Tài liệu ôn thi Tốt nghiệp 2012 môn Toán theo chuyên đề

8/3/2019 Ti liu n thi Tt nghip 2012 mn Ton theo chuyn

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TTTRRRNNNGGG TTTHHHPPPTTT CCCHHHUUU VVVNNN AAANNNTTT TTTOOONNN

GGGVVV::: DDDnnng gg PPPhhhccc SSSaaannnggg

On tap Tot nghiep

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Dng Phc Sang - 1 - THPT Chu Vn An

PPPhhhnnn III... KKKHHHOOO SSST TT HHHM MM SSS VVV BBBIII TTTOOONNN LLLIIINNN QQQUUUAAANNN

1. Hm s bc ba, hm s trng phng v cc vn lin quana) Kho st s bin thin v v th hm s

1 Tp xc nh: D = 2 Tnh y

3 Cho 0y = tm cc nghim0

x (nu c).

4 Tnh hai gii hn: lim ; limx x

y y +

5 V bng bin thin ca hm s.6 Nu s ng bin, nghch bin v cc tr (nu c) ca hm s.7 Tm im un (i vi hm s bc ba).8 Lp bng gi tr.9 V th hm s v nu nhn xt.

3 2 ( 0)y ax bx cx d a = + + +

S nghim ca phngtrnh 0y = 0a > 0a <

0y = c 2 nghimphn bit

0y = c nghim kp

0y = v nghim

th hm s bc ba lun i xng qua im un

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01688559752 [email protected]

Ti liu tham kho - 2 - n tp tt nghip mn Ton

4 2 ( 0)y ax bx c a = + +

S nghim ca phngtrnh 0y = 0a > 0a <

0y = c 3 nghimphn bit

0y = c 1 nghim duy

nht

th hm s trng phng lun i xng qua trc tungb) Vit phng trnh tip tuyn (dng 1 bit to tip im M0)

1 Ch r0

x v0

y (honh & tung ca im M0)

2 Tnh0

( )f x

3

Cng thc: 0 0 0( )( )y y f x x x

= c) Vit phng trnh tip tuyn (dng 2 bit trc h s gc k)

1 Lp lun c c0

( ) f x k = (*)

2 Thay0

( )y x vo (*) tm0

x

3 C0

x , tm0

y v dng cng thc

0 0 0( )( )y y f x x x =

Lu : Tip tuyn song song vi y ax b= + c h s gc k= a Tip tuyn vung gc vi ( 0)y ax b a = + c h s

gc 1a

k =

d) Bin lun s nghim ca phng trnh bng th (C):y=f(x)1 a phng trnh v dng: ( ) ( ) f x BT m = 2 Lp lun: s nghim ca phng trnh cho bng vi s giao

im ca th ( ) : ( )C y f x = v ng thng : ( )d y BT m = .3 V 2 ng ln cng 1 h trc to v lp bng kt qu

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Lu : nu bi ton ch yu cu tm cc gi tr ca m phng

trnh c ng 3 nghim, 4 nghim, ta khng cn lp bng ktqu nh trn m ch cn ch r cc trng hp tho .e) S tng giao gia th (C):y=f(x) v ng thng d: y= ax+ b

1 Lp phng trnh honh giao im ca ( )C v d:

( ) f x ax b= + (*)2 Lp lun: s giao im ca ( )C v dbng vi s nghim ca (*)

3m s nghim ca (*) suy ra s giao im ca ( )C v d

V D MINH HOBi 1: Cho hm s 3 26 9 1y x x x = + +

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti giao im ca ( )C vi

trc tung.c) Tm cc gi tr ca tham sm phng trnh sau y c

nghim duy nht: 3 26 9 0x x x m + + =

Bi giiCu a: Hm s 3 26 9 1y x x x = + + Tp xc nh: D= R

o hm: 23 12 9y x x = +

Cho 20 3 12 9 0 1y x x x = + = = hoc 3x =

Gii hn: lim ; limx x

y y +

= = +

Hm s ng bin trn cc khong (;1) v (3;+)Hm s nghch bin trn khong (1;3) th hm s c im cc i (1;5)D , im cc tiu (3;1)T

Cho6 12. 0 2 3y x y x y = = = = . im un (2;3)I

Bng bin thin:(ch : do a> 0) x 1 3 + y + 0 0 +

y5 +

1

m BT(m) S giao im S nghim pt . .

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Bng gi tr: x 0 1 2 3 4y 1 5 3 1 5

th hm s l mt ng cong i xngqua im (2;3)I nh hnh v bn y:

Cu b: Cho 0 (0) 1x y= = .Giao im ca ( )C vi trc tung l: (0;1)A

(0) 9f = Phng trnh tip tuyn ca ( )C ti A l:

1 9( 0) 9 1y x y x = = +

Cu c: Ta c, 3 2 3 26 9 0 6 9x x x m x x x m + + = + = 3 2

6 9 1 1x x x m + + = (*) Phng trnh (*) c nghim duy nht khi v ch khi th ( )C vng thng : 1d y m= ct nhau ti 1 im duy nht

1 5 4

1 1 0

m m

m m

> < < >

Bi 2: Cho hm s 2 33 2y x x= a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti cc giao im ca ( )C

vi trc honh.

c) Bin lun theo as nghim phng trnh: 3 24 6 3 0x x a = Bi gii

Cu a: Hm s 2 33 2y x x= Tp xc nh: D =

o hm: 26 6y x x =

Cho 20 6 6 0 0y x x x = = = hoc 1x =


y y +

= + =

Hm s ng bin trn khong (0;1)

Bng bin thin:(ch : do a< 0)

x 0 1 + y 0 + 0

y+ 1

0

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Hm s nghch bin trn cc khong ( ; 0) v (1; )+ th hm s c im cc i (1;1)D , im cc tiu (0; 0)O

Cho 1 12 2

6 12 . 0y x y x y = = = = . im un 1 12 2

( ; )I

Bng gi tr: x 12 0 12 1 12 y 1 0 1

21 0

th hm s l mt ng cong i xngqua im 1 1

2 2( ; )I nh hnh v bn y:

Cu b: Cho 2 30 3 2 0y x x= = 3

2

0x

x

= =

Giao im ca ( )C vi trc honh l: (0;0)O v 32

( ;0)B

Ti (0;0)O : (0) 0f = , phng trnh tip tuyn l: 0y =

Ti 32

( ;0)B : 3 92 2

( )f = , phng trnh tip tuyn l:

279 3 92 2 2 4

0 ( )y x y x = = +

Cu c: Ta c,3 2 2 3 2 34 6 3 0 6 4 3 3 2x x a x x a x x = = 3

2a= (*)

S nghim phng trnh (*) bng vi s giao im ca th ( )C

v ng thng 32

:d y a= , do ta c bng kt qu sau y:

a 32

a S giao imca ( )C v d

S nghim caphng trnh (*)

2

3a < 3

2 1a > 1 123

a = 32

1a = 2 2

23

0a < < 32

0 1a< < 3 3

0a = 32 0a = 2 2

0a > 32

0a < 1 1

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Bi 3:a) Kho st v v th ( )C ca hm s 3 23 32

x x xy

+ +=

b) Vit phng trnh tip tuyn vi th ( )C bit tip tuyn song

song vi ng thng

3

2:y x

= c) Tm to cc giao im ca ( )C vi ng thng

23y 2x= +

Bi gii

Cu a:3 23 3

2

x x xy

+ += Tp xc nh: D =

o hm23 6 3

0,2

x xy x

+ + = do hm s lun ng

bin trn v khng t cc tr. Gii hn: lim ; lim

x xy y

+= = +

Bng bin thin:

12

3 3 0 1y x x y = + = = =

im un 12

( 1; )I

Bng gi tr: x 3 2 1 0 1y 9

2 1 1

20 7

2

th hm s l ng cong i xng qua im12( 1; )I

Cu b: Tip tuyn ca ( )C song song vi ng thng 32

: y x = c h

s gc 30 2( )k f x= =

20 0

3 6 3

2

x x+ + = 3

22 00 0

0

03 6 0

2

xx x

x

= + = =

Vi0

0x = th0

(0) 0y y= = , tip tuyn tng ng l3 32 2

0 ( 0)y x y x = = (trng vi )

x 1 + y + 0 +

y

+

12

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Vi0

2x = th 0 ( 2) 1y y= = , tip tuyn tng ng l

3 32 2

1 ( 2) 2y x y x + = + = + (song song vi )

Vy, tip tuyn tho l2

3y 2x= +

Cu c: Honh giao im (nu c) ca ( )C v23y 2x= + l nghim

phng trnh3 23 3

2

x x x+ += 3 2

32 3 3 3 4

2x x x x x + + + = +

3 2 21

3 4 0 ( 1)( 4 4) 02

xx x x x x

x

= + = + + = =

721x y= = v 2 1x y= =

Vy, ( )C v 32

: 2d y x= + ct nhau ti 2 im:

( )721;A v ( 2; 1)B

Bi 4:a) Kho st v v th ( )C ca hm s: 4 22 3y x x= b) Vit phng trnh tip tuyn vi th ( )C ti im trn ( )C

c honh xl nghim ca phng trnh ( ) 20f x = c) Tm cc gi tr ca tham sm phng trnh sau y c nhiu

hn hai nghim: 4 22 0x x m + = Bi gii

Cu a:Hm s 4 22 3y x x= Tp xc nh: D =

34 4y x x = Cho 30 4 4 0 0; 1y x x x x = = = =


y y +

= + = +

Bng bin thin:

x 1 0 1 +

y 0 + 0 0 +y + 3 +

4 4

Hm s ng bin trn cc khong trn (1;0), (1;+) v nghch

bin trn cc khong (;1), (0;1).

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th hm s c im cc i (0; 3)D

v hai im cc tiu1 2( 1; 4), (1; 4)T T

Bng gi tr:

x 2 1 0 1 2 y 3 4 3 4 3 th hm s l ng cong i xngqua trc tung nh hnh v

Cu b:Ta c, 2 2 212 4 20 12 24 2 2y x x x x = = = = =

p s: 4 2 11y x= v 4 2 11y x= (hc sinh t gii)

Cu c:Ta c, 4 2 4 22 0 2 3 3x x m x x m + = = (*) Phng trnh (*) c nhiu hn 2 nghim khi v ch khi ( )C v

: 3d y m= ct nhau ti nhiu hn 2 im (3 hoc 4 im)

3 3 00 1

3 4 1

m mm

m m

4 m3 > 1 0 0m= 4 m3 = 1 2 2

0 < m< 4 3 < m3 < 1 4 4m= 0 m3 = 3 3 3m< 0 m3 < 3 2 2

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BI TP V HM S BC BA V HM S TRNG PHNGBi 6: Cho hm s 3 3 1y x x= + c th l ( )C

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti im thuc ( )C c honh bng 2.c) Vit pttt vi ( )C bit tip tuyn c h s gc bng 9.d) Tm iu kin ca m phng trnh sau c 3 nghim phn bit:

3 3 1 2 0x x m+ + = .

