .

4. Gii phng trnh log3(x + 1) - log 13

(x + 3) = 1.

5. Gii phng trnh: 2 4log log ( 3) 2x x = 6. Gii phng trnh: 26log 1 log 2xx = +

7. Gii phng trnh ( ) ( )22

2 12

log 1 log 1 5 0x x+ + =

8. Gii bt phng trnh 13

3 1log 1

2

x

x

>

+

9. Gii bt phng trnh1 1

11 log logx x

+ >

10. Gii bt phng trnh 20,2 0,2log 5log 6x x

V. HM SM

1. Gii phng trnh:2 222 2 3x x x x + =

HD gii : t 22 0x xt t= > .

Khi phng trnh trthnh: 24

3 3 4 0 ( 1)( 4) 0 4t t t t t t t

= = + = = (v 0t> ).

2 22 4 2 1 2x x x x x hay x = = = = . Do phng trnh c 2 nghim l: 1 ; 2x x= = .


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2. Gii h phng trnh:

3 2

1

2 5 4

4 2

2 2

x

x x

x

y y

y+

= +

= +

HD gii: h phng trnh cho3 2 3 22 5 4 5 4 0

2 2 0

x

x x

y y y y y

y y

= + =

= = >

0 1 4 1 4

2 0 0 2x

y hay y hay y y yhay

y x x

= = = = =

= > = =

3. Tm a bt phng trnh ( ) 2.9 1 .3 1 0x xa a a++ + > c nghim ng vi mi x .

HD gii : t 3 0xt= > . BPT 2 29( 1) 1 0 ( 9 1) 9 1at a t a a t t t + + > + + > +

( )29 1

19 1

ta

t t

+ >

+ +. Bt phng trnh cho sc nghim ng ( )1x ng 0t > .

Xt hm s ( ) 29 1

9 1

tf t

t t

+=

+ +. Ta c : ( )

( )

2

22

9 2' 0, 0

9 1

t tf t t

t t

= < >

+ +

Do xt bng bin thin ta c ( )1 ng ( )0 max 1t a f t a > .

4. Gii phng trnh:3 1

125 50 2x x x+

+ = HD gii :

125 50 125 252 2 0

8 8 8 4

x x x x

PT + = + =

t5

02

x

t = >

. PT thnh 3 2 2 0t t+ = . Gii phng trnh trn ta c 1t= suy ra 0x = .

5. Tm m bt phng trnh2 2 22 2 2.9 (2 1)6 .4 0x x x x x xm m m + + nghim ng vi mi x tha

mn iu kin1

2x

HD gii : BPT( )

( )

2 22 2 2

3 3. (2 1) 102 2

x x x x

m m m

+ +

t

223

2

x x

t

=

do iu kin1

2x ( )

223 3

' 4 1 .ln2 2

x x

t x

=

lun cng du vi 4 1x .

t ly cc gi tr trong [1; )+ . ( ) ( )2 21 2(2 1) 0 ( 2 1) 1mt m t m m t t + + +

( )1 ng ( )1

22

x ng [1; )t +( )

2

1, 1 0

1m t m

t >

6. Gii phng trnh: 3 5 6 2x x x+ = + HD gii : t

( )3 5 6 2x xf x x

= + . Phng trnh tng ng vi:

( )0f x

=

D thy phng trnh c 0; 1x x= = l nghimTa c ( )' .ln 3 .3 6ln55x xf x += v ( ) 2 2" .ln 3 .3 n 5 05 lx xf x += > vi x

( ) ( )min ' ; min ' 6x x

f x f x+

= + =

Suy ra ( )'f x l hm lin tc, ng bin v nhn c gi tr m, c gi tr dng trn nn

phng trnh ( )' 0f x = c nghim duy nht ox .

T bng bin thin ca hm ( )f x ( ) 0f x = c khng qu hai nghim.Vy phng trnh c ng hai nghim : 0; 1x x= =

Ch : C th chng minh phng trnh ( )' 0f x = c nghim nh sau :Ta c : ( )' 0 ln 3 ln 5 5 0f = + < v ( )' 1 3ln 3 5ln 5 6 0f = + >

Suy ra phng trnh ( )' 0f x = c nghim duy nht ( )0;1ox .


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7. Gii phng trnh:1

5 .8 500x

x x

=

HD gii : ( )1

3( 1) 33

3 2 3 35 .2 5 .2 5 2 5 2 xx x

xx x xx xPT

= = =

( )1

1 1

33

3

5

3 0 315 5.2 1

log 22 5.2 1x

x x

xx

xx x

x

= = = = = =

8. Gii phng trnh:2 25 1 54 12.2 8 0x x x x + =

HD gii : t2

25

2

32 5 12 ( 0) 9

4 5 2 4

x xxt x x

t tt xx x

== = = > = = =

9. Gii phng trnh: ( ) ( )2 3 2 3 4x x

+ + =

HD gii : t ( )2 3x

=t (t>0). phng trnh trthnh :2 3 21

422 3

t xt

xt t

= =+ = = = +

10. Gii phng trnh: ( ) ( )7 5 2 ( 2 5) 3 2 2 3(1 2) 1 2 0x x

x+ + + + + + = .

HD gii : t (1 2) ; 0xt t= + > 3 2 2( 2 5) 3 1 2 0 ( 1)( ( 2 4) 2 1) 0PT t t t t t t + + + = + + =

1 0

3 2 2 2

11 2

t x

t x

xt

= = = = == +

11. Gii phng trnh: ( )3 2 ( 3 2) ( 5)x

x x + + =

HD gii: PT3 2 3 2

15 5

x x +

+ =

. t3 2 3 2

, 0 1; , 15 5

u u v v +

= < < = >

+ Nu 0 : 0; 1 1x xx u v VT > > + Nu 0 : 1; 0 1x xx u v VT < > > . Vy PT v nghim.

12. Gii phng trnh: 2 23.16 (3 10)4 3x xx x + + HD gii : t 24 , ( 0).x t t = > PT trthnh : 23 (3 10) 3 0t x t x+ + =

24

2

114 2 log 3

332

3 4 3

x

x

xt

xt x x

= = = == =

13. Tm m phng trnh .2 2 5 0x x

m

+ = c nghim duy nht.HD gii: t 2 , .xt t o= > Pt trthnh : ( )2

15 0 ( ) 5 1 0 *mt f t mt t

t+ = = + =

+ Nu1

0 :5

m t= = (t.m) ;

+ Nu 0 :m PT cho c nghim duy nht khi v ch khi phng trnh ( )* c duy nht 1nghim dng.

Xt 3 trng hp :1 2

1 2

1 2

0 0 00 khng c 25

0 v 00 4

t t m mt t m

m

mt t

< <


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HD gii : PT5 1 5 1

12 2

x x +

+ =

t t =5 1

2

x +

(t>0) phng trnh trthnh : 21 0a

t t t at

+ = + =

p s :1

04

a hay a = .

15. Tm m phng trnh .16 2.81 5.36x x xm + = c nghim duy nht.

HD gii: t9

; 04

x

t t = >

. Phng trnh trthnh 22 5 0 (*)t t m + = 22 5m t t = +

Phng trnh cho c nghim duy nht khi v ch khi phng trnh (*) c ng mt nghim

dng. Kho st hm s 22 5y t t = + trn (0 : +) ta c25

; 08

m m= .

VI. S PHC

1. Xc nh cc s phc biu din bi cc nh ca mt lc gic u c tm l gc ta O trongmt phng phc bit rng mt nh biu din s i.

2. Chng minh: a) S phc Z l s thc khi v ch khi Z Z= .

b) S phc Z l so khi v ch khi Z Z= .3. Chng minh rng mi s phc Z1 , Z 2 ta c:

1 2 1 2Z Z Z Z + = + ; 1 2 1 2.Z Z Z Z = 4. Tm s phc Z tha mn mt trong cc iu kin sau:

a) 2Z = v Z l so.

b) 5Z = v phn thc bng hai ln phn o.

5. Chng minh:

3

(1 ) 3 2 5i i i+ + = + .6. Chng minh: 7 7

1 1 12

ii i

= .

7. Trn mt phng phc, hy tm tp hp im biu din s phc z tha mn bt ng thc:

1 1z i < .8. Trn mt phng phc, hy tm tp hp im biu din s phc z tha mn bt ng thc

1 2z< .9. Tm m un s phc: 34 3 (1 )z i i= + .

10. Cho s phc z tha 2 8z z i+ = + . Hy tm2

z .11. Tm 2 s thc x, y tha mn 3(3 5 ) (1 2 ) 9 14x i y i i+ + = + .12. Gii phng trnh sau trn tp hp s phc: (3 + 4i)z + (1 3i) = 2 + 5i

13. Gii phng trnh sau trn tp hp s phc: z 2 4z + 6 = 0.

14. Gii phng trnh sau trn tp hp s phc: z 2 8(1 i)z + 63 16i = 0

15. Gii phng trnh sau trn tp hp s phc: z 3 8 = 0.

16. Gii phng trnh sau trn tp hp s phc: z 4 + 4z 2 5 = 0.

17. Gi z1 v z 2 l hai nghim ca phng trnh: z 2 + 2z + 10 = 0. Tnh 21 2A z z= + .

18. Tm tp hp im biu din s phc z bit rng z l acqumen3

.


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19. Chng minh rng s phc 2 3z i= + + c 1 acqumen l .212

k + .

20. Tnh gi tr 0 2 4 2008 20102010 2010 2010 2010 2010...A C C C C C = + + .

VII. HNH HC KHNG GIAN

1. Cho hnh chp SABC c SA (ABC) SA= 3a , ABC u cnh bng a. M, N ln lt l hnh

chiu ca A trn SB, SCa) CMR MN song song mp(ABC)b) Tnh th tch khi chp ABCNM

2. Cho hnh chp SABC c ng cao SA = a; ABC vung cn, AB = BC = a, gi B ltrung im cnh SB , C l chn ng cao h t A ca SAC .

a) CMR SC (ABC).b) Tnh th tch khi chp S. ABC.

3. Cho hnh chp SABC c y l tam gic cn, AB = AC ;

BAC= 2 ; hai mt bn SAB, SACcng vung gc vi y, cnh bn SB= b to vi y gc . Tnh th tch khi chp SABC.

4. Cho hnh chp t gic u S.ABCD c cc cnh bn to vi y 1 gc .a) V thit din qua AC vung gc SDb) Tnh t s th tch 2 phn ca hnh chp b chia ct bi thit din trn.

5. Tnh th tch khi hp ABCD.ABCD bit AABD l khi t din u cnh a.6. Cho hnh chp tam gic u c cnh y bng a mt bn to vi y 1 gc 060 .

a) Tnh th tch khi chp.b) Tnh din tch xung quanh v th tch khi nn ngoi tip khi chp.

7. Cho hnh chp SABCD c y l hnh vung cnh a, SA (ABCD), SC hp vi mt phng y1 gc 060 . Gi H, I , K ln lt l hnh chiu ca A trn SB, SC, SD.

a) Chng minh 7 im A, B, C, D, H, I, K thuc mt mt cu. Tnh th tch khi cu .b) Tnh th tch khi chp SABCD

8. Cho t din ABCD c AD=AC = a , AB = 2a, AD (ABC) , ABC vung C.a) Tnh th tch khi t din ABCDb) Tnh din tch mt cu ngoi tip t din ABCD

9. Cho hnh chp SABC c SA = SB= SC = a ,

ASB =

BSC= 060 ,

ASC= 090a) CMR ABC vung . Tnh th tch khi chp S.ABCb) Xc nh tm v bn knh mt cu ngoi tip hnh chp SABC

10. Cho hnh chp SABC c hnh chiu ca nh S trn (ABC) l trung im ca cnh AC, SA =

SB = SC = a , SB to vi y 1 gc 060 v

ABC= 030 .a) CMR ABC vung . Tnh th tch khi chp S.ABC

b) Xc nh tm v bn knh mt cu ngoi tip hnh chp SABC11. Cho hnh chp S.ABC c y ABC l tam gic u, cc cnh bn u bng a, gc gia cnhbn v mt y bng 030 . Tnh th tch khi chp S.ABC theo a.

12. Cho lng tr tam gic ABC.ABC c y ABC l tam gic u cnh a, AA = 2a, ng thngAA to vi mt phng (ABC) mt gc 060 . Tnh th tch ca khi lng tr.

13. Cho hnh chp u S.ABC c di cnh y bng a, cnh bn to vi mt phng y mtgc 060 . Tnh th tch khi chp trn.

14. Cho hnh lng trng ABC.ABC c y ABC l tam gic vung ti B,

060ACB = , cnhBC = a, ng cho AB to vi mt phng (ABC) mt gc 300. Tnh th tch khi lng trABC.ABC .

15. Cho hnh chp S.ABCD c y ABCD l hnh ch nht, cnh BC=2a, SA= a, SAmp(ABCD), SB hp vi mt y mt gc 450. Tnh th tch ca khi cu ngoi tip hnh chpS.ABCD.


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16. Cho khi chp S.ABC c y ABC l tam gic vung ti B, cnh AB = a, BC = 2a; SA vung

gc vi mt phng (ABC) v SA = 2a .Gi A v B ln lt trung im ca SA v SB. Mt phng(CAB) chia hnh chp thnh hai khi a din. Tnh th tch ca hai khi a din .

17. Cho hnh chp u S.ABCD c cnh y bng a, gc gia cnh bn v mt y bng 600. Tnhdin tch xung quanh v th tch ca hnh nn c nh S v y l ng trn ngoi tip y hnh chp cho.

18. Cho khi chp tam gic u S.ABC cnh y AB = a, gc gia mt bn v mt y l 60o .Tnh th tch khi chp theo a.

19. Cho hnh tr c thit din qua trc l mt hnh vung cnh a. Tnh din tch xung quanh, dintch ton phn ca hnh tr v th tch ca khi tr.

20. Mt hnh nn c nh S, khong cch t tm O ca y n dy cung AB ca y bng a,

30SAO = o ,

60SAB = o . Tnh di ng sinh theo a .

VIII. PHNG PHP TA TRONG KHNG GIAN

Ch 1: H TA TRONG KHNG GIAN

1. (1; 2;5); (3; 4;1); (1; 2; 5)a b c= = =

a) Tm ta 2 3 4d a b c= + .Tnh di ca d

b) Tnh ( 2 )a b c+ Hng dn gii:

a) ( 3;0; 13) ; 178d d= = ; b) ( 2 ) 2 16 2( 20) 24a b c ab ac+ = + = + =

2. Cho A(1;2;3) ; B(-1;3;4) ; C(0;4;1)a) Chng minh rng : A; B; C khng thng hngb) Tm ta trng G ca tam gic ABCc) Tnh chu vi v din tch tam gic ABCd) Tnh gc A ca tam gic ABC.

Gii: a) ( 2;1;1) ; ( 1;2; 2)AB AC = = Hai vec tny khng cng phng nn 3 im A ; B ; C khng thng hng.

b) Gi G l trng tm tam gic ABC.Ta c :1

( ).3

OG OA OB OC = + +uuur uuur uuur uuur

O l gc ta .

T ta c8

(0;3; )3

G

c) Chu vi tam gic ABC : 6 3 11AB AC BC + + = + +

Din tch tam gic ABC : ( )2

2 21 5 2. .2 2

S AB AC AB AC = =uuur uuur

d). 2

cos cos( ; ). 3 6

AB ACA AB AC

AB AC= = =

uuur uuur

3. Tm tm v bn knh mt cu ( S ) c phng trnh :a) 2 2 2 4 2 2 1 0x y z x y z+ + + + = b) 2 2 22 2 2 6 8 4 8 0x y z x y z+ + + + =

Gii: a) Phng trnh mt cu cho tng ng phng trnh: 2 2 2( 2) ( 1) ( 1) 5x y z + + + =

Vy mt cu cho c tm I(2;-1;1); bn knh R = 5 .

b) Phng trnh mt cu cho tng ng PT:2

2 23 45( 2) ( 1)2 4

x y z + + + + =

Vy mt cu cho c tm3

; 2; 12

I

, bn knh R =3 5

2.

4. Vit phng trnh mt cu (S ) trong mi trng hp sau :a) ng knh AB vi A(1;2;-3) ; B(5;4;1)b) Mt cu (S) c tm M(2;1;4) v tip xc mt phng ( P ) : : 3x + 4y+ z 5 = 0c) Mt cu ( S ) ngoi tip t din ABCD vi A(1;-2;-1); B(4;3;3); C(5;6;-1); D(-3;3;-4).


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Gii: a) Mt cu ( S) c ng knh AB nn tm I ca mt cu (S) l trung im I ca AB.

Do I ( 3;3;-1) ; R =2

AB= 3. Phng trnh mt cu ( S ) : 2 2 2( 3) ( 3) ( 1) 9x y z + + + = .

b) Mt cu (S) tm M, tip xc mt phng (P) nn bn knh: R=d(M;(P))=6 4 4 5 9

9 16 1 26

+ + =

+ +

Phng trnh mt cu (S) : 2 2 281

( 2) ( 1) ( 4)26

x y z + + =

c) Phng trnh mt cu ( S) c dng :2 2 2

2 2 2 0x y z ax by cz d + + + + + + = Mt cu ( S) i qua 4 im A ; B ;C ; D nn ta c :

6 2 4 2 0

34 8 6 6 0

62 10 12 2 0

34 6 6 8 0

a b c d

a b c d

a b c d

a b c d

+ + = + + + + = + + + = + + +

Gii h phng trnh trn ta c : a = -1; b= -3; c = 1; d= -14.

Vy phng trnh mt cu (S) ngoi tip t din ABCD l: 2 2 2 2 6 2 14 0x y z x y z+ + + = .5. Vit phng trnh mt cu ( S ) trong mi trng hp sau :

a) Mt cu ( S ) c tm B(-1;4;5) v i qua A(1;2;3)

b) Mt cu ( S) c tm A(1;2;5) v tip xc ng thng ( d) :1 32

3 2

x ty t

z t

= + = = +

c) Mt cu ( S) c tm A(1;4;3), ct mt phng Oxz theo mt ng trn ng knh bng 8.

Gii: a) Mt cu tm B, i qua A nn c bn knh R = BA = 4 4 4 2 3+ + = .Phng trnhmt cu ( S ) : 2 2 2( 1) ( 4) ( 5) 12x y z+ + + =

b) Gi H l hnh chiu ca I trn d. H thuc (d) nn ta im H (1+ 3t; 2- t; 3 + 2t).