Bi 7: Cho hm s 3 21 32 2

2y x x= +

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

b) Vit pttt vi ( )C song song vi ng thng d: 92

2y x= +

c) Tm cc gi tr ca k phng trnh sau y c nghim duy

nht: 3 23 4 0x x k = Bi 8: Cho hm s 3 22 3 1y x x= +

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti giao im ca ( )C vi trc honh.c) Vit pttt vi ( )C bit tip tuyn song song vi : 12 1d y x=

d) Bin lun theo ms nghim phng trnh: 3 22 3 2 0x x m+ + =

Bi 9: Cho hm s 3 21 3 53 2 2

y x x= + c th l ( )C


b) Vit pttt vi ( )C ti im trn ( )C c honh xtho 1y = c) Tnh din tch hnh phng gii hn bi ( )C v : 2 0d y = .d) Tm cc gi tr ca m phng trnh sau c nghim duy nht

3 22 9 6 0x xe e m + =

Bi 10: Cho hm s 3 213

y x x=

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt ca ( )C ti im trn ( )C c tung bng 0.c) Vit pttt ca ( )C song song vi ng thng 8 3y x= d) Tm cc gi tr ca a phng trnh sau y c nghim duy

nht: 3 23 log 0x x a =

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Bi 11: Cho hm s 3 22 3 1y x x= (*)a) Kho st s bin thin v v th ( )C ca hm s.b) Tm to giao im ca ( )C vi ng thng d: 1y x=

c) Bin lun theo ms nghim ca phng trnh3 24 6 1 0x x m + =

Bi 12: Cho hm s 3 23 2y x x= + , m l tham s.a) Kho st s bin thin v v th ( )C ca hm s.

b) Vit pttt ca ( )C vung gc vi ng thng d: 1 13 3

y x=

c) Tm cc gi tr ca a ng thng 2y ax= + ct ( )C ti ba

im phn bit.Bi 13: Cho hm s 3 23 2y x x= + c th ( )C a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti im A(0; 2)c) Vit pttt ca ( )C bit tip tuyn song song vi 9 4 4 0x y = d) Bin lun theo ms giao im ca ( )C v : 2d y mx =

Bi 14: Cho hm s 34 3 1y x x= , c th l ( )C a) Kho st s bin thin v v th ( )C ca hm s.

b) Tm m phng trnh 34 3 1x x m = c ng 3 nghim.c) Vit pttt vi ( )C ti giao im ca ( )C vi trc honh.

d) Vit pttt vi ( )C bit tip tuyn vung gc vi 172

:d y x=

Bi 15: Cho hm s 3 22 6 6 2y x x x = + a) Kho st s bin thin v v th ( )C ca hm s.

b) Tnh din tch hnh phng gii hn bi ( )C , Ox, 1, 2x x= =

Bi 16: Cho hm s 2 2(2 )y x x= a) Kho st s bin thin v v th ( )C ca hm s.

b) Vit pttt vi ( )C ti im trn ( )C c honh bng 2 c) Vit pttt vi ( )C bit tip tuyn c h s gc bng 24.d) Tm cc gi tr ca tham sm phng trnh sau c 4 nghim

4 22 0x x m + =

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Bi 17: Cho hm s 4 22 3y x x= + a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt ca ( )C ti im trn ( )C c tung bng 5.

c) Tm iu kin ca m phng trnh sau y c ng 2 nghim:4 22 3 2 0x x m+ + + =

Bi 18: Cho hm s 12

y = 4 23x x + 32

c th ( )C .

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C bit tip tuyn c h s gc bng 8.

c) Tm m phng trnh sau c 4 nghim: 4 26 log 0x x m + =

Bi 19: Cho hm s 2 2(1 ) 6y x= c th ( )C a) Kho st s bin thin v v th ( )C ca hm s.

b) Bin lun theo ms nghim ca phng trnh : 4 22x x m = c) Vit pttt ca ( )C bit tip tuyn vung gc vi 1

24:d y x=

Bi 20: Cho hm s 14

y = 4 22 1x x+


b) Tm m phng trnh 4 28 4x x m + = c nhiu hn 2 nghimc) Vit phng trnh tip tuyn ca th ( )C ti im trn ( )C

c honh l nghim ca phng trnh ( ) 10y x =

Bi 21: Cho hm s 14

y = 4 22x x


b) Vit pttt ca ( )C song song vi1: 15 2012d y x= + .

c) Vit pttt ca ( )C vung gc vi2

:d 845

2012y x= +

d) Tm m phng trnh 4 28x x m + = c 4 nghim phn bit.

Bi 22: Cho hm s 4 2 ( 1)y x mx m = + c th ( )Cm a) Tm m th hm s i qua im ( 1; 4)M b) Kho st v v th ( )C ca hm s khi 2m = .c) Gi ( )H l hnh phng gii hn bi ( )C v trc honh. Tnh th

tch vt th trn xoay to ra khi quay ( )H quanh trc honh.

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2. Hm s nht bin v cc vn lin quana) Kho st s bin thin v v th hm s( 0, 0)c ad cb

ax by

cx d

+=

+

1 Tp xc nh: { }\ dc

D =

2 Tnh2( )

ad cby

cx d

=+

v khng nh y dng hay m, dc

x

3Suy ra hm s ng bin hay nghch bin trn mi khong xc

nh ( ; ),( ; )d dc c

+ v khng t cc tr.

4 Tnh cc gii hn v tm hai tim cn:Tnh lim

xy

a

c= v lim

xy

a

+ c= , suy ra

ay

c= l TCN

Tnh( )

limd

cx

y

v( )

limd

cx

y+

, suy rad

xc

= l TC

5 V bng bin thin ca hm s.6 Lp bng gi tr.

7V th hm s (c 2 tim cn) v nu nhn xt.

( 0, 0)ax b

y c ad cb cx d

+=

+

0y > 0y <

th hm s nht bin gm hai nhnh ring bitlun i xng nhau qua giao im ca hai ng tim cn

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b) Vit phng trnh tip tuyn (dng 1 bit to tip im M0)

1 Ch r0

x v0

y (honh & tung ca im M0)

2 Tnh0

( )f x

3 Cng thc: 0 0 0( )( )y y f x x x = c) Vit phng trnh tip tuyn (dng 2 bit trc h s gc k)

1 Lp lun c c0

( ) f x k = (*)

2 Thay0

( )y x vo (*) tm0

x

3 C0

x , tm0

y v dng cng thc0 0 0

( )( )y y f x x x =

Lu : Tip tuyn song song vi y ax b= + c h s gc k= a Tip tuyn vung gc vi ( 0)y ax b a = + c h s

gc 1a

k =

d) S tng giao gia th (C):y=f(x) v ng thng d: y= ax+ b1 Lp phng trnh honh giao im ca ( )C v d:

( ) f x ax b= + (*)2 Lp lun: s giao im ca ( )C v dbng vi s nghim ca (*)

3 m s nghim ca (*) suy ra s giao im ca ( )C v dV D MINH HO

Bi 23: Cho hm s 2 11

xy

x

+=

+

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti im trn ( )C c tung

bng 5

2

c) Chng minh rng ng thng : 2d y x m = + lun ct th( )C ti 2 im phn bit.

Bi gii

Cu a: Hm s2 1

1

xy

x

+=

+ Tp xc nh: \{ 1}D =

o hm:2

10, 1

( 1)

y x

x

= >

+

, do hm s ng bin

trn cc khong ( ; 1) , ( 1; ) + v khng t cc tr.

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Gii hn v tim cn:

lim 2 ; lim 2x x

y y +

= = y= 2 l tim cn ngang.

( 1) ( 1)

lim ; limx x

y y +

= + = 1x = l tim cn ng.

Bng bin thin:x 1 +

y + +

y+

22

Bng gi tr:

x 2 32

1 12

0

y 3 4 0 1 th hm s gm hai nhnh i xngnhau qua im ( 1;2)I nh hnh v

Cu b: Vi 52

y = th2 1 5

2(2 1) 5( 1) 31 2

xx x x

x

+= + = + =

+

Ta c2

1 14( 2)

( 3)f

= =

Vy, tip tuyn ca ( )C ti 52( 3; )M l:5 1 1 132 4 4 4

( 3)y x y x = + = +

Cu c:Honh giao im (nu c) ca ( )C v dl nghim phng trnh2 1

2 2 1 ( 2 )( 1)1

xx m x x m x

x

+= + + = + +

+, 1x

22 (4 ) 1 0x m x m + + = (*) ( 1x = khng tho (*))

Bit thc ca phng trnh (*):2 24 12 ( 2) 8 0,m m m m = + = + >

Do 0 > nn (*) lun c 2 nghim phn bit, t ( )C v dlun c 2 im chung phn bit.

Bi 24:a) Kho st v v th ( )C ca hm s 32

xy

x

=

b) Vit pttt ca ( )C bit tip tuyn song song vi :d y x=

c) Tm cc gi tr ca m ng thng :d y x m = + ct th( )C ti 2 im phn bit.

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Cu a: Hm s3 3

2 2

x xy

x x

= =

+ Tp xc nh: \ {2}D =

o hm:2

10, 2

(2 )y x

x

= <

, do hm s nghch bin

trn cc khong ( ;2) , (2; )+ v khng t cc tr. Gii hn v tim cn:

lim 1 ; lim 1x x

y y +

= = 1y = l tim cn ngang.

2 2

lim ; limx x

y y +

= = + 2x = l tim cn ng.

Bng bin thin:x 2 +

y

y1

+

1 Bng gi tr:

x 0 1 2 3 4y 3

22 0 1

2

th hm s gm hai nhnh i xng

nhau qua im (2; 1)I nh hnh v

Cu b: V tip tuyn song song vi ng thng y x= nn c h s

gc0

( ) 1k f x= =

20

11

(2 )x

=

2

0(2 ) 1x = 0 0

0 0

2 1 1

2 1 3

x x

x x

= = = =

p s:c 2 tip tuyn tho l 1y x= v 3y x= + Cu c: Phng trnh honh giao im ca ( )C v d:

3

2

xx m

x

= +

2 ( 3) 2 3 0x m x m + + + = (*)

( )C v d ct nhau ti 2 im phn bit khi v ch khi phng

trnh (*) c 2 nghim phn bit 20 2 3 0m m > > ( ; 1) (3; )m +

Vy vi ( ; 1) (3; )m + th th ( )C v ng thng:d y x m = + ct nhau ti 2 im phn bit.