VTCP ca ( d) : (3; 1;2)u = ; (3 ; ;2 2)AH t t t = .

. 0 9 4 4 0AH u AH u t t t = + + =

2

7t = .

Bn knh R = AH =36 4 100 140

49 49 49 49+ + = .

Phng trnh mt cu ( S) : 2 2 2140

( 1) ( 2) ( 5)49

x y z + + =

(Hc sinh ban t nhin c th p dng cng thc khong cch t mt im n 1 ng thng tnh bn knh).

c) Gi H l hnh chiu vung gc ca A trn mt phng oxz, ta c AH = 4 .Gi R l bnknh mt cu ( S), ta c 2 2 16 16 16 32R AH = + = + = .

Phng trnh mt cu (S) :2 2 2

( 1) ( 4) ( 3) 32x y z + + = Ch 2: NG THNG

A. Kin thc cbn:1. Phng trnh tng qut ca mt phng:

Ax + By +Cz + D = 0 vi 2 2 2 0A B C + + > , VTPT ca (P) ( ; ; )n A B C =

2. Phng trnh mt phng i qua mt im M0(x0; y0; z0) v c VTPT ca (P) ( ; ; )n A B C = A(x x0) + B(y-y0) +C(z-z0) = 0, khai trin a pt v dng: Ax + By +Cz + D = 0

3. Mt phng (P) qua A(a; 0; 0), B(0; b; 0), C(0; 0; c) c phng trnh dng: 1x y z

a b c+ + = , vi a,

b, c khc 0B. Cc dng ton:Dng 1: Vit phng trnh mt phng ( )P i qua 1 im M0(x0; y0; z0) v song song vi 1 mt phng

( ) : 0Q Ax By Cz D+ + + = cho trc.* Phng php gii:


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Cch 1: Thc hin theo cc bc sau:

1. VTPT ca ( ) l ( ); ;n A B C =

2. ( )P //( )Q nn VTPT ca mt phng ( )P l ( ); ;P Qn n A B C = =

3. Phng trnh mt phng ( )P : A(x x0) + B(y-y0) +C(z-z0) = 0,

Cch 2:1. Mt phng ( )P //( )Q nn phng trnh ( )P c dng: Ax + By + Cz + D = 0 (*), vi D D.

2. V ( )P qua 1 im M0(x0; y0; z0) nn thay ta M0(x0; y0; z0) vo(*) tm c D.* Bi tp p dng :

1. Vit phng trnh mt phng ( )P

a) qua A( 1; 3; -2) v song song mt phng ( )Q : 2x + 5y 3z 1 = 0

b) qua B(-1; 4; 0) v song song mt phng ( )Q ( ) : 2x + 3y 4z + 4 = 0

2. Cho 4 im A(2; 3; 1), B(1; 1; -2), C(2; 1; 0), D(0; -1; 2).a) CMR A, B, C, D l bn nh ca mt t dinb) Vit phng trnh mt phng i qua D v song song mt phng (ABC)

Dng 2: Vit phng trnh mt phng ( ) i qua 3 im A, B, C khng thng hng

* Phng php gii:* Tm ta cc vect: ;AB AC

* Vectphp tuyn ca ( ) l : , n = AB AC =( m; n; p)

* im thuc mt phng: A (hoc B hoc C).* Vit phng trnh mt phng qua 1 im v c VTPT n =( m; n; p)

* Bi tp p dng :3. Vit phng trnh mp i qua 3 im A,B,C trong cc trng hp sau

1. A(3,-2,1) B(1,-1,2) C(1,3,4)2. A(1,-1,4) B(2,5,-3) C(1,-3,7)

4. Cho hnh hp ABCD.ABCD c B(-1; 0; 3), D(1; 2; 1), A(2; -1; 1), C(-2; 5; -3).a) Tm ta cc nh cn li ca hnh hp ABCD.ABCD.b) Vit phng trnh mt phng (ABD) v (CBD), chng minh 2 mt phng ny song song.c) Vit phng trnh 2 mt phng (AACC) v (BBDD).

5. Cho 4 im A(-1; 2; 0); B(-3; 0; 2), C(1; 2; 3), D(0; 3; -2). Vit phng trnh mt phng (ABC)v suy ra 4 im A, B, C, D to thnh 1 t din.

Dng 3: Vit phng trnh mt phng ( ) i qua im M v vung gc vi ng thng * Phng php gii:

1. Tm VTCP ca l u 2. V ( ) nn ( ) c VTPT n u=

3. p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT* Bi tp p dng :

6. Vit phng trnh mt phng i qua M v vung gc vi ng thng trong cc trng hpsau:

a) M(3; -2; 1) v 1 2

3

3 2

x t

y t

z t

= + = = +

b) M(0; 2; 3) v 2 2 1

3 1 1

x y z + = =

7. Cho ng thng d:1 22

3

x ty t

z t

= = + =

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

a) Tm ta giao im A ca d v ( )


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b) Vit phng trnh mt phng (Q) qua A v vung gc vi d.8. Vit phng trnh mt phng i qua M(-1; 2; 1) v vung gc vi giao tuyn ca 2 mt phng

(P): 3x + 2y 2z + 8 = 0 v (Q): 2x y + 3z + 7 = 0Dng 4: Vit phng trnh mt phng ( ) cha ng thng , vung gc vi mt phng ( ) * Phng php gii

1. Tm VTPT ca ( ) l n

2. Tm VTCP ca l u

3. VTPT ca mt phng ( ) l: ;n n u =

4. Ly mt im M trn 5. p dng cch vit phng trnh mt phng i qua 1 i m v c 1 VTPT

* Bi tp p dng :9. Vit phng trnh mp cha ng thng d v vung gc vi mt phng (P) trong cc trng hp

sau :

a)

1 3

3 9

5

x t

y t

z t

= + = =

(P) : x + 7y 2z + 12 = 0

b) d : 1 2 33 1 2

x y z + += =

(P) : 2x + y + 3z 1 = 0

c) d l giao tuyn ca hai mt phng (R): 2x y + 3z + 1 = 0 v (Q): x + y z + 5 = 0; ( P): 3x y + 1 = 0

Dng 5: Vit phng trnh mt phng ( ) qua hai im A, B v vung gc vi mt phng ( )

* Bi tp p dng :10. Vit phng trnh mt phng ( ) :

a) i qua A(1; 2; 10) , B(2; 1; 3) v vung gc vi (P): x 3y + 2z - 6 = 0b) i qua C(2; -1; 4) , D(3; 2; -1 ) v vung gc vi (Q): x + y + 2z + 1 = 0

Dng 6: Vit phng trnh mt phng ( ) cha ng thng v song song vi ( , chonhau).* Phng php gii:

1. Tm VTCP ca v l u v 'u

2. VTPT ca mt phng ( ) l: 'n u u = 3. Ly mt im M trn 4. p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT

* Bi tp p dng :

11. Cho hai ng thng 1 2

11

: ;2 1 1

x tx y z

d d y t z t

=

= = = =

a. Chng minh d1 v d2 cho nhaub. Vit phng trnh mt phng (P) cha d1 v song song d2;c. Vit phng trnh mt phng (Q) cha d2 v song song d1

12. Cho 4 im A(-1; 2; 0); B(-3; 0; 2), C(1; 2; 3), D(0; 3; -2). Vit phng trnh mt phng chaAD v song song vi BC.

Dng 7: Vit phng trnh mt phng ( ) cha ng thng v 1 im M* Phng php gii:

1. Tm VTCP ca l u , ly 1 im N trn . Tnh ta MN

2. VTPT ca mt phng ( ) l: ;n u MN =

3. p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT* Bi tp p dng :

13. Vit phng trnh mt phng i qua Av cha ng thng d trong cc trng hp sau:


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a) A(1; 2; 1) v d:1

33 4

x yz

= = +

b) A(2; 3; 1) v cha ng thng d:

5 3

2

x t

y t

z t

= + = =

c) A(2; 1; -1) v d l giao tuyn ca 2 mp (P): x y + z 4 = 0, (Q): 3x y + z 1 = 0

Dng 8: Vit phng trnh mt phng ( ) cha 2 ng thng ct nhau v * Phng php gii:

1. Tm VTCP ca v l u v 'u

2. VTPT ca mt phng ( ) l: ';n u u =

3. Ly mt im M trn 4. p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT.

* Bi tp p dng :14. Cho 2 ng thng 1d v 2d . Chng minh d1, d2 ct nhau, vit phng trnh mt phng cha hai

ng thng trong cc trng hp sau:

a) 1 21 4

: 1 2 ; :1 2 5

3

x tx y z

d y t d

z t

= = = =

=

b) 1 21 6 5 2 6

: ; :1 2 5 4 1 7

x y z x y zd d

+ = = = =

c) d1: 2

1 3 '

1 2 ; : 1 '

3 2 3 2 '

x t x t

y t d y t

z t z t

= + = = + = + = = +

Dng 9: Vit phng trnh mt phng ( ) cha 2 song song v * Phng php gii:

1. Tm VTCP ca v l u v 'u , ly , 'M N

2. VTPT ca mt phng ( ) l: ;n u MN =

3.p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT.* Bi tp p dng :

15. Cho 2 ng thng 1d v 2d . Chng minh d1 song song d2 ct nhau, vit phng trnh mt

phng cha hai ng thng trong cc trng hp sau:

a)

5 2 3 2 '

: 1 , ' : 3 '

5 1 '

x t x t

d y t d y t

z t z t

= + = + = = = =

.

b) 1 21 2 1 4 2

: ; :3 2 1 3 2 1

x y z x y zd d

+ + = = = =

c) 1 21 4 3

: 8 4 ; :1 4 3

3 3

x tx y z

d y t d

z t

= + + = = =

=

Dng 10: Vit phng trnh mt phng ( ) tip xc vi mt cu (S).