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BI TP V HM S NHT BIN


xy

x

+=


b) Vit pttt vi ( )C bit tip tuyn c h s gc bng 3.c) Vit pttt vi ( )C ti im trn ( )C c tung bng 7

2

d) Tm m : ( 1) 2d y m x = + + ct ( )C ti 2 im phn bit.


xy

x

+=

+

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

b) Lp phng trnh tip tuyn ca ( )H bit tip tuyn song songvi ng phn gic ca gc phn t th nht.c) Vit pttt vi ( )H ti im trn ( )H c honh bng 3 .d) Tm m ng thng 1y mx= + ct ( )C ti 2 im phn bit.


xy

x

=


b) Vit pttt vi ( )C bit tip tuyn c h s gc bng34

c) Chng minh rng vi mi gi tr ca tham s m ng thngy x m= lun ct th ( )C ti hai im phn bit.

Bi 28: Cho hm s 32y1x

= +

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi th ( )C ti giao im ca ( )C vi trc honh.

c) Tm m ng thng :d y m x = ct ( )C ti 2 im phn bit

Bi 29: Cho hm s 23

xy

x

+=

c th ( )C .

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti im trn ( )C c honh bng 1.c) Vit pttt vi ( )C ti im trn ( )C c tung bng 3

2

d)Vit pttt vi ( )C bit tip tuyn c h s gc bng 5

4

e) Xc nh to giao im ca ( )C v 3 2y x= +

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Bi 30: Cho hm s 21

yx

=+

c th l ( )C .

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca th ( )C ti cc giao im

ca ( )C vi ng thng : 2 1d y x= c) Tm gi tr ln nht ca hm s trn on [0;2]

d) Vit pttt ca ( )C bit tip tuyn song song vi 1 32 2

y x= +

e) Tnh din tch hnh phng gii hn bi ( )C trc honh v haing thng x= 0, x= 2.

Bi 31: Cho hm s 11

xy

x

=

+

c th ( )C .

a) Kho st s bin thin v v th hm s.b) Tm im Mtrn trc honh m tip tuyn ca ( )C i qua im

Msong song vi ng thng d: y= 2x

Bi 32: Cho hm s 21

xy

x

=

+

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti giao im ca ( )C vi : 2 3d y x= .

c) Vit pttt ca ( )C vung gc vi ng thng21y 2012x= +

d) Tm m ng thng d: 2y mx= + ct c hai nhnh ca ( )C .


xy

x

=

a) Kho st s bin thin v v th ( )C ca hm s.b) Tnh din tch hnh phng gii hn bi ( )C , Ox v 2x = .

c) Vit phng trnh cc ng thng song song vi ng thng3y x= + ng thi tip xc vi th ( )C


xy

x

+=

a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti giao im ca ( )C vi trc tung.c) Vit pttt vi ( )C ti cc giao im ca ( )C vi : 2 4d y x=

d) Tm a ng thng : 3y ax = + th ( )C khng giao nhaue) Tm tt c cc im trn ( )C c to u l cc s nguyn.

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20/74



3. Tm GTLN, GTNN ca hm s y= f(x) trn on [a;b]1Hm s ( )y f x= lin tc trn on [a;b].

2 Tnh ( )y f x = .

3

Cho 0y = tm cc nghim [ ; ]ix a b (nu c) v cc s[ ; ]

jx a b lm cho y khng xc nh (nh loi cc s [ ; ]x a b

l)

4 Tnh cc gi tr ( )i

f x , ( )j

f x v ( ), ( )f a f b

(khng c tnhf ca cc xl

b loi)

5Chn kt qu ln nht v kt qu nh nht t bc 4 kt lunv gi tr ln nht v gi tr nh nht ca hm s trn on [a;b].

4. iu kin hm s c cc tr(tm tt)

Nu 0

0

( ) 0

( ) 0

f x

f x

= th hm s ( )y f x= t cc tiu ti

0x

Hm s 3 2y ax bx cx d = + + + c cc i, cc tiu 0y >

Hm s 4 2y ax bx c = + + c cc i, cc tiu . 0a b < 5. iu kin hm s n iu trn tng khong xc nh

Hm s 3 2y ax bx cx d = + + + ng bin trn

00,

0yy x

a

>

Hm s 3 2y ax bx cx d = + + + nghch bin trn

00,

0yy x

a

> (khng c du =)

Hm sax b

y

cx d

+=

+

nghch bin trn tng khong xc nh

0, 0y x D ad cb < < (khng c du =)

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21/74


V D MINH HOBi 35: Tm gi tr ln nht v gi nh nh nht ca hm s:

a) 3 28 16 9y x x x = + trn on [1;3]

b) 2 4 ln(1 )y x x= trn on [3;0]c) 3 22 ln 3 ln 2y x x= trn on 2[1; ]e

d) 2( 1)xy e x x = trn on [0;2]

Bi giiCu a: Hm s 3 28 16 9y x x x = + lin tc trn on [1;3]

o hm: 23 16 16y x x = +

Cho2

0 3 16 16 0y x x = + =

loai

nhan43

4 [1; 3] ( )

[1; 3] ( )

x

x

= =

Trn on [1;3] ta c: ( ) ; ;43 27 (1) 0 (3) 613 f f f = = =

Do 1327

6 0 < < nn[1;3]

min (3) 6x

y f

= = v[1;3]

maxx

y

( )4 133 27f= = Cu b: Hm s 2 4 ln(1 )y x x= lin tc trn on [3;0]

24 2 2 4

2 1 1

x x

y x x x

+ +

= + =

Cho(nhan)

(loai)2

1 [ 3;0]0 2 2 4 0

2 [ 3;0]

xy x x

x

= = + + = =

Trn on [2;0]: ; ;( 1) 1 4 ln 2 ( 3) 9 8 ln 2 (0) 0 f f f = = =

Do16

1 4 ln 2 ln 0e = < v2

9 8 ln 2 1 8 ln 0e = + > nn

[ 3;0]

min ( 1) 1 4 ln 2x

y f

= = v[ 3;0]

max ( 3) 9 8 ln 2x

y f

= =

Cu c:Hm s 3 22 ln 3 ln 2y x x= lin tc trn on 2[1; ]e t lnt x= th 2[1; ] [0;2]x e t , hm s tr thnh

3 2( ) 2 3 2y g t t t = = c 20 [0;2]

( ) 6 6 01 [0;2]

tg t t t

t

= = = =

Trn on [0;2]: (0) 2 ; (1) 3 ; (2) 2g g g= = =

Do 3 2 2 < < nn 2[1; ]min (1) 3x e y g = = v 2[1; ]max (2) 2x e y g = =

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22/74



Cu d: p s:[0;2]min (1)y f e= = v 2

[0;2]max (2)y f e= =

Bi 36: Tm iu kin ca tham sm hm s 3 2 4 3y x mx x = + + + a) ng bin trn b) C cc i v cc tiu

Bi giiCu a: 3 2 4 3y x mx x = + + + (*)

Tp xc nh: D= R

o hm: 23 2 4y x mx = + + c 2 12y

m =

Hm s (*) ng bin trn 0,y x

2

3 00 2 30 12 0y

am

m

>>

Vy, vi 2 3 ; 2 3m th hm s (*) ng bin trn

Cu b:Hm s (*) c cc i v cc tiu 0y = c 2 nghim phn

bit 20 12 0 ( ; 2 3) (2 3; )y

m m > > +

Vy vi ( ; 2 3) (2 3; )m + th hm s (*) c cc i vcc tiu.

Bi 37: Tm iu kin ca m hm s 3 2 23 ( 1) 2y x mx m x = + + t cc i ti

02x =

Bi gii

Cu a: 3 2 23 ( 1) 2y x mx m x = + + (*)

Tp xc nh: D=R

o hm: 2 2( ) 3 6 ( 1)y f x x mx m = = +

( ) 6 6y f x x m = =

Hm s (*) t cc i ti0

2x = khi v ch khi2(2) 0 {1;11}12 11 0

11(2) 0 212 6 0

f mm mm

f mm

= + = = < >


23/74


Bi 38: Chng minh rng nu sinx

xy

e= th 2 2 0y y y + + =

Bi gii

Hm s

sin

.sin

x

x

x

y e xe

= = c tp xc nh D = ( ) . sin .(sin ) (cos sin )x x xy e x e x e x x = + =

( ) (cos sin ) (cos sin ) 2 cosx x xy e x x e x x e x = + =

2 2 2 cos 2 (cos sin ) 2 sin 0x x xy y y e x e x x e x + + = + + =

Vy, vi .sinxy e x= th 2 2 0y y y + + =

BI TP V CC VN KHC LIN QUAN HM SBi 39: Tm gi tr ln nht, gi tr nh nht ca cc hm s sau y

a) 3 2( ) 2 3 12 10 f x x x x = + trn on [ 2; 0]

b) 5 4 3( ) 5 5 1 f x x x x = + + trn on [1;2]

c) 4 3 2( ) 2 1 f x x x x = + trn on [1;1]

d) 5 3( ) 5 10 1 f x x x x = + trn on [2;4]

e) 2( ) 25 f x x = trn on [3;4]

f) 2( ) 2 5 f x x x = + trn tp xc nh.

g)4

( ) 12

f x x x

= + +

trn on [1;2]

h) 3( ) 3 sin 2 sin 1 f x x x = + trn on [0; ] i) ( ) cos 2 sin 3 f x x x = +

j) ( ) 2 sin sin 2 f x x x = + trn on [ ]32

0;

Bi 40: Tm gi tr ln nht, gi tr nh nht ca cc hm s sau y:a) 2( ) x x f x e e = + trn on [ 1;2]

b) 2( ) ( 1) x f x x e = trn on [0;2]

c) 2( ) ( 1) x f x x x e = trn on [ 1;1]

d)

2

( ) 2 2

x

f x xe x x = trn on [0;1] e) 2( ) 2( 2) 2x f x x e x x = + trn on [0;2]

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24/74



f) 2( ) ln(1 2 ) f x x x = trn on [ 2; 0]

g) 2( ) 2 4 ln f x x x x = trn on [1;2]

h) 2( ) ln( 1) f x x x = + trn on [0;2]

i) ( ) ln 2 2 f x x x x = + trn on 2[1; ]e

j) 2 2( ) 2 ln 3 f x x x x = trn on [1;2 ]e

k)2ln

( )x

f xx

= trn on [ ]31;e

l)ln

( )x

f xx

= trn on 12

[ ;e 2 ]e

Bi 41: Tm cc gi tr ca tham sm hm s sau y lun ng bina) 3 2 ( 6) 2y x mx m x = + +

b) 3 2 22( 1) (2 2) 3y x m x m m x m = + + + Bi 42: Tm cc gi tr ca tham sa hm s sau y lun nghch bin

a) 3 2( 1) (2 1) 3y x a x a x = + + + b)7

5 3

ax ay

x a

+ =

+

Bi 43: Tm cc gi tr ca m hm s sau y c cc i v cc tiua)

3 2 2

2( 1) ( 3 2) 2y x m x m m x = + + + + b)

2 2 4

2

x mx m y

x

+ =

+

c) 4 2( 1) 2 3y m x mx = Bi 44: Tm cc gi tr ca tham sm hm s:

a) 3 2 22 ( 1) ( 4) 1y x m x m x m = + + + + t cc i ti0

0x =

b) 2 3 2(2 1) (2 3) 2y m x mx m x = + + t cc tiu ti0

1x =

c)2 63

my = 3 1x mx+ + t cc tiu ti0

2x =

d) 12

y = 4 2x mx n + t cc tiu bng 2 ti0

1x =

Bi 45: Chng minh rnga) Nu (cos2 sin2 )xy e x x = + th 2 5 0y y y + =

b) Nu 4 2x xy e e= + th 13 12y y y =

c) Nu ln xyx

= th 23 0y xy x y + + =

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25/74


PPPhhhnnn IIIIII... PPPHHHNNNGGG TTTRRRNNNH HH --- BBBT TT PPPHHHNNNGGG TTTRRRNNNH HH MMM! !! &&& LLLGGGAAARRRIIITTT

1. Phng trnh m (n gin)Cc tnh cht v lu tha cn lu : vi 0, 0a b> > v ,m n ta c

( ).