* Phng php gii:1. Tm ta tm I v tnh bn knh ca mt cu (S)2. Nu mt phng ( ) tip xc vi mt cu (S) ti M (S) th mt phng ( ) i qua im M

v c VTPT l MI


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3. Khi bi ton khng cho tip im th ta phi s dng cc d kin ca bi ton tm cVTPT ca mt phng v vit phng trnh mt phng c dng: Ax + By + Cz + D = 0 (D cha bit).

S dng iu kin tip xc: ( )( ),d I R = tm D.* Bi tp p dng :

16. Vit phng trnh mp tip xc vi mt cu (S) ti M trong cc trng hp sau:a) (S): x2 + y2 + z2 6x 2y + 4z + 5 = 0 ti M(4; 3; 0)b) (S): x2 + y2 + z2 +2x y - 6z + 1 = 0 ti M(-1; 0; 0)

17. Cho mt cu (S) c ng knh l AB bit A(6; 2; -5), B(-4; 0; 7)a) Tm ta tm I v tnh bn knh ca mt cu (S)b) Vit phng trnh mt cu (S)c) Vit phng trnh mt phng tip xc mt cu (S) ti A.

18. Vit phng trnh mt phng tip xc vi mt cu (S): x2 + y2 + z2 2x 4y 6z 2 = 0 vsong song mt phng (P): 4x + 3y 12z + 1 = 0

19. Vit phng trnh mt phng tip xc vi mt cu (S): x2 + y2 + z2 + 2x y - 6z + 1 = 0 vsong song mt phng (P): 2x + 2y +z 1 = 0

20. Cho mt cu (S): x2 + y2 + z2 - 2x + 2y + 4z - 3 = 0 v 2 ng

thng ( ) ( )1 2

21

: 1 ; : 1 1 1

x tx y z

y tz t

=

= = = =

a) Chng minh ( ) ( )1 2; cho nhaub) Vit phng trnh mt phng (P) tip xc vi mt cu (S), bit rng (P) song song vi 2

ng thng ( ) ( )1 2;

Ch 3: MT THNGA. Kin thc cbn:

1. ng thng d i qua im M0(x0; y0; z0), nhn ( ); ;u a b c= lm VTCP c phng trnh thams l:

0

0

0

;x x at y y bt t

z z ct

= + = + = +

Khi a, b, c khc 0 th ta c phng trnh chnh tc ca d l: 0 0 0x x y y z z

a b c

= =

2. Nu ng thng d l giao tuyn ca hai mp (P) v (P) ln lt c phng trnh :Ax + By +Cz + D = 0 Ax + By +Cz + D = 0

Th ng thng d c VTCP: '; ; ;' ' ' ' ' 'P PB C C A A B

u n nB C C A A B

= =

r uur uur

Mun tm mt im thuc d th ta cho x = x0 (thng cho x = 0), gii h phng trnh tm y, zB. Cc dng ton:

Dng 1: Vit phng trnh ng thng di qua mtim M0(x0; y0; z0) v c VTCP* Phng php gii:

1. Tm VTCP ( ); ;u a b c= 2. Vit phng trnh ng thng d dng tham s hoc chnh tc

Ch : Cch tm VTCP

- Nu ng thng d qua A, B th VTCP u AB= - Nu ng thng ( )d P th d c VTCP Pu n= ( Pn l VTPT ca (P))

- Nu / /d th d v c cng VTCP

- Nu ;d a d b th d c VTCP ;u u ua b =

r uur r

- Nu ; / /( )d d P th d c VTCP Pu u n=


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d

AB

* Bi tp p dng :1. Vit phng trnh tham s, phng trnh chnh tc( nu c) ca ng thng d i qua 2 im A

v B trong cc trng hp sau:a) A(1,3,7), B(2,1,2)b) A(5,1,3), B(2,5,1)

2. Vit phng trnh ng thng qua A v song song vi ng thng d:

a) A(1,5,2) , d:3 2 3

2 1 1

x y z + = =

b) A(-2,4,1) , d l giao tuyn ca hai mt phng ( ) 3 0P x y z+ + = v ( )2 5 3 0Q x y z+ + =

c) A(7,3,-5), d:

6

5 8

5

x t

y t

z t

= = = +

3. Vit phng trnh ng thng qua A v vung gc vi mt phng ( ) trong cc trng hpsau:

a) A(3, 5, -2), ( ) : x + 5y 3z + 1 = 0b) A(1, 7, 3), ( ) :x +z = 0

c) A(0, 6, 1), ( ) : y + z 3 = 04. Vit phng trnh ng thng qua A v vung gc vi 2 ng thng d1 v d2 trong cc

trng hp sau:

a) A(1,2,-3), d1:

1 2

3 4

1 5

x t

y t

z t

= + = = +

, d2:3 1 5

2 3 1

x y z + += =

b) A(1; 2; 3), d1:23 10

8 4 1

x y z+ += =

, d2:

1 1

1 1 2

x y z+ = =

5. Vit phng trnh ng thng d qua A v vung gc ng thng v song song vi mt

phng (P) trong cc trng hp sau

a) A(1; -2; 3) , ng thng

1 3

: 3 2

2

x t

y t

z t

= + = + =

; mt phng (P): 2x + y + 3z 5 = 0

b) A(1; 1; -2) , ng thng1 1 2

:2 1 3

x y z+ = = ; mt phng (P): x y z 1 = 0

6. Vit phng trnh ng thng (d) trong cc trng hp sau

a) (d) i qua im M(1; 4; -2) v song song vi 2 mt phng c cc phng trnh:(P): 6x + 2y + 2z + 3 = 0, (Q): 3x 5y 2z 1 = 0

b) (d) i qua trng tm ca tam gic ABC v vung gc vi mt phng cha tam gic, bit A(1;3; 2), B(1; 2; 1), C(-1; 1; 3)

7. Cho im A(2; 3; 5) v mt phng (P): 2x + 3y + z 17 = 0a) Vit phng trnh ng thng d qua A, vung gc mt phng (P)b) Tm giao im ca d vi trc Oz.

Dng 2: Vit phng trnh ng thng i qua im A, ct v vung gc vi ng thng d.* Phng php gii:

1. Vit phng trnh mt phng (P) i qua A v vung gc vi d2. Tm giao im B ca d v mt phng (P)3. ng thng cn tm l ng thng qua A, B

* Bi tp p dng :8. Vit phng trnh ng thng


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d d'

BA

Q

d

P

M

B

A

a) Qua im A(3; 2; 1), ct v vung gc vi ng thng3

:2 4 1

x y zd

+= =

b) Qua im A(0; 1; -1), ct v vung gc vi ng thng : 1 3

1 2

x t

d y t

z t

= = =

Dng 3: Vit phng trnh ng thng i qua im A, vung gc vi ng thng d, ctngthng d

* Phng php gii:1. Vit phng trnh mt phng (P) i qua A v vung gc vi d2. Tm giao im B ca d v mt phng (P)3. ng thng cn tm l ng thng qua A, B.

* Bi tp p dng :9. Vit phng trnh ng thng i qua M v vung

gc vi ng thng dv ct ng thng 'd trong cc trng hp sau:

a) M(0; 1; 1);

1

:

1

x t

d y t

z

= = =

;1 1

' :2 7 9

x y zd

= =

b) M(0; 1; 1),1 2

:3 1 1

x y zd

+= = ,

1

' : 2

3

x

d y t

z t

= = + = +

c) M(-4; -5; 3),

2

: 4 3

6 5

x t

d y t

z t

= = + =

,

1 3 '

' : 3 2 '

2 '

x t

d y t

z t

= = + =

Dng 4: Vit phng trnh ng thng d qua M, ct 2 ng thng a, b

* Phng php gii:- L lun: Gi (P) l mt phng cha a, M v (Q) l mt phng cha b, Mng thng d cn tm l giao tuyn ca (P) v (Q)

1. Tm VTCP ca a, b l ;a bu u . Ly ;A a B b , tnh ;AM BM 2. Tnh VTPT ca (P) v (Q): ; ; ;P a Q bn AM u n BM u = = 3. Vit phng trnh ng thng (d) c VTCP l ;d P Qu n n =

v qua im M.- Cch gii khc:

1. Chuyn phng trnh ca hai ng thng v dng phng trnh tham s t v u2. Gi ( ) ( )1M d d = v ( ) ( )2N d d = 3. T PTTS ca ( )1d v ( )2d ta suy ra to hai im M v N theo hai tham s t v u4. (d) qua A M , N , A thng hng

,AM AN cng phng t, t t suy ra toM, N.