1 1

nm n m n m mn

mmnm n m n

n

n n

n n

a a a a a

aa a a

a

a aa a

+

= =

= =

= =

i i

i i

i i

( )

( ) ( )

( ) .

n

n

n n n

na a

b bn n

a b

b a

ab a b

=

=

=

i

i

i

a) Phng trnh m c bn: vi 0a > v 1a , ta c

x

a b= v nghim nu 0b logx

aa b x b= = nu 0b >

b) Phng php a v cng c s: vi 0a > v 1a , ta c( ) ( ) ( ) ( ) f x g x a a f x g x = =

c) Phng php t n s ph: Phng php gii chung:

0

Bin i phng trnh theo

( )f x

a , chng hn: 2 ( ) ( ). . 0 f x f x m a n a p+ + =

( )

( ) 1. . 0f x

f x

am a n p+ + =

1 t ( )f xt a= (km iu kin cho t) v thay vo phng trnh2 Gii phng trnh mi theo t tm nghim

0t (nu c)

3 i chiu nghim0

t tm c vi iu kin bc 1 ri tm x.

Lu 1: gp dng ( ) ( ). . 0 f x f x m a n a p+ + = , ta dng bin i

( )

( ) 1f x

f x

aa =

Lu 2: gp dng 2 ( ) ( ) 2 ( ). .( ) . 0 f x f x f x m a n ab p b+ + = , ta chia 2 v

phng trnh cho 2 ( )f xb d) Phng php lgarit ho: vi 0 1a< v 0 1b< , ta c

( ) ( ) ( ) ( )log log f x g x f x g x

a a

a b a b = =

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26/74



2. Phng trnh lgarit (n gin)Phng php chung: t iu kin xc nh ca phng trnh

Bin i phng trnh tm x(nu c) i chiu xtm c vi iu kin kt lun

Cc cng thc v quy tc tnh lgarit: vi 0 1a< v b > 0, 0 : log 1 0

a= log ( ) logn

m maa n

b b= ( 0n )

log ( )a

a = .log ( ) log loga a a

m n m n = + ( , 0m n > )

logaba b= ( )log log logma a an m n= ( , 0m n > )

log ( ) . loga a

b b = log

loglog c

c

b

a ab = ( 0 1c< )

1log logaa b b = 1loglogb

a ab = ( 1b )

a) Phng trnh lgarit c bn: vi 0a > v 1a , ta clog b

ax b x a = =

b) Phng php a v cng c s: vi 0a > v 1a , ta c log ( ) log ( ) ( ) ( )

a a f x g x f x g x = = (km iu kin ( ) 0f x > )

log ( ) ( ) ba

f x b f x a = =

Lu : Nu c ( ) 0f x > th2

log ( ) 2 log ( )n

a a f x n f x =

Nu ch c ( ) 0f x th2

log ( ) 2 log ( )n

a a f x n f x =

Bin i sau y rt dsai st(khng nn s dng):

a ra ngoi: log ( )a

f x thnh . log ( ) a f x

Tch log ( ). ( )a

f x g x

thnh log ( ) log ( )a a

f x g x +

Tch ( )( )

logf x

a g x

thnh log ( ) log ( )a a

f x g x

(ch c dng cc bin i trn khi ( ) 0, ( ) 0 f x g x > > ) Nn dng bin i di y:

a vo trong: . log ( )a

f x thnh log ( )a

f x

Nhp log ( ) log ( )a a

f x g x + thnh log ( ). ( )a

f x g x

Nhp log ( ) log ( )a a

f x g x thnh ( )( )

logf x

a g x

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27/74


c) Phng php t n s ph:

0 Bin i phng trnh theo log ( )a

f x , chng hn:2. log ( ) . log ( ) 0a a

m f x n f x p+ + =

1 t log ( )at f x= v thay vo phng trnh.2 Gii phng trnh mi theo t tm nghim

0t (nu c)

3 T0

t t= ta gii phng trnh lgarit c bn tm x.d) Phng php m ho: vi 0 1a< v 0 1b< , ta c

log ( ) log ( )log ( ) log ( ) a b

f x g x

a a f x g x a a = =

3. Bt phng trnh m lgarit (n gin) Cng c cc cch gii nh cch gii phng trnh m, lgarit. Tuy nhin khi gii bt phng trnh m v bt phng trnh lgaritcn ch so snh c savi 1 s dng tnh ng bin, nghch binca hm s m v hm s lgarit.

Hm s m xy a= ng bin khi a> 1, nghch bin khi 0 1a< <

Hm s lgarit loga

y x= cng ng bin khi a> 1 v nghch binkhi 0 1a< <

V D MINH HOBi 1: Gii cc phng trnh sau y:

a)2 35 625x x+ = b) ( )

15 7 23

(1, 5)x

x+ = c) 12 .5 200x x+ =

Bi gii

Cu a:2 23 3 45 625 5 5x x x x + += = 2 23 4 3 4 0x x x x + = + =

hoac1 4x x = =

Vy, phng trnh cho c 2 nghim: va1 4x x= = Cu b: ( ) ( ) ( )

1 5 7 15 7 2 3 33 2 2

(1,5) 5 7 1 1x x x

x x x x+ = = = =

Vy, phng trnh cho c nghim duy nht: x= 1

Cu c: 12 .5 200 2.2 .5 200 10 100 2x x x x x x+ = = = = Vy, phng trnh cho c nghim duy nht: x= 2

Bi 2: Gii cc phng trnh sau y:a) 9 5.3 6 0

x x

+ = b)1 1

4 2 21 0x x +

+ = c) 25 2.5 5 0x x + = d) 6.9 13.6 6.4 0x x x + =

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28/74



Hng dn gii v p s

Cu a: 29 5.3 6 0 3 5.3 6 0x x x x + = + =

t 3xt = (t> 0), phng trnh trn tr thnh:

(nhan so vi )(nhan so vi )2 3 05 6 0

2 0t tt tt t

= > + = = >

3t = th 3 3 1x x= = 2t = th 33 2 log 2x x= =

Vy, phng trnh cho c 2 nghim: x= 1 v3

log 2x =

Cu b: 1 14 2 21 0x x ++ = 44

x

2.2 21 0 4 8.2 84 0x x x+ = + =

Hng dn: t 2 ( 0)

x

t t= > . p s: 2log 6x = Cu c: 2

505 2.5 5 0 5 5 0

5

x x x

x

+ = + =

Hng dn: t 5 ( 0)xt t= > . p s: 1x =

Cu d: 6.9 13.6 6.4 0x x x + = .Chia 2 v ca phng trnh cho 4x ta

c: ( ) ( ) ( ) ( )2

9 6 3 34 4 2 2

6 13 6 0 6 13 6 0x x x x

+ = + =

Hng dn: t ( )32 ( 0)x

t t= > . p s: 1x =

Bi 3: Gii cc phng trnh sau y:a)

2 2log 4 log 1 1x x + = b)

5 25 0,2log log log 3x x+ =

c) 24 82

log 2 log log 13x x x+ + = d) 233

log ( 2) log ( 4) 0x x + =

Hng dn gii v p s

Cu a:2 2

log 4 log 1 1x x + =

(1)

iu kin:4 0 4

41 0 1

x xx

x x

> > > > > . Khi ,

(1)2

log ( 4)( 1) 1 ( 4)( 1) 2x x x x = = 2( 4)( 1) 4 5 0 0x x x x x = = = hoc 5x =

So vi iu kin x> 4 ta ch nhn nghim x= 5 Vy, phng trnh cho c nghim duy nht l x= 5

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29/74


Cu b:5 25 0,2

log log logx x+ = 13

(2) .

Vi iu kin x> 0, ( )2 11

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

+ =

p s: 3 3x = Cu c: 2

4 82log 2 log log 13x x x+ + = (3).

iu kin: x> 0, khi 2 12 2 23

(3) 2 log log log 13x x x + + =

p s: 8x = Cu d: 2

33log ( 2) log ( 4) 0x x + = (4).

iu kin: 22 0 2

4( 4) 0

x x

xx

> >

(I). Khi ,

23 3

(4) 2 log ( 2) log ( 4) 0x x + = 22 2

3 3 3log ( 2) log ( 4) 0 log ( 2)( 4) 0x x x x + = =

2 ( 2)( 4) 1( 2)( 4) 1

( 2)( 4) 1

x xx x

x x

= = =

p s: x= 3 v 3 2x = + Bi 4: Gii cc phng trnh sau y:

a) 22 2

log log 6 0x x = b) 22 2

4 log log 2x x+ =

c) 1 25 log 1 log

1x x +

+ = d) 2log (5 2 ) 2x x =

Hng dn gii v p s

Cu a:2

2 2log log 6 0x x = (5) iu kin: x> 0, t

2logt x= , phng trnh cho tr thnh:

2 6 0 3t t t = = hoc 2t = Vi 3t = th

2log 3 8x x= = (tho x> 0)

Vi 2t = th 22

log 2 2x x = = (tho x> 0)

Vy, tp nghim ca phng trnh (5) l: 14

{ ;8}S =

Cu b:22 24 log log 2x x+ = (6)

iu kin: x> 0, khi

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30/74



1/22 22 2 22

(6) 4 log log 2 4 log 2 log 2 0x x x x + = + =

Hng dn: t2

logt x= . p s: 12

x = v 2x =

Cu e:

1 2

5 log 1 log 1x x ++ = (7) iu kin: 0; log 1x x> v log 5x (I). t logt x= ,

(7) tr thnh 1 25 1

1 1 2(5 ) (5 )(1 )t t

t t t t +

+ = + + = +

2 5 6 0 3t t t + = = hoc 2t = Vi 3t = th log 3 1000x x= = (tho iu kin (I))

Vi 2t = th log 2 100x x= = (tho iu kin (I))

Vy, tp nghim ca phng trnh (7) l: {100;1000}S = Bi 5: Gii cc bt phng trnh sau y:

a)26 3 77 49x x+ b)( )

2 7 23 95 25

x x + +> c)4 3.2 2 0x x + <

Bi gii

Cu a: 2 26 3 7 (8) 6 3 7 27 49 7 7x x x x + + 26 3 7 2x x +

2

6 3 9 0x x + 3

2[ ;1]x (gii bng bng xt du) Vy, tp nghim ca bt phng trnh (8) l S= 3

2[ ;1]

Cu b:( ) ( ) ( )2 27 2 7 2 2(9)3 9 3 3

5 25 5 5

x x x x + + + +> > 2 7 2 2x x + + <

2 7 0 ( ; 0) (7; )x x x + < + (gii bng bng xt du)

Vy, bt phng trnh (9) c tp nghim: S= (;0)(7;+)

Cu c:4 3.2 2 0

x x

+ < (10)

t 2xt = (t> 0), (10) tr thnh: 2 3 2 0t t + < vi t> 0

Bng xt du: cho 2 3 2 0 1; 2t t t t + = = = t 0 1 2 +

2 3 2t t + + 0 0 +

Nh vy,1

2

t

t

> > <


31/74


Bi 6: Gii cc bt phng trnh sau y:a) 2

0,5log ( 5 6) 1x x + b) 2 2ln( 2) ln(2 5 2)x x x+ +

c) 1 13 3

2log (2 4) log ( 6)x x x+

Bi gii

Cu a: 20,5

log ( 5 6) 1x x +

iu kin: hoac2 5 6 0 2 3x x x x + > < > (I). Khi ,2 2 1

0,5log ( 5 6) 1 5 6 (0,5)x x x x + +

2 5 4 0 1 4x x x +

Kt hp vi iu kin (I) ta nhn cc gi tr: [1;2) (3;4]x Vy, tp nghim ca bt phng trnh l: [1;2) (3;4]S =

Cu b: 2 2ln( 2) ln(2 5 2)x x x+ +

iu kin:hien nhien

2

2

2 5 2 0

2 0 :

x x

x

+ > + >

hoac12

x 2x < > (I)

Khi , 2 2 2 2ln( 2) ln(2 5 2) 2 2 5 2x x x x x x + + + + 2 5 0 0 5x x x

Kt hp vi iu kin (I) ta nhn cc gi tr: 12

[0; ) (2;5]x

Vy, tp nghim ca bt phng trnh l: 12

[0; ) (2;5]S =

Cu c: 1 13 3

2log (2 4) log ( 6)x x x+

iu kin:

hoac2 2 36 0322 4 0

x xx x

xxx

< > > > > + >

Vi iu kin x> 3 ta c

1 13 3

2 2log (2 4) log ( 6) 2 4 6x x x x x x + +

2 3 10 0 2 5x x x Kt hp vi iu kin x> 3 ta nhn cc gi tr 3 5x< Vy, tp nghim ca bt phng trnh l:

(3;5]S

=

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BI TP V PHNG TRNH MBi 7: Gii cc phng trnh sau y:

a) 27 8.7 7 0x x + = b) 22.2 2 1 0x x+ =

c) 9 3 6 0x x = d) 25 2.5 15 0x x+ =

e) 2 12 2 6x x+ = f) 2 38 2 56 0x x =

g) 33 3 12x x+ = h) 32 2 2 0x x + =

i) 2 3 25 5 20x x = j) 17 2.7 9 0x x+ =

k) 2 24. 3x xe e = l) 16 2.6 13 0x x+ + =

m)3.4 2.6 9x x x = n) 2 125 10 2x x x++ =

o) 25 15 2.9x x x

+ = p) 5.4 2.25 7.10 0x x x

+ = q) 6 33. 2 0x xe e + = r) 4 12 15.4 8 0x x+ =

s) 2 15 5.5 250x x + = t) 2 13 9.3 6 0x x+ + =

u) 2 6 72 2 17x x+ ++ = v) 1 1 12 (2 3 ) 9x x x x + = Bi 8: Gii cc phng trnh sau y:

a) 2 5 2 32 2 12x x+ ++ = b) 4 2 12 2 5 3.5x x x x + + ++ = +

c)2 1 2

3 3 108x x

+ = d)2 2

5 7 .17 7 5 .17x x x x

+ = + e) 1 22 .5 0,2.10x x x = f)

25 5 11 1 2 212 .4 48.3x x x x + =

g) 3 1 3 28.4 2x x = h) 3 3 1 12 .3 2 .3 192x x x x + =

i)2 2 13 .2 72x x x x + = j) 1 2 3(0,25) 2 0,125.16x x =

Bi 9: Gii cc phng trnh sau y:a) 13.2 4 1 0x x++ = b) 2 4 15 110.5 75 0x x+ + =

c) ( ) ( ) 15 7 231, 5xx + = d) ( )

52 22 16

9(0, 75) 0xx x =

e) 2 1 23 3 108x x + = f) 2( 1)16 2 12 0x x++ =

g) 4.9 12 3.16 0x x x+ = h) 4 8 2 53 4.3 27 0x x+ + + =

i) 13 (3 30) 27 0x x+ + = j) 3 2 1 32 2 2 0x x x+ + =

k) 2 22 9.2 2 0x x+ + = l) 1 3 21 3.2 2 0x x + =

m)2 1 2

3 2.3 5 0x x

+ = n) 4.9 12 3.16 0x x x

+ = o)

2 222 2 3x x x x + = p) 4 2 22.16 2 4 15x x x =

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q) ( ) ( )2 33 24. 2. 6 0x x

+ = r) ( ) ( )2 3 2 3 4x x

+ + =

s) 12 .4 64 5 0x x x + = t) 14 4 .4 3 0x x x+ + =

u)

1

36 3 .2 4 0

x x x+

= v)

1

4 2 .4 3 0

x x x

= BI TP V PHNG TRNH LGARITBi 10: Gii cc phng trnh sau y:

a) 2log( 6 5) log(1 )x x x + = b) 4 22

ln . log ( 2 ) 3 lnx x x x =

c) 17

27

log ( 2) log (8 ) 0x x+ + = d) 13

23

log ( 10) log (3 ) 0x x + =

e) ln(4 4) ln( 1) lnx x x = f)22

log ( 1) log (7 )x x =

g) 2 4log 2 log ( 1) 1x x + + = h) 13

3log ( 2) log ( 4) 1x x =

i)2 2

log ( 1) log (2 11) 1x x = j)2 4 0,5

log (2 ) log logx x x+ =

k)2 0,5

log ( 3) log ( 1) 3x x + = l)5 0,25

log log log 2x x x+ =

m)3 9 27

log log log 11x x x+ + = n) 4 3log log(4 ) 2 logx x x+ = + Bi 11: Gii cc phng trnh sau y

a)25 5log 4 log 3 0x x + = b)

22 22 log log 1 0x x+ =

c) 25 0,2

log log 12 0x x+ = d) 2ln ln( ) 1 0x ex =

e) 22 0,5

log 5 log 4 0x x+ + = f) 22 0,5 2

3 log log log (2 )x x x =

g) ( )22 4 8log 6 log 7xx = h) 2

0,2 5log 5 log 6 0x x+ + =

i) 2 2log 3 log log 4x x x = j) 2log (10 ) 9 log(0,1. )x x=

k) 3log log 9 3xx + = l) 3log 27 3 log 8x x = m)

22 log 2 log 5

xx+ = n) ( )6 62 log 5 log 6

xx

x =

Bi 12: Gii cc phng trnh sau ya) 2

3 3log ( 5) log (2 5)x x x = + b) log (2 ) log (10 3 )x x

=

c) 3 3log log4 5.2 4 0x x + = d) 2log (10 ) 3 log 1 0x x =

e) 55log ( 2) log (4 5)x x+ = + f)2

3 3log (3 ) log 1 0x x+ = g) 2

2 0,52log 3 log log 2x x x+ + = h) 2 3log log 2 0x x + =

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i) log 1 log 2 1log 2 log 1 2

x x

x x

+ +

= j) 2 84 16

log log (4 )

log (2 ) log (2 )

x x

x x=

k) 13 3log (3 1). log (3 3) 6x x+ = l)

5 5log ( 2) log ( 6)x x x+ = +

m) 3log(10 ).log(0,1. ) log 3x x x= n)

4 22log 4 log log (4 ) 12x x x+ + =

o) 2 14 22log ( 2) log (3 1) 1x x + =

p)2 2

1log log ( 1)( 4) 2

4

xx x

x

+ + = +

BI TP V BT PHNG TRNH M - LGARITBi 13: Gii cc bt phng trnh sau y

a)22 3(0, 5) 2x x b)2 2 3 0x x+ < c)

2 32 4x x + <

d) 2 13 3 28x x+ + e)4 3.2 2 0x x + > f)2

3 9x x < Bi 14: Gii cc bt phng trnh sau y

a) 2 6 72 2 17x x+ ++ > b) 2 3 25 2.5 3x x c)4 2 3x x> +

d) 4 4 2 22.2 2 4 15x x x e)5.4 2.25 7.10x x x+ f) 14 16 3x x+ Bi 15: Gii cc bt phng trnh sau y

a)2 2

log ( 5) log (3 2 ) 4x x+ b) 13

52

log log 3 x

x >

c) 28 8 32 log ( 2) log ( 3)x x > d) 1

3

3 1log 1

2

x

x

>

+

e)4 4

log ( 7) log (1 )x x+ > f) 22 2

log log 0x+ Bi 16: Gii cc bt phng trnh sau y

a) 1 12 2

2log (5 10) log ( 6 8)x x x+ < + +

b)2 2

log ( 3) log ( 2) 1x x + c) 1 12 2

log (2 3) log (3 1)x x+ > +

d)0,2 0,2

log (3 5) log ( 1)x x > + e)3 3

log ( 3) log ( 5) 1x x + <

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PPPhhhnnn IIIIIIIII... NNNGGGUUUYYYNNN HHHM MM --- TTTCCCH HH PPPHHHNNN VVV (((NNNGGG DDD***NNNGGGI. TM TT CNG THC V PHNG PHP GII1. Bng cng thc nguyn hm v nguyn hm m rng

11

2 2

1. .( )1

. ( ) .1 1

1 1 ln. ln .

1 1 2. 2 .

1 1 1 1 1. .