5. (d) qua A nhn MN lm VTCP kt qu.* Bi tp p dng :

10. Vit phng trnh ng thng d qua M, ct 2 ng thng a, b trong cc trng hp sau:

a) M(1; -1; 1) v

1 2

:

3

x t

a y t

z t

= + = =

2 3:1 2 1

x y zb + = =


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P

d

Q

a b

d

b) M(2; 3; 1) v

1

: 1

4

x t

a y t

z

= + = =

v

1 3 '

: '

2 '

x t

b y t

z t

= + = = +

c) M(-4; 5; 3) v1 3 2

:3 2 1

x y zd

+ + = =

v

2 1 1' :

2 3 5

x y zd

+ = =

Dng 5: Vit phng trnh ng thng d vung gc vi mt phng ( ) v ct 2 ng thng a, b chotrc.* Phng php gii:L lun: Gi (P) l mt phng cha a, vung gc vi( ),

(Q) l mt phng cha b, vung gc vi( )ng thng d cn tm l giao tuyn ca (P) v (Q)

1. Tm VTCP ca a, b l ;a bu u . Ly ;A a B b .Tm VTPT ca ( ) : n

2.Mt phng (P) c VTPT ;P an u n = v qua A.Vit phng trnh mt phng (P).

3. Mt phng (Q) c VTPT ;Q bn u n = v qua B.Vit phng trnh mt phng (Q)

4. Ly M thuc giao tuyn ca (P) v (Q)5. Vit phng trnh ng thng d c VTCP l n v qua M.

* Bi tp p dng :11. Vit phng trnh ng thng d trong cc trng hp sau

a) Vung gc vi mt phng (Oxz) v ct 2 ng thng : 43

x t

a y tz t

=

= + = v

2 3 4:

2 1 5

x y zb

+ = =

b) Vung gc vi mt phng (P): x + y + z +2 = 0 v ct hai ng thng2 1 2 '

: 1 ; 3

2 1 '

x t x t

a y t b y

z t z t

= + = = = = = +

c) Vung gc vi mt phng (P): x + y + z 1 = 0 v ct hai ng thng

1 1 2 7: ; : ; ;2 1 1 3 3

x y za b x t y z t += = = + = =

Dng 6: Vit phng trnh ng thng d song song vi ng thng , ct 2 ng thng.12. Vit phng trnh ng thng d trong cc trng hp sau

a) Song song vi

3

: 1

5

x t

y t

z t

= = = +

v ct hai ng thng 11 2 2

:1 4 3

x y zd

+ = = ,

2

1 2 1:

5 9 1

x y zd

= =

b) Song song vi 1 3 2:3 2 1

x y z+ + = =

v ct hai ng thng 1 2 2 1: 3 4 1x y zd + = =

, 27 3 9

:1 2 1

x y zd

= =

Dng 7: Vit phng trnh ng thng i qua A(P),


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P

d

Q

d'

nm trong (P) v vung gc vi ng thng d.* Phng php gii:

1. Tm VTCP ca d : du v VTPT ca (P): Pn

2. ng thng c VTCP l ;d Pu u n =

3. Vit phng trnh ng thng qua A vc VTCP va tm c trn.

* Bi tp p dng :13. Vit phng trnh ng thng d i qua im A(1; 1; 0), nm trong mt phng (P):

3x 2y 1 = 0 v vung gc vi ng thng11 16

:1 2 1

x y zd

+ = =

14. Cho mt phng (P): 2x + 5y + z + 17 = 0 v ng thng

111

52

: 2757 15

x t

y t

z t

= = +

= +

a) Tm giao im A ca (P) v .b) Vit phng trnh ng thng d qua A, vung gc vi v nm trong mt phng (P)

15. Cho mt phng (P): x + y + z -1= 0 v ng thng6 12 3

:3 5 1

x y zd

+ + += =

a) Tm giao im A ca (P) v d.b) Vit phng trnh ng thng d qua A, vung gc vi d v nm trong mt phng (P)

16. Cho mt phng (P):2x + y -2z + 9 = 0 v ng thng1 3 3

:1 2 1

x y zd

+ = =

a) Tm giao im A ca (P) v d.b) Vit phng trnh ng thng d qua A, vung gc vi d v nm trong mt phng (P)

Dng 8: Vit phng trnh hnh chiu vung gc ca ng thng d ln 1 mt phng (P).* Phng php gii:

1. Vit phng trnh mt phng (Q) cha dv vung gc (P) ( c cch gii).

2.Gi d l hnh chiu ca d ln (P) th d lgiao tuyn ca (P) v (Q) ( c cch gii).

* Phng php gii:

17. Vit phng trnh hnh chiu vung gc ca ng thng d:2 1 3

3 1 2

x y z +

= =

ln mt phng (P): x + 5y + 7z + 1 = 0

18. Vit phng trnh hnh chiu vung gc ca ng thng d:

2

3

1 3

x t

y t

z t

= = + =

ln mt phng (P): 2x + 3y z + 1 =0

19. Vit phng trnh hnh chiu vung gc ca ng thng d:2 2

1 2 1

x y z+ = =

a) ln mt phng Oxy. b) ln mt phng (Oxz) c) ln mt phng (Oyz)

20. Vit phng trnh hnh chiu vung gc ca ng thng d: 2 2 13 4 1

x y z + = =

ln mt phng (P): x + 2y +3 z + 4 = 0.

Ch 3: MT PHNG V MT CU


22/37


1. Vit phng trnh mt phng (P) i qua 3 im A(5;1;3); B(5;0;4); C(4;0;6).

Gii: (0; 1;1); ( 1; 1;3).AB AC = =

Vectphp tuyn ca mt phng (P) : , ( 2; 1; 1)n AB AC = =

Phng trnh mt phng (P) : -2(x - 5) - (y 1) ( z 3)= 0 2 14 0x y z + + = 2. Vit phng trnh mt phng ( P ) i qua 3 im A ; B ; C ln lt l hnh chiu vung gc ca

M(2; 3; 5) trn 3 trc ta .Gii: A, B, C ln lt l hnh chiu ca M trn trc ox, oy, oz nn A(2;0;0), B(0;3;0); C(0;0;5)

Suy ra phng trnh mt phng (ABC) : 1 15 10 6 30 02 3 5x y z x y z+ + = + + = .

3. Cho 2 im M ( 1;3;4);N(4;2;1) v mt phng ( Q ) : 2x+ 3y+ 4z - 1 = 0a) Vit phng trnh mt phng trung trc on MN.b) Vit phng trnh mt phng ( P ) i qua 2 im M ,N v vung gc mt phng (Q).c) Vit phng trnh mt phng ( R ) i qua M v song song mt phng (Q).

Gii: a) Gi ( ) l mt phng trung trc on MN.Ta c ( ) i qua trung im I ca on MN v vung gc vi MN.5 5 5

; ;2 2 2

I

, (3; 1; 3)MN= l vectphp tuyn ca mt phng ( ):

5 5 53 3 02 2 2

x y z =

6 2 6 5 0x y z + = .

b) Mt phng (Q) c vectphp tuyn 1 (2;3;4)n = mt phng (P) i qua 2 im M v N v

vung gc mt phng (Q) 1, (5; 18;11)Pn MN n = = l vectphp tuyn ca mt phng (P)

Phng trnh mt phng ( P ) : 5(x -1 ) 18(y 3) + 11(z 4 ) = 0 5 18 11 5 0x y z + + = .c) Mt phng ( R ) song song mt phng (Q) vect php tuyn ca mt phng (Q):

(2;3;4)Qn = l vectphp tuyn ca mt phng (R). Mt phng (R) i qua M nn phng trnh mtphng ( R) l: 2(x 1)+ 3(y 3) + 4(z 4) = 0 2x + 3y+ 4z - 27 = 0.

4. Cho hai mt phng : (P): 2x + my + 3z 5 = 0 v (Q): nx 8y - 6z + 2 = 0.a) Tm m v n hai mt phng trn song song vi nhaub) Vi m v n tm c cu a).Hy tnh khong cch gia hai mt phng trn

Gii: a) Mt phng (P) c vectphp tuyn 1 (2; ;3)n m= .

Mt phng (P) c vectphp tuyn 2 ( ; 8; 6)n n= , ta c :2 1

2 1

4( ) / /( )

4

mn knP Q

nD kD

== =

b) Ta c5

0; 0; ( )3

A P

. d( (P); (Q)) = d(A;(Q)) =

50 0 6 2

3 8

16 64 36 116

+ =

+ +.

5. Cho A(3;-2;-2) ; B(3;2;0);C(0;2;1);D(-1;1;2)a) Vit phng trnh mt phng ( BCD).T suy ra ABCD l t din.b) Tnh ng cao AH ca t din ABCD.c) Vit phng trnh mt phng ( P) cha AB v song song CD.d) Vit phng trnh mt cu ( S ) tm A, tip xc mt phng (BCD). Tm tip im.

Gii: a) ( 3;0;1) ; ( 4; 1;2)BC BD= =

Vectphp tuyn ca mt phng (BCD): , (1;2;3)n BC BD = =

Phng trnh mt phng ( BCD): 1(x 3) + 2(y 2) + 3(z- 0) = 0 x+ 2y + 3z 7 = 0.Ta im A khng thomn PT mt phng (BCD) ( )A BCD ABCD l t din.

b) 3 4 6 7 14( ;( )) 141 4 9 14

AH d A BCD = = = =+ +

c) (0;4;2); ( 1; 1;1)AB CD= = . Mt phng ( P ) cha AB v song song CD nn ( P) c

vectphp tuyn , (6; 2;4)pn AB CD = = Phng trnh mt phng (P):


23/37


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

d) Mt cu (S) tm A, tip xc mp (BCD) nn c bn knh R = d(A,(BCD)) = 14 Phng trnh mt cu ( S ) : 2 2 2( 3) ( 2) ( 2) 14x y z + + + + = .

* Tip im ca mt cu ( S) v mt phng ( BCD) l im H. AH vung gc mt phng(BCD) nn vectphp tuyn ca mt phng (BCD) l vctch phng ca ng thng AH, suy ra

phng trnh tham s ca AH l :

3

2 2

2 3

x t

y t

z t

= + = +

= +

; (3 ; 2 2 ; 2 3 )H AH H t t t + + + ;

( ) 3 2( 2 2 ) 3( 2 3 ) 7 0H BCD t t t + + + + + = 14 14 0 1 (4;0;1)t t H = = Vy tip im ca mt cu ( S) v mt phng (BCD) l H ( 4;0;1).

6. Cho mt cu ( S) : 2 2 2( 3) ( 2) ( 1) 100x y z + + + = v mt phng ( P ) : 2x-2y-z + 9 = 0a) Tm tm I v bn knh R ca mt cu ( S)b) Chng minh rng ( P ) ct ( S) theo mt ng trn ( C )c) Tm tm v bn knh ng trn ( C ).

Gii: a) Tm mt cu ( S) : I(3 ; -2 ; 1). Bn knh mt cu ( S) : R = 10.

b) d =6 4 1 9 18

( ; ( )) 634 4 1

d I P R+ +

= = =

.

2. Tnh tch phn: J = ( )4

2

0

cos3 .s inx tan 3x x dx

+ .

Cu3. (2,0 im)Cho hnh chp SABC c ng cao SA = a; ABC vung cn, AB = BC = a; B l trung im

cnh SB, C l chn ng cao h t A ca SAC .1. CMR SC (ABC).2. Tnh th tch khi chp S. ABC.

PHN RING (3,0 im)Th sinh chc lm mt trong hai phn A hoc B

A. Theo chng trnh nng cao

Cu 4a. (2,0 im)1. Vit phng trnh mt cu (S ) tm M(2;1;4) v tip xc mt phng (P): 3x + 4y+ z 5 = 0

2. Cho bn im (1;2; 1), (3;4; 1), (1;4;1), (3;2;1)S A B C . Vit phng trnh ng vung gcchung ca SA v BC.Cu 5a. (1,0 im)

Tm 2 s thc x, y tha mn 3(3 5 ) (1 2 ) 9 14x i y i i+ + = + .

B. Theo chng trnh chun

Cu 4b. (2,0 im)1. Vit phng trnh mt cu (S) ng knh AB vi A(1; 2; -3) ; B(5; 4; 1).2. Cho bn im (1; 2; 3), (2; 2; 3), (1; 1; 3), (1; 2; 5)S A B C . Vit phng trnh cc hnh

chiu ca SB trn mt phng (ABC).

Cu 5b. (1,0 im)Gii phng trnh sau trn tp hp s phc: (3 + 4i)z + (1 3i) = 2 + 5i.

8PHN CHUNG CHO TT C CC TH SINH (7.0 im)Cu 1. (3,0 im)

Cho hm s2 1

( )1

xy f x

x

+= =

.

1. Kho st s bin thin v v th hm s.2. Tm cc gi tr m ng thng 2y mx= + ct th hm s cho ti 2 im phn bit.

Cu 2. (2,0 im)1. Gii phng trnh:

2 222 2 3x x x x + =

2. Tnh tch phn : J = 2

1

( 1) lne

x x x dx+ + Cu3 . (2,0 im)

Cho hnh chp SABCD c y l hnh vung cnh a, SA (ABCD) SC hp vi y 1 gc060 . Gi H, I , K ln lt l hnh chiu ca A trn AB, SC, SD.

1. Chng minh 7 im A, B, C, D, H, I, K thuc 1 mt cu. Tnh th tch khi cu .2. Tnh th tch khi chp S.ABCD.

PHN RING (3.0 im)Th sinh chc lm mt trong hai phn A hoc B

A. Theo chng trnh nng cao

Cu 4a. (2,0 im)


28/37


1. Vit phng trnh tham s ca ng thng i qua im ( 4; 2;4)A , vung gc v ct

ng thng

3 2

: 1

1 4

x t

d y t

z t

= + = = +

2. Cho hai mt phng ( P) : 3x 2y + 2z + 1 = 0 ( Q) : 5x 4y + 3z 1 = 0. Vi t phng trnhmt phng ( R ) i qua M ( 1 ; 2 ; 3) v vung gc hai mt phng (P) v (Q).

Cu 5a. (1,0 im)

Gii phng trnh sau trn tp hp s phc: z 2 8(1 i)z + 63 16i = 0.


Cu 4b. (2,0 im)1. Vit phng trnh tham s ca ng thng i qua im (2; 1; 3)A , vung gc v ct ng

thng

1 3

: 1

2 2

x t

y t

z t

= + = + = +

2. Cho hai mt phng ( P) : 5x 4y + 3z 1 = 0 ( Q) : 3x 2y + 2z + 1 = 0. Vi t phng trnh

mt phng ( R ) i qua M ( 2; 1 ; 3) v vung gc hai mt phng (P) v (Q).Cu 5b. (1,0 im)

Gii phng trnh sau trn tp hp s phc: z 4 + 4z 2 5 = 0

9

PHN CHUNG CHO TT C CC TH SINH (7.0 im)

Cu 1. (3,0 im)

Cho hm s2

1y

x=

; c th l (H)

1. Kho st s bin thin v v th hm s (H)2. Vit phng trnh tip tuyn vi th (H) bit tip tuyn song song vi ng thng d:2 5 0x y+ = .

Cu 2. (2,0 im)

1. Gii h phng trnh:

3 2

1

2 5 4

4 2

2 2

x

x x

x

y y

y+

= +

= +


2

0

.cos .x x dx

Cu3 . (2,0 im)

Cho t din ABCD c AD=AC = a, AB = 2a, AD (ABC) , ABC vung C.1. Tnh th tch khi t din ABCD.2. Tnh din tch mt cu ngoi tip t din ABCD.

PHN RING (3.0 im)

Th sinh chc lm mt trong hai phn A hoc BA. Theo chng trnh nng cao

Cu 4a. (2,0 im)

1. Chng t rng cp ng thng sau y cho nhau 12 1

: 3 2 2

x y z

d

+= = v

2

1 1:

1 2 4

x y zd

+= = . Vit phng trnh ng vung gc chung ca chng.


29/37


2. Cho A(3; -2; -2); B(3; 2; 0); C(0; 2; 1); D(-1; 1; 2). Vit phng trnh mt cu ( S ) tm A,tip xc mt phng (BCD). Tm tip im.Cu 5a. (1,0 im)

Trn mt phng phc, hy tm tp hp im biu din s phc z tha mn bt ng thc:1 1.z i <


Cu 4b. (2,0 im)

1. Chng t rng cp ng thng sau y cho nhau 12 1 3

:2 1 2

x y zd

= =

v

2

3 1 1:

2 2 1

x y zd

+ = =

. Vit phng trnh ng vung gc chung ca chng.

2. Cho A(3; -2; -2); B(3; 2; 0); C(0; 2; 1); D(-1; 1; 2). Vit phng trnh mt cu ( S ) tm D,tip xc mt phng (ABC). Tm ta tip im.

Cu 5b. (1,0 im) Trn mt phng phc, hy tm tp hp im biu din s phc z tha mn btng thc 1 2z< .

10PHN CHUNG CHO TT C CC TH SINH (7.0 im)

Cu 1. (3,0 im)Cho hm s 3 23 4y x x= + ; c th l (C)1. Kho st s bin thin v v th hm s (C)2. Trn (C) ly im A c honh 2. Vit phng trnh ng thng d qua A v tip xc vi

(C).

Cu 2. (2.0 im)

1. Gii phng trnh:2 25 1 54 12.2 8 0x x x x + =


cos( ).sin .x x dxxe

+ Cu3 . (2,0 im)

Cho hnh chp SABC c SA = SB= SC = a ,

ASB =

BSC= 060 ,

ASC= 090 .1. CMR ABC vung . Tnh th tch khi chp S.ABC.2. Xc nh tm v bn knh mt cu ngoi tip hnh chp S.ABC.

PHN RING ( 3,0 im).

Th sinh chc lm mt trong hai phn A hoc B

A. Theo chng trnh nng caoCu 4a. (2,0 im)

1. Tm m hai ng thng d1 v d2 ct nhau. Khi tm to giao im ca chng:

1 2

2 4 0 2 3 0: ; :

3 0 2 6 0

x y z x y mzd d

x y x y z

+ = + + = + = + + =

2. Cho hai mt phng ( P ) : x- 2y + 3z + 1 = 0 ( Q) : x - 2y + 3z + 5 = 0. Vit phng trnh mtphng (R) song song v cch u hai mt phng (P) v (Q).

Cu 5a. (1,0 im) Tm GTLN, GTNN ca hm s ( )x

x

ef x

e e=

+trn on [ ln 2 ; ln4] .