( )

. .ax b

x x ax b

dx x C a dx ax C

ax bxx dx C ax b dx C

aax b

dx x C dx C x ax b a

ax bdx x C dx C

ax ax b

dx C dx C x a ax bx ax b

ee dx e C e dx

a

++

++

= + = ++

= + + = ++ +

+= + = +

++

= + = ++

= + = +++

= + =

i i

i i

i i

i i

i i

i i

2 2

2 2

sin( )cos . sin cos( ).

cos( )

sin . cos sin( ).tan( )1 1

. tan .cos cos ( )

cot( )1 1. cot .

sin sin ( )

C

ax bx dx x C ax b dx C

aax b

x dx x C ax b dx C a

ax bdx x C dx C

ax ax b

ax bdx x C dx C

ax ax b

+

+= + + = +

+

= + + = ++

= + = ++

+= + = +

+

i i

i i

i i

i i

2. Cng thc tch phnVi ( )F x l mt nguyn hm ca hm s ( )f x trn on [ ; ]a b th

( ) ( ) ( ) ( )b b

aaf x dx F x F b F a = =

3. Phng php i bin s (loi 2): xt ( ) . ( ).b

aI f t x t x dx =

1 t ( )t t x= ( ).dt t x dx = (v 1 s biu thc khc nu cn)2 i cn: ( )x b t t b= =

( )x a t t a = =

3 Thay vo:( )

( ).( )t b

t aI f t dt = v tnh tch phn mi ny (bin t)

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Vi dng tch phn i bin thng dng:Dng tch phn Cch t c im nhn dng

( )

( )

t xdx

t x

( )t t x= mu

( ). ( )t xe t x dx ( )t t x= m( )( ) . ( ).f t x t x dx ( )t t x= ngoc

( )n ( ) . ( )f t x t x dx n ( )t t x= cn

( )ln dxf xx

lnt x= ln x

(sin ).cos f x xdx sint x= cos .x dx i km biu thc theo sin x (cos ).sin f x xdx cost x= sin .x dx i km biu thc theo cos x

2(tan )

cos

dxf x

x tant x= 2cos

dx

xi km biu thc theo tan x

2(cot )

sin

dxf x

x

cott x= 2sin

dx

x

i km biu thc theo cotx

( ).ax ax f e e dx axt e= axe dx i km biu thc theo axe i khi thay cch t ( )t t x= bi . ( )t m t x n = + ta s gp thun li hn

4. Phng php tch phn tng phn

( ). . .b bb

aa au dv u v v du =

Vi dng tch phn i bin thng dng:Vi ( )P x l mt a thc, ta cn ch cc dng tch phn sau y

( ).sin .P x ax dx , ta t( )

sin .

u P x

dv ax dx

= =

( ).cos .P x ax dx , ta t( )

cos .

u P x

dv ax dx

= =

( ). .axP x e dx , ta t ( ).axu P xdv e dx = =

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. sin .axe bx dx , ta t sin .axu e

dv bx dx

= =

(khong co )

( ). ln . ,n f x x dx

dx

x

ta t ln( ).

n

u xdv f x dx

= =

5. Tnh din tch hnh phngCho hai hm s ( )y f x= v

( )y g x= u lin tc trn on

[ ; ]a b ,Hl hnh phng gii hn

bi cc ng: 1 2( ) : ( ),( ) : ( ),C y f x C y g x x a = = = v x b= Khi , din tch ca hnh phngHl: ( ) ( )

b

aS f x g x dx =

Lu 1: nu2

( )C l trc honh th ( ) 0g x = v ( )b

aS f x dx =

Lu 2: Khi tnh tch phn ( )b

as x dx ta cn lu nh sau:

Nu ( ) 0, [ ; ]s x x a b th

( ) ( ).b b

a as x dx s x dx =

Nu ( ) 0, [ ; ]s x x a b th

( ) ( ).b b

a as x dx s x dx =

Nu ( )s x khng c nghim trn khong ( ; )a b th

( ) ( ).b b

a a

s x dx s x dx =

Nu ( )s x c nghim 1 2 nc c c< <


38/74



V D MINH HO

Bi 1: Tnh 30 2

3

1

xA dx

x=

+ 2

3

1 cos

sin (1 cos )

xC dx

x x

=

+

22

13 . .xB x e dx

=

4

2ln 1

. lnxD dx

x x+=

Bi gii

Cu a:3

0 2

3

1

xA dx

x=

+ t

2 2 21 1t x t x = + = + 2 . 2 . . .t dt x dx t dt x dx = =

i cn: 3x = 2t = 0x = 1t =

Vy, ( )2 2 211 1

3. 3. 3 6 3 3tdtA dt t t

= = = = =

Cu b:22

13 . .xB x e dx

= t 2t x= 122dt xdx xdx dt = =

i cn: 2x = 4t = 1x = 1t =

Vy, ( )44 43 3 3

2 2 21 1

3 .

2

tte dtB e e e = = =

Cu c: 2 23 3

2

1 cos sin

sin (1 cos ) (1 cos )

x xC dx dx

x x x

= =

+ +

t 1 cos sin .t x dt x dx = + = sin .x dx dt =

i cn:2

x = 1t =

3x = 3

2t =

Vy, ( )33

2232

1

12 21 11 . t

dtC dt

t t= = = ( )2 1 13 1 3= =

Cu d:4

2

ln 1

. ln

xD dx

x x

+= t

1lnt x dt dx

x= =

i cn: 4x = 2ln2t = 2x = ln 2t =

Vy, ( )ln 4ln 4 ln 4

ln2 ln2 ln2

1 11 ln

tD dx dt t t

t t

+ = = + = +

( ) ( )ln 4 ln ln 4 ln 2 ln ln 2 ln 4 = + + =

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Bi 2: Tnh cc tch phn sau y: 20

( 1) sinE x xdx =

2

13 . xF x e dx

=

22

1(3 1) ln .G x x dx =

Bi gii

Cu e: 20

( 1) sinE x xdx =

t1

sin cos

u x du dx

dv xdx v x

= = = =

Suy ra, ( ) ( )22 20 00

( 1)cos cos 0 1 sinE x x xdx x = + = +

21 sin sin 0 0= + =

Cu f:2

13 . xF x e dx

= t 3 3x xu x du dx dv e dx v e

= = = =

Nh vy, ( ) ( )2 22

2 1

11 13 . 3 6 3 3x x xF x e e dx e e e

= = +

2 2 1 2 2 23 3 3 66 3( ) 6 3 3e e e e e e e e e e

= + = + + = +

Cu g:2 2

1(3 1) ln .G x x dx = t 2

3

1ln(3 1)

u x du dx x

dv x dx v x x

= = = =

( ) ( )2 223 2 31 4

3 311 1ln ( 1). 6 ln 2 6 ln 2G x x x x dx x x = = =

Bi 3: Tnh cc tch phn sau y2

1

1xH x e dx x

=

22

0 ( 1).I x x xdx = + + 3

21

2 1e t tJ dt

t

+= 20 (1 2 sin )sinK a ada

= +

Bi gii

Cu h:2 2 2 2

1 1 1 1

1( 1) 1.x x xH x e dx xe dx xe dx dx

x

= = =

Xt 21 1:xH xe dx = t x xu x du dx dv e dx v e = = = =

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( ) ( )2 22

2 21 11 1

. 2x x xH xe e dx e e e e = = = =

Xt ( )2 2

2 111 2 1 1H dx x = = = =

Vy, 21 2 1H H H e = =

Cu i:2 2 2

2 2 2

0 0 0( 1). . 1.I x x x dx x dx x xdx = + + = + +

Xt ( )22 2 31 8

1 3 30 0I x dx x = = =

Xt2

22 0

I 1.x xdx = + . t 2 1t x tdt xdx = + =

i cn: 2x = 5t = 0x = 1t =

( )55 5 2 31

2 31 1 1.I t tdt t dt t = = = 5 5 13

=

Vy, 5 5 71 2 3I I I+= + =

Cu j:3 2

2 22 1 2 1 1

1 1 2 12 ln

e

e et t tt tt t

J dt t dt t + = = + =

( ) ( )2 21 1 1 1 32 2 1 2 22 ln 2 ln 1e ee ee= = Cu k: 2 2 2

0 0(1 2 sin )sin (sin 2 sin )K a ada a a da

= + = + 2

0(sin 1 cos 2 )a a da

= + ( ) 2sin22 0cosaa a

= +

( ) ( )sin sin 02 2 2 2 2cos cos 0 0 1= + + = +

Bi 4: Tnh din tch hnh phng gii hn bi cc ng sau y:a) 3 3 2y x x= + , trc honh, 1x = v 3x =

b) 24y x= v 2 42y x x=

c) 3 2y x x= v tip tuyn ca n ti im c honh bng 1

d) 3y x x= v 2y x x=

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Hng dn gii v p s

Cu a: Xt3

3( ) 3 2 ( ) ( ) 3 2( ) 0

f x x x f x g x x x

g x

= + = + =

Din tch cn tm l 2 31

3 2S x x dx

= + Bng xt du ca 3 3 2x x + trn on [ 1;2]

x 1 1 23 3 2x x + + 0 +

Vy, ( )2

3

13 2S x x dx

= + ( )

4 22

3 214 2 4

12x x x

= + =

Cu b: Xt2

4 22 4

( ) 4( ) ( ) 3 4

( ) 2

f x x f x g x x x

g x x x

= = =

Cho 4 23 4 0x x = 2x =

Din tch cn tm l2

4 2

23 4S x x dx

=

Bng xt du ca 4 23 4x x trn on [ 2;2]

x 2 24 23 4x x

( )22 4 2 5 31 96

5 52 2( 3 4) 4S x x dx x x x

= = =

Cu c: HD: vit phng trnh tip tuyn tho (p s: 2y x= + )

Xt3

3( ) 2 ( ) ( ) 3 2

( ) 2

f x x x f x g x x x

g x x

= =

= +

Cho 3 3 2 0 1x x x = = hoc 2x =

Din tch cn tm l:2

3

13 2S x x dx

=

Bng xt du ca 3 3 2x x trn on [ 1;2] x 1 2

3 3 2x x

( )22 3 4 2 271 34 2 41 1( 3 2) 2S x x dx x x x = = =

1

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42/74



Cu d:Xt3

3 22

( )( ) ( ) 2

( )

f x x x f x g x x x x

g x x x

= = + =

Cho 3 2 2 0 2; 0; 1x x x x x x + = = = = .

Din tch cn tm l1 3 2

22S x x x dx

= +

HD: xt du 3 2 2x x x+ v a n cng thc0 1

3 2 3 2

2 0( 2 ) ( 2 )S x x x dx x x x dx

= + +

( ) ( )0 1

4 3 2 4 3 2 371 1 1 14 3 4 3 122 0

x x x x x x

= + + =

Bi 5: Tnh th tch vt th trn xoay sinh ra khi quay hnh (H)quanhtrc Oxbit (H) gii hn bi: siny x= ,Ox, 0x = v 3

2x =

Bi gii

Ta c, ( ) sin f x x = . Xt on [ ]32

0;

Th tch cn tm l:32 2

0(sin )V x dx

= 3 3 32 2 22

0 0 0

1 cos 2 1 cos 2sin2 2 2

x xV xdx dx dx

= = =

( ) ( )3

221 1 3 1 32 4 4 4 40

sin 2 sin 3 .0x x

= = =

BI TP V TCH PHNBi 6: Tnh cc tch phn sau ya)

12

0

.(2 1)x x dx

b)

ln2

0

(3. 5)x xe e dx

c)

13

1

(2 3 )x dx

d)2

1

1 tte tdt

t

+ e)

2

1

(1 ) x

x

x e xdx

xe

+ f)

23

1

3 2t tdt

t

+

g) ( )22

11 t

t dt h) ( )21

22

x x dx x

+ i)1

3

0(1 )x x dx

j) 46

cos4 .cos3x xdx

k) 64

sin 3 . sin .t t dt

l)4 2

0tan xdx

m) 120

1 .cos

xx ee dxx

+ n)2 1ln2

0

1xx

e dxe

+ + o)2

01 x dx

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p)32

1

2 5t tdt

t

q)

22

0

3 1

1

x xdx

x

+ r) 12

1 3 1

( 1)

xdx

x x

++

m) 3

6

2 2

2

tan cos

sin

x xdx

x

n) 3

20

2 cos2 1

cos

xdx

x

o) 4 2

0

sin .x dx

Bi 7: Tnh cc tch phn sau ya) 2

0

sin

1 3 cos

xdx

x

+ b)2

21

1

2 3

xdx

x x

c)

21 1

0. xx e dx

d)1/2

21

xedx

x e) 2

62

cos

(1 sin )

xdx

x

+f)

20

41 (1 )

xdx

x

g) 20sin .