Cu 4b. (2,0 im)1. Tm m hai ng thng d1 v d2 ct nhau. Khi tm to giao im ca chng:


30/37


1 2

1 3 2 1 2: ; :

1 2 1 1 1 3

x y z m x y zd d

= = = =

2. Cho mt phng (P) : x+ 2y + 3z + 4 = 0. Vit phng trnh mt phng ( Q) song song mtphng ( P ) v cch ( P) mt khong bng 3.Cu 5b. (1,0 im)

Tm GTLN, GTNN ca hm s 2( ) ln( 5 )f x x x= + + trn on [-2;2].

11 THI TT NGHIP TRUNG HC PH THNG NM 2008

I. PHN CHUNG CHO TH SINH C 2 BAN (8 im)Cu 1 (3,5 im)

Cho hm s y = 2x3 + 3x2 - 11) Kho st s bin thin v v th ca hm s.2) Bin lun theo m s nghim thc ca phng trnh 2x3 + 3x2 1 = m.

Cu 2 (1,5 im)Gii phng trnh: 2 13 9.3x x+ + 6 = 0 .

Cu 3 (1,0 im)

Tnh gi tr ca biu thc: P (1 3 i)2

(1 - 3 i)2

.Cu 4 (2,0 im)Cho hnh chp tam gic u S.ABC c cnh y bng a, cnh bn bng 2a. Gi I l trung im

ca cnh BC.1) Chng minh SA vung gc vi BC.2) Tnh th tch khi chp S.ABI theo a.

II. PHN DNH CHO TH SINH TNG BAN (2 im)

A. Th sinh Ban KHTN chn cu 5a hoc cu 5bCu 5a (2,0 im)

1) Tnh tch phn1

2 3 4

1

(1 )I x x dx=

2) Tm gi tr ln nht v gi tr nh nht ca hm s f(x) =x + 2 cosx trn on [0;2

].

Cu 5b (2,0 im)Trong khng gian vi h to Oxyz, cho im A(3; -2; -2) v mt phng (P) c phng trnh 2x

- 2y + z - 1 = 0.1) Vit phng trnh ca ng thng i qua im A v vung gc vi mt phng (P).2) Tnh khong cch tim A n mt phng (P). Vit phng trnh ca mt phng (Q) sao cho

(Q) song song vi (P) v khong cch gia (P) v (Q) bng khong cch tim A n (P).

B. Th sinh Ban KHXH-NV chn cu 6a hoc cu 6b

Cu 6a (2,0 im)

1) Tnh tch phn2

0

(2 1) cosI x xdx

= 2) Tm gi tr ln nht v gi tr nh nht ca hm s f(x) = x4 2x2 + 1 trn on [0; 2].

Cu 6b (2,0 im)Trong khng gian vi h to Oxyz, cho tam gic ABC vi A(1; 4; 1), B(2; 4; 3) v

C(2; 2; 1) .1) Vit phng trnh mt phng i qua A v vung gc vi ng thng BC.2) Tm toim D sao cho t gic ABCD l hnh bnh hnh.


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


31/37


Cu 1. (3,0 im). Cho hm s2 1

2

xy

x

+=

.

1) Kho st s bin thin v v th (C) ca hm s cho.2) Vit phng trnh tip tuyn ca th (C),bit h s gc ca tip tuyn bng -5.

Cu 2. (3,0 im)1) Gii phng trnh .

2) Tnh tch phn

0

(1 cos )I x x dx

= + .

3) Tm gi tr nh nht v gi tr ln nht ca hm s 2( ) ln(1 2 )f x x x= trn on [-2; 0].Cu 3. (1,0 im). Cho hnh chp S.ABC c mt bn SBC l tam gic u cnh a, cnh bn SA vunggc vi mt phng y. Bit gc BAC = 1200, tnh th tch ca khi chp S.ABC theo a.

II. PHN RING (3,0 im) Th sinh hc chng trnh no th chc chn phn dnh ringcho chng trnh (phn 1 hoc phn 2)

1. Theo chng trnh Chun :Cu 4a (2,0 im). Trong khng gian Oxyz, cho mt cu (S) v mt phng (P) c phng trnh:

( ) ( ) ( )2 2 2

( ) : 1 2 2 36 ( ) : 2 2 18 0S x y z v P x y z + + = + + + = .1) Xc nh ta tm T v tnh bn knh ca mt cu (S). Tnh khong cch t T n mt phng

(P).2) Vit phng trnh tham s ca ng thng d i qua T v vung gc vi (P). Tm ta giao

im ca d v (P).Cu 5a. (1,0 im). Gii phng trnh 28 4 1 0z z + = trn tp s phc.2. Theo chng trnh Nng cao:Cu 4b. (2,0 im). Trong khng gian Oxyz, cho im A(1; -2; 3) v ng thng d c phng trnh

1 2 3

2 1 1

x y z+ += =

1) Vit phng trnh tng qut ca mt phng i qua im A v vung gc vi ng thng d.2) Tnh khong cch tim A n ng thng d. Vit phng trnh mt cu tm A, tip xc vi d.

Cu 5b. (1,0 im). Gii phng trnh 22 1 0z iz + = trn tp s phc.BI GII

Cu 1: 1) MX: R\{2}; y =2

5

( 2)x

< 0,x 2. Hm s lun nghch bin trn cc khong xc nh

2

limx

y

= ;2

limx

y+

= + x = 2 l tim cn ng.

lim 2x

y ++

= ; lim 2x

y

= y = 2 l tim cn ngang

BBT :x 2 + y'

y 2- +

- 2+

Giao im vi trc tung (0;1

2 ); giao im vi trc honh (

1

2 ; 0)

th :

x

y

-

-

0 2

2


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2) Tip tuyn ti im c honh x0, c h s gc bng 5

20

55

( 2)x

=

x0 = 3 hay x0 = 1 ; y0 (3) = 7, y0 (1) = -3

Phng trnh tip tuyn cn tm l: y 7 = -5(x 3) hay y + 3 = -5(x 1) y = -5x + 22 hay y = -5x + 2

Cu 2: 1) 25x 6.5x + 5 = 0 2(5 ) 6.5 5 0x x + = 5x = 1 hay 5x = 5 x = 0 hay x = 1.

2)0 0 0

(1 cos ) cosI x x dx xdx x xdx

= + = + =2

0

cos2

x xdx

+

t u = x du = dx; dv = cosxdx v = sinx

I =2

00

sin sin2

x x xdx

+ =2 2

0cos 2

2 2x

+ =

3) Ta c : f(x) = 2x +22 4 2 2

1 2 1 2

x x

x x

+ +=

f(x) = 0 x = 1 (loi) hay x = 12 (nhn); f(-2) = 4 ln5, f(0) = 0, f( 12 ) = 1 ln 24

v f lin tc trn [-2; 0] nn[ 2;0]max ( ) 4 ln 5f x

= v

[ 2;0]

1min ( ) ln 2

4f x

=

Cu 3: Hnh chiu ca SB v SC trn (ABC) l AB v AC,m SB = SC nn AB = AC. Ta c:

BC2 = 2AB2 2AB2cos1200

a2 = 3AB2 =3

aAB

Nn :

22 2 2

= a SA =3 3

a a

SA 2 2

01 1 3 a 3= . .sin120 = =2 2 3 2 12ABC

aS AB AC .

2 31 2 3 2= =

3 12 363

a a aV (vtt)

Cu 4.a:1) Tm mt cu: T (1; 2; 2), bn knh mt cu R = 6

d (T, (P)) =1 4 4 18 27

931 4 4

+ + += =

+ +

2) (P) c vectphp tuyn (1;2;2)n =

Phng trnh tham s ca ng thng (d) :

1

2 2

2 2

x t

y t

z t

= + = + = +

(t R)

Th vo phng trnh mt phng (P) : 9t + 27 = 0 t = -3 (d) (P) = A (-2; -4; -4)Cu 5.a: 28 4 1 0z z + = ; / 24 4i = = ; Cn bc hai ca / l 2i

Phng trnh c hai nghim l1 1 1 1

4 4 4 4z i hay z i= + =

Cu 4.b:1) (d) c vc tch phng (2;1; 1)a =

Phng trnh mt phng (P) qua A (1; -2; 3) c ve1to7 php tuyn a , nn :2(x 1) + 1(y + 2) 1(z 3) = 0 2x + y z + 3 = 0

B

A

S

a

a

a

C


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2) Gi B (-1; 2; -3) (d). BA = (2; -4; 6), ,BA a = (-2; 14; 10)

d(A, (d)) =, 4 196 100

5 24 1 1

BA a

a

+ + = =+ +

r

Phng trnh mt cu tm A (1; -2; 3), bn knh R = 5 2 :(x 1)2 + (y + 2)2 + (2 3)2 = 50

Cu 5.b: 22 1 0z iz + = 2 8 9i = = = 9i2; Cn bc hai ca l 3i

Phng trnh c hai nghim l1

2z i hay z i= = .


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

Cu 1 (3,0 im). Cho hm s 3 21 3

54 2

y x x= +

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

2) Tm cc gi tr ca tham sm phng trnhx

3

6x

2

+ m = 0 c 3 nghim thc phn bit.Cu 2 (3,0 im).