8 cos 1

x dx

x

+ h)19

0 3 2

3

8

xdx

x + i)2

1

1e ln xdxx

+

j)1

1

(1 ln )e

e

dxx x k)

3

1 . 4 ln

e dx

x x

l)1

ln .

.(ln 3)e

e x dx

x x +

m)1

2012

0( 1)x x dx n)

12

01x x dx + o)

73

0. 1x x dx +

p)2

2

3

sin .cos .x x dx

q) 40

sin2

.cos2

x

e xdx r)0

5 4 .x x dx s)

22

sin2

1 cos

xdx

x

+t)

1

2 20

4

(2 1)

xdx

x +u)

ln 3

0 1 xdx

e+

Bi 8: Tnh cc tch phn sau ya)

1

0( 1) xx e dx + b)

1

0(2 1) xx e dx c)

12 1

0. xx e dx

d)

ln 5

ln2 2 ( 1)x

x e dx e)ln2

0 ( 1)x

x e dx

f)2

x x dx 0 2 .cos .

g) 4

0(2x 1)cos

xdx h)0

(1 )cosx xdx

i) 20 2 .sinx xdx

j) 40

(x 1)sin2

xdx+ k) 40 x xdx sin2

l) 1 ln .e

x dx

m)1

2 .(ln 1)e

x x dx n)3

22 ln( 1)x x dx o)

2

21

ln xdx

x

p)3

2 2

0( 1). xx e dx + q) 40 sin

xe xdx

r)4

1

xe dx

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Bi 9: Tnh cc tch phn sau ya)

1

0(3. 5 )x xe e x dx b) ( )0 x x cos

x dx+ c)2

2

0( )xx x e dx +

d)

2

1

lnx x

dxx

+

e)4

1

xx e

dxx

+

f) 211e lnx x

dxx

+

g) ( )1

ln 1e

x x dx + h) 40 (x x xdx cos )sin

+ i)2

1( 2 )xx xe dx +

j)1

0

1

1

x

x

xe xdx

e

+ +

+k) 2

0

1 sin

1 cos

xdx

x

+ l)

2

21

( 1). lnx xdx

x

Bi 10: Tnh cc tch phn sau y1) ( )0 2 11 xx ee dx 2)

2

1 ( 1)dx

x x + 3)6

0cos

2 sin 1xdxx

+

4)1

03 1.x dx+ 5)

2

1(2 1) ln .x x dx + 6) 1 ln( 1)

e

x dx+

7)22

1

1 ln xdx

x

+ 8)

4

1

lne .x dx

x 9)2 22

1

lnx xdx

x

+

10)

1

0

2 1

1

x

dxx

+ 11)4

1 ( 2)

dx

x x + 12)32 2

0 2 1

x dx

x +

13) 4tan

20 cos

xe dx

x

14) 20

cos sin

1 cos

x xdx

x

+ 15)

2ln2

30 ( 4)

x

x

e dx

e +

16)0

ln63.x xe e dx + 17) 0 ( cos )

xx e

x dx+ x xdx 18) 0 2 sin

19) 34

30

cos sin

cos

x xdx

x

+

20)

21 (ln 1)

e dx

x x +21)

2

1

ln .

(ln 2)

e x dx

x x +

22) 2 20

sin 2 .sin .x x dx

23) 2 20

sin .cosx xdx

24)1

0(4 1) xx e dx +

25)2

1

ln 1e x xdx

x

+ 26) 20

sin2 .

1 cos

x dx

x

+ 27)2

0

sin 2 .

3 sin 1

x dx

x

+

28)0

(1 cos ) cos .x x dx

29) ( )2

04 1x x dx + 30)

1

0( 3)xxe dx +

31) ( cos 2)x x dx

32) 1 ( ln 2)

ex x x dx + 33)

21 3

0

xx e dx

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45/74


BI TP V NG DNG HNH HC CA TCH PHNBi 11: Tnh din tch hnh phng gii hn bi cc ng sau y

a) 3 21 23 3

y x x= + , trc honh, x= 0 v x= 2.

b)2

1, 1, 2y x x x = + = = v trc honh.c) 3 12y x x= v 2y x= .

d) 2 2y x x= + v 2y x+ = .

e) 3 1y x= v tip tuyn ca n ti im c tung bng 2.

f) 3 3 2y x x= + v trc honh.

g)2

2y x x= v2

4y x x= + h) 2 2y x x= v y x=

i) 3 2y x x= v ( )91y 1x=

j) ( ) : 1 , 1C xy x x = + = v tip tuyn vi ( )C ti im ( )322; .

k)3 1

, , 01

xy Ox x

x

+= =

l) 1ln , ,e

y x x x e = = = v trc honh.

m)ln

1x

y xx

= + , 1y x= v x e=

Bi 12: Tnh th tch cc vt th trn xoay khi quay cc hnh phng giihn bi cc ng sau y quanh trc km theoa) 2 4 ,y x x= trc honh, 0, 3x x= = ( l trc honh)

b) cos ,y x= trc honh, 0,x x = = ( l trc honh)c) tan ,y x= trc honh,

40,x x = = ( l trc honh)

d) ,xy e x= trc honh v 1x = ( l trc honh)

e)2

,2

yx

=

trc honh, 0, 1x x= = ( l trc honh)

f) 22 , 1y x y= = ( l trc honh)

g) 22y x x= v y x= ( l trc honh)h) 3 2 1y x= + , 3y = v trc tung ( l trc tung)

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BI TP V NGUYN HM

Bi 13: Chng minh rng hm s 2( ) ( 1)xF x e x = + l mt nguyn hmca hm s 2( ) ( 1)x f x e x = + trn .

Bi 14: Chng minh rng hm s ( ) ln 3F x x x x = + l mt nguynhm ca hm s ( ) ln f x x = trn .

Bi 15: Chng minh rng 4 4( ) sin cosF x x x = + v 1 cos 4( )4

xG x

= l

nguyn hm ca cng mt hm s vi mi xthuc

Bi 16: Tm gi tr ca tham sm 3 2( ) (3 2) 4 3F x mx m x x = + + + l mt nguyn hm ca hm s 2( ) 3 10 4 f x x x = + trn

Bi 17: Tm a,b v c 2( ) ( ) xF x ax bx c e = + + l mt nguyn hm cahm s ( ) ( 3) x f x x e = trn

Bi 18: Tm nguyn hm ( )F x ca hm s ( ) cos (2 3 tan ) f x x x = bitrng ( ) 1F =

Bi 19: Tm nguyn hm ( )F x ca hm s 21 2( ) xf xx

+= tha mn iu

kin ( 1) 3F = .

Bi 20:Tm nguyn hm ( )F x ca hm s2

1 ln( )

xf x

x

+= tha mn

iu kin ( ) 0F e = .

Bi 21: Tm nguyn hm ( )F x ca hm s 2( ) (2 ) f x x x = tha mniu kin ( 1) 3F = .

Bi 22: Tm nguyn hm ( )F x ca hm s 2(1 2 )( ) xf xx

= tha mn

iu kin ( 1) 1F = .

Bi 23: Tm nguyn hm ( )F x ca hm s ( ) (4 1) x f x x e = + tha mniu kin (1)F e= .

Bi 24: Tm nguyn hm ( )F x ca hm s (1 ln )( ) xx x ex

f x+

= tha mn

iu kin (1)F e= .

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PPPhhhnnn IIIVVV... SSS PPPHHH(((CCC

1. Cc khi nim v php ton lin quan n s phc n v o i: 2 1i = i 3i i= i 4 1i =i S phc z a bi = + l s c phn thc l a v phn o b Mun ca s phc z a bi = + l: 2 2z a b= + S phc lin hp ca s phc z a bi = + l: z a bi =

Hai s phc bng nhau:a c

a bi c di b d

=+ = + =

Php cng hai s phc: ( ) ( ) ( ) ( )a bi c di a c b d i + + + = + + +

Php tr hai s phc: ( ) ( ) ( ) ( )a bi c di a c b d i + + = + Php nhn hai s phc: ( ).( ) ( ) ( )a bi c di ac bd ad bc i + + = + +

Php chia hai s phc: 1 1 2

2 2 2

.

.

z z z

z z z= (nhn c t ln mu cho

2z )

S phc nghch o ca zl: 1.z

z z z=

Mi s thc am c 2 cn bc hai phc l: .a i

Ch :s phc ch c phn o (phn thc bng 0) gi l s thun o2. Gii phng trnh bc hai h s thc ( < 0) trn tp s phc

Cho phng trnh bc hai va2 0 ( , , 0)az bz c a b c a + + =

Tnh 2 4b ac = v ghi kt qu di dng 2( . )i Kt lun phng trnh c 2 nghim phc:

1z =

2

b i

a

v2

z =2

b i

a

+

Lu :

Ch c dng cng thc nghim nu trn khi < 0 Trng hp 0 ta gii pt bc hai trn tp s thc (nh trc).