1) Gii phng trnh: 22 42log 14log 3 0x x + =

2) Tnh tch phn:1

2 2

0

( 1)I x x dx=

3) Cho hm s 2( ) 2 12f x x x= + . Gii bt phng trnhf(x) 0.Cu 3 (1,0 im). Cho hnh chp S.ABCD c yABCD l hnh vung cnh a, cnh bn SA vung

gc vi mt phng y, gc gia mt phng (SBD) v mt phng y bng 60o

. Tnh th tch khi chp

S.ABCD theo a.II. PHN RING - PHN TCHN (3,0 im)Th sinh chc lm mt trong hai phn (phn 1 hoc phn 2).1. Theo chng trnh Chun:Cu 4.a (2,0 im).

Trong khng gian vi h toOxyz, cho 3 imA(1; 0; 0),B(0; 2; 0) v C(0; 0; 3).1) Vit phng trnh mt phng i quaA v vung gc vi ng thngBC.2) Tm to tm mt cu ngoi tip t din OABC.

Cu 5.a (1,0 im). Cho hai s phc v Xc nh phn thc v phn o ca s phc2. Theo chng trnh Nng cao:

Cu 4.b (2,0 im). Trong khng gian vi h toOxyz, cho ng thng c phng trnh:1 1

2 2 1

x y z+ = =

1) Tnh khong cch tim On ng thng .2) Vit phng trnh mt phng cha im Ov ng thng .

Cu 5.b (1,0 im). Cho hai s phc z1 = 2 + 5i v z2 = 3 - 4i. Xc nh phn thc v phn o ca sphc z1.z2.

14 THI TT NGHIP GDTX TRUNG HC PH THNG NM 2009

Cu 1 (3,0 im)Cho hm sy =x3 3x2 + 4.1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm to cc giao im ca th (C) v ng thng y = 4


34/37


Cu 2 (2,0 im)

1. Tnh tch phn:1

0

(2 )xI x xe= + dx.

2. Tm gi tr ln nht v gi tr nh nht ca hm s2 1

( )1

xf x

x

+=

trn on [2; 4].

Cu 3 (2,0 im). Trong khng gian vi h to Oxyz, cho ba im A(1; 0; 0), B(0; 3; 0) vC(0; 0; 2).

1. Vit phng trnh tng qut ca mt phng (ABC).2. Vit phng trnh ca ng thng i qua im M(8; 5; -1) v vung gc vi mt phng

(ABC); t, hy suy ra to hnh chiu vung gc ca im M trn mt phng (ABC).

Cu 4 (2,0 im)1. Gii phng trnh: log 2 (x + 1) = 1 + log 2 x2. Cho s phc z = 3 2i. Xc nh phn thc v phn o ca s phc z2 + z.

Cu 5 (1,0 im). Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AB = a v AC = a 3 ;

cnh bn SA vung gc vi mp (ABC) v SA = a 2 . Tnh th tch ca khi chp S.ABC theo a.

p n v thang im (ca B GD&T)

CU P N IM1. (2,0 im)a) Tp xc nh: D = R 0,25

b) S bin thin: Chiu bin thin: y' = 3x2 6x; y = 0 x = 0 ; x = 2

y > 0 x < 0 ; x > 2 v y < 0 0 < x < 2Suy ra, hm s nghch bin trn mi khong (; 0), (2; +) v

nghch bin trong khong (0; 2).

Cc tr: Hm st cc i ti x = 0 v yC= 4; t cc tiu ti x =2 v yCT = 0.

0,50

Gii hn: lim , limx x

y y +

= = + 0,50

Cu 1

(3,0 im)

Bng bin thin:

x - 0 2 + y + 0 - 0 +

y 4 +

- 0

0,25

c) th (C): y

4

x

0,50


35/37


CU P N IM

O 2

Lu :Nu th sinh chvng dng ca th (C) th cho 0,25 im

2. (1,0 im)

I =

1 1 1

0 0 0(2 ) 2x xx xe dx xdx xe dx+ = + = I1 + I2 0,25

Tnh I1 =1

12

00

2 1xdx x= = 0,25Tnh I2: t u = x v dv = e

xdx, ta c du = dx v v = ex. Do :

I2 = xex

11 1

0 00

1x x xxe e dx e e = = .Vy : I1 + I2 = 2

0,50

2. (1,0 im)

Ta c: [ ]' 23( ) 0, 2;4(1 )f x x

x= >

f(x) ng bin trn on [2;4] 0,50

Cu 2(2,0 im)

V vy:[2;4][2;4]

max ( ) (4) 3;min ( ) (2) 5f x f f x f = = = = 0,50

1. (0,75 im)V A(1; 0; 0) Ox, B(0; 3; 0) Oy, C(0; 0; 2) Oz nn phng

trnh on chn ca mt phng (ABC) l: 11 3 2

x y z+ + = 0,25

Suy ra, phng trnh tng qut ca mp(ABC) l: 6x + 2y + 3z 6 = 0 0,252. (1,25 im)

Phng trnh ca ng thng di qua M v vung gc mp(ABC):V d (ABC) nn vect php tuyn n ca (ABC) l vect chphng ca d. T phng trnh tng qut ca d ta c: n = (6; 2; 3).

0,25

Do , phng trnh tham s ca d l:

8 6

5 2

1 3

x t

y t

z t

= + = + = +

0,25

Ta hnh chiu vung gc ca im M trn mp(ABC):V d i qua im M v (ABC) nn giao im H ca d v (ABC)

l hnh chiu vng gc ca im M trn (ABC).

Do H d nn ta ca H c dng (8 + 6t; 5 + 2t; -1 + 3t).V H (ABC) nn : 6(8 + 6t) +2(5 + 2t) + 3(-1 + 3t) 6 = 0.

0,50

Cu 3(2,0 im)

Do H (2; 3; -4) 0,251. (1,0 im)iu kin xc nh: x > 0 0,25Vi iu kin , phng trnh cho tng ng vi phng trnh:

log2(x + 1) = log22x x + 1 = 2x0,50

x = 1Vy phng trnh cho c nghim duy nht x = 1.

0,25

Cu 4(2,0 im)

2. (1,0 im)

Ta c: z2

+ z = (3 2i)2

+ 3 2i = 9 12i + 4i2

+ 3 2i = 8 14iV vy, s phc z2 + z c phn thc bng 8 v phn o bng -14.Cu 5

(1,0 im)Xt tam gic vung ABC, S

ta c:2 2 2BC AC AB a= =

0,50


36/37


CU P N IM

Suy ra:21 2

.2 2ABC

aS AB AC = = A C

BV SA (ABC) nn SA l ng cao ca khi chp S.ABC.

Do d, th tch ca khi chp S.ABC l:3

.

1.

3 3S ABC ABC

aV S SA= =

0,50

15 THI TT NGHIP GDTX TRUNG HC PH THNG NM 2010

Cu 1. (3,0 im)

Cho hm s3 1

2

xy

x

+=

+

1) Kho st s bin thin v v th (C) ca hm s cho.2) Vit phng trnh tip tuyn ca th (C) ti im c honh x = 1.

Cu 2. (2,0 im)1) Tm gi tr ln nht v gi tr nh nht ca hm s f(x) = x4 8x2 + 5 trn on [1; 3].

2) Tnh tch phn:1

3

0

(5 2)I x dx= Cu 3. (2,0 im)

Trong khng gian vi h ta Oxyz, cho hai imM(1; 2; 3),N(3; 4; 1) v mt phng (P) cphng trnhx + 2y z + 4 = 0.1) Vit phng trnh mt phng trung trc ca on thngMN.2) Tm ta giao im ca ng thngMNv mt phng (P).

Cu 4. (2,0 im)1) Gii phng trnh: 9

x

3x

6 = 0.

2) Gii phng trnh: 2z2

+ 6z + 5 = 0 trn tp s phc.

Cu 5. (1,0 im)Cho hnh chp S.ABCD c y ABCD l hnh ch nht tm O; SA = SB = SC = SD. Bit

AB = 3a, BC= 4a vSAO = 45

o

. Tnh th tch khi chp S.ABCDtheo a.


37/37

MC LC

Phn thc nhtTm tt l thuyt v bi tp rn luyn

I. Kho st v v th hm s ........................................................................... trang 1

II. Tm gi tr nh nht v gi tr ln nht ca hm s ........................................ trang 4III. Nguyn hm v tch phn ............................................................................. trang 5

IV. Hm s lgarit .............................................................................................. trang 7

V. Hm s m .................................................................................................... trang 7

VI. S phc ......................................................................................................... trang 10

VII. Hnh hc khng gian ................................................................................... trang 11

VIII. Phng php ta trong khng gian........................................................ trang 12

Ch 1: Ta trong khng gian ....................................................... trang 12Ch 2: ng thng .......................................................................... trang 13

Ch 3: Mt phng ............................................................................. trang 17

Ch 4: Mt phng v mt cu ............................................................ trang 21

Phn thc nhtB rn ruyn (15 )

* 1: ................................................................................................................ trang 24

* 2: ................................................................................................................ trang 24* 3: ................................................................................................................ trang 24

* 4: ................................................................................................................ trang 25

* 5: ................................................................................................................ trang 25

* 6: ................................................................................................................ trang 26

* 7: ................................................................................................................ trang 26

* 8: ................................................................................................................ trang 27

* 9: ................................................................................................................ trang 28* 10: ............................................................................................................... trang 29

* 11: .............................................................................................................. trang 30

* 12: .............................................................................................................. trang 31

* 13: .............................................................................................................. trang 33

* 14: .............................................................................................................. trang 34

* 15: ............................................................................................................... trang 36

tai lieu on tap toan lop 12 nam 2011

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