Khi gii phng trnh trng phng trn C, ta t 2t z= (khng cn

iu kin cho t)Nu dng bit thc th cng thc tm hai nghim phc l

1z = b i

a v

2z = b i

a +

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V D MINH HOBi 1: Thc hin cc php tnh

a)(2 4 )(3 5 ) 7(4 3 )i i i+ + b) 2(3 4 )i c) 23 2

i

i

++

Bi giiCu a: 2(2 4 )(3 5 ) 7(4 3 ) 6 10 12 20 28 21i i i i i i i + + = + +

6 10 12 20 28 21 54 19i i i i = + + + = Cu b: 2 2(3 4 ) 9 24 16 9 24 16 7 24i i i i i = + = =

Cu c:2

2 2 2

(2 )(3 2 )2 6 4 3 2 6 2 8 13 2 (3 2 )(3 2 ) 13 133 4 3 4

i ii i i i i

i i i ii

+ + + ++ + +

= = = =

Bi 2: Tm mun ca s phc sau ya) 23 2 (1 )z i i= + + + b) 3

(1 )(2 )i

i iz +

+ =

Bi gii

Cu a: 23 2 (1 ) 3 2 3 221 2 1 2 1z i i i i i i i = + + + = + + + + = + + + 2 2 2 23 4 3 4 5z i z a b = + = + = + =

Cu b:2

3 3 3 3

(1 )(2 ) 2 2 1 32 2

1 1i i i i

i i i i i i i i

z z+ + + +

+ + + + +

= = = = = =

Bi 3: Tm s phc nghch o ca s phc: 2(1 ) (2 )z i i= + Bi gii

2 2 2(1 ) (2 ) (1 2 )(2 ) ( 2 )(2 ) 4 2 2 4z i i i i i i i i i i = + = + + = + = =

Suy ra2

2 4 2 4 2 41 1 1 12 4 (2 4 )(2 4 ) 20 10 54 16

i i i

z i i i ii+ + +

+ = = = = = +

Bi 4: Gii phng trnh sau trn tp s phc: 2 3 5 4iz z i + = + Bi gii

2 3 5 4 5 2 3 4 (5 2 ) 3 4iz z i z iz i i z i + = + = = 2

2 2

(3 4 )(5 2 ) 15 6 20 83 4 23 145 2 (5 2 )(5 2 ) 29 295 4

i i i i ii

i i i iz i

+ + +

= = = =

Bi 5: Gii cc phng trnh sau y trn tp s phc:a) 2 2 0z z + = b) 4 22 3 0z z+ = c) 3 1 0z + =

Bi giiCu a: 2 22 0 2 0z z z z + = + = (1)

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Ta c, 2 21 4.1.2 7 0 ( 7. )i = = < = Vy, phng trnh (1) c 2 nghim phc phn bit

1 7 711 2 2 2

iz i= = v 1 7 712 2 2 2iz i+= = +

Cu b: 4 22 3 0z z+ = (2)

t 2t z= , phng trnh (2) tr thnh:

2 321

03

tt t

t

=+ = = . T ,

2

2

11

3.3

zz

z iz

= = = =

Vy, phng trnh (2) c 4 nghim phc phn bit :

1 2 3

1, 1, 3.z z z i = = = v4

3.z i=

Cu c: 3 (3) 2 2 (*)1

1 0 ( 1)( 1) 01 0

zz z z z

z z

= + = + + = + =

Gii (*) : ta c 2 2( 1) 4.1.1 3 0 ( 3 )i = = < =

(*) c 2 nghim phc phn bit: 1 31 2iz += ; 1 32 2

iz =

11

Vy, phng trnh (3) c 3 nghim phc phn bit

z = , 312 2 2z i= + v31

3 2 2z i=

Bi 6: Tm mun ca s phc zbit:a) 3 (3 )(1 ) 2iz i i + + = b) 5 11 17iz z i + =

Bi gii

Cu a: 23 (3 )(1 ) 2 3 3 3 2iz i i iz i i i + + = + + =

3 3 3 1 2 3 2 2iz i i iz i + + + = = 2 2

3

i

i

z =

2 23 3

z i = + 2 2z a b = + = ( ) ( )2 2

2 22 23 3 3

+ =

Cu b: Vi ( , )z a bi a b= + ta c z a bi = , do 5 11 17 ( ) 5( ) 11 17iz z i i a bi a bi i + = + + =

2 5 5 11 17 (5 ) ( 5 ) 11 17ia bi a bi i a b a b i i + + = + =

2 25 11 3

3 4 3 4 55 17 4

a b a

z i za b b

= = = + = + = = =

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III. BI TP V S PHCBi 7: Thc hin cc php tnh

a) 2(1 )i+ b) 2(3 4 )i c) 2( 2 )i + d) 3(2 3 )i+

e) 3(1 3 )i f) 2012(1 )i g) 2012(1 )i+ h) 2012(1 3 )i

i) 2 33

i

i

++

j) 4 21

i

i

+

k) 2 4ii

+ l) 12 i

m)2

1

1

i

i

+n)

51

1

i

i

+ o)

2

2

(2 )ip)

(2 1)

1

i i

i

+

Bi 8: Xc nh phn thc, phn o v mun ca cc s phc sau y:a)(2 4 )(3 5 ) 7(4 3 )i i i+ + b)(1 4 )(2 3 ) 5( 1 3 )i i i +

c) 2(1 2 ) (2 3 )(3 2 )i i i + d) 2(2 3 ) (1 3 )(5 2 )i i i + e) 2 2(1 2 ) (1 2 )i i+ + f) 2 2(1 3 ) (1 3 )i i+

g)5

(4 5 ) (4 3 )i i + + h)3

(5 ) (2 7 )i i +

i)(2 ) (1 )(4 3 )

3 2

i i i

i

+ + + +

j)(2 ) (1 )(1 3 )

3 9

i i i

i

+

k)(3 4 )(1 2 )

4 31 2

i i

ii

++ l)

(2 3 )(1 2 )(2 4 )1

i i

ii

+ + +

Bi 9: Gii cc phng trnh sau trn tp s phc:a) 3 8 5 4z i i+ = + b) 22 (2 ) 2 3iz i i + = +

c) (3 ) (1 )(4 2 )i z i i = + d) 2(1 ) (1 ) 2 3i z i i + + =

e)2 1 3

1 2

i iz

i i

+ +=

+f)

2 1 3

1 2 2

i iz

i i

+ =

+ +

g) (2 ) 3 2i z i i + = + h)2 . 1 5. 2i z z i = i) 2 3 5 4iz z i + = + j) 3 . 5 3z i z i = k) 2 6 2z z i+ = + l) 3 7 5iz z i + = + m)3 2 5 2z z i+ = + n) . 2 2 5i z z i + =

Bi 10: Tnh z , bit rng a) 2(1 2. )z i= + b) 34

(1 )

(1 )

i

iz

+

=

Bi 11: Tm s phc nghch o ca cc s phc sau y:a) 3 4z i= b) (4 )(2 3 )z i i= + c) 2(2 )z i i=

Bi 12: Cho1 2

2 3 , 1z i z i = + = + . Tnh 21 2.z z ;

1 2z z v

1 23z z

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Bi 13: Cho 2 3z i= + . Tm phn thc, phn o v mun ca 75

z i

iz

++

Bi 14: Cho ( )3311 2 2z i= + v ( )3

311 2 2

z i= + . Tnh1 2.z z

Bi 15: Gii cc phng trnh sau trn tp s phc:a) 2 2 0z + = b) 24 9 0z + = c) 2 4 8 0z z + =

d) 22 2 5 0z z+ + = e) 2 2 17 0z z+ + = f) 2 3 3 0z z + =

g) 3 4 0z z+ = h) 3 27 4z z z+ = i) 3 8 0z + =

j) 4 22 3 0z z+ = k) 4 22 3 5 0z z+ = l) 49 16 0z =

m) 22 4 9 0z z+ + = n) 2 1 0z z + = o) 2 4 11 0z z+ =

Bi 16: Tm s phc zc phn thc v phn o i nhau v 2 2z = Bi 17: Cho

1 2,z z l hai nghim phc ca phng trnh 25 2 1 0z z + =

Chng minh rng tng nghch o ca1

z v2

z bng 2.

Bi 18: Cho1 2,z z l hai nghim phc ca phng trnh 23 2 4 0z z + =

Chng minh rng1 2 1 2

. 2z z z z + + =


Chng minh rng

2 2

1 2z 6z+ = Bi 20: Cho1 2,z z l hai nghim phc ca phng trnh 25 2 2 0z z + =

Chng minh rng1 2 1 2

.z z z z + =


v2

z c phn o l mt s m. Tnh1 2

2z z+

Bi 22: Tm s phc z c phn thc v phn o bng nhau v 2 2z = Bi 23: Cho hai s phc ( 1)z m m i = + v 2 (2 3 )z n n i = + , vi

,m n . Tm z v z bit rng 1 7z z i+ = + .

Bi 24: Cho s phc ( 1) ,z m m i m = + + . Tm zbit rng 5z = .Bi 25: Cho s phc ( 1) ( 1) ,z m m i m = + + . Tm zbit . 10z z = .Bi 26: Cho s phc 2 ( 2) ,z m m i m = + + . Tm z bit rng 2z l

mt s phc c phn thc bng 5 .Bi 27: Gii cc phng trnh sau y trn tp cc s phc

a)5( 1)( 1) 2(4 5) 0z z z + + + = b) 22(2 1) (17 6) 0z z z + + =

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PPPhhhnnn VVV... PPPHHHNNNGGG PPPHHHPPP TTTOOO+ ++ TTTRRROOONNNGGG KKKHHHNNNGGG GGGIIIAAANNN

1. H to OxyzGm 3 trc Ox,Oy,Ozi mt vung gc nhau

c vct n v ln lt l: , ,i j k 2. To ca ima) nh ngha

( ; ; ) . . .M M M M M M

M x y z OM x i y j z k = + +

b) To ca cc im c bit Trung im Ica on AB

Trng tm Gca tam gic ABC

2

2

2

A BI

A BI

A BI

x x

x

y yy

z zz

+= +

= + =

3

3

3

A B CG

A B CG

A B CG

x x x

x

y y yy

z z zz

+ += + +

= + + =

Hnh chiu vung gc ca im ( ; ; )M M M

M x y z ln:

Trc Ox:1

( ;0;0)M

M x mp( )Oxy :12

( ; ; 0)M M

M x y

Trc Oy:2(0; ;0)

MM y mp( )Oxz :

13( ; 0; )

M MM x z

Trc Oz:3(0;0; )

MM z mp( )Oyz :

23(0; ; )

M MM y z

3. To ca vcta) nh ngha:

1 2 3 1 2 3( ; ; ) . . .a a a a a a i a j a k = = + +

b) Cng thc to ca vct

Nu ( ; ; ), ( ; ; )A A A B B B

A x y z B x y z th ( ; ; )B A B A B A

AB x x y y z z =

Nu1 2 3

( ; ; )a a a a = ,1 2 3

( ; ; )b b b b= th

1 1 2 2 3 3

( ; ; )a b a b a b a b+ = + + +

1 1 2 2 3 3

( ; ; )a b a b a b a b =

1 2 3. ( ; ; )k a ka ka ka =

, k c) iu kin cng phng ca hai vct

Cho 1 2 3( ; ; )a a a a =

, 1 2 3( ; ; )b b b b=

v 0b

. Khi ,a

cng phng vi b

tn ti s thc t sao cho .a t b=

1 1

2 2

3 3

a b

a b a b

a b

== = =

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4. Tch v hng ca hai vct

a) Cng thc: Nu 1 2 3

1 2 3

( ; ; )

( ; ; )

a a a a

b b b b

= =

th1 1 2 2 3 3

. . . .a b a b a b a b= +

+

b) ng dng: 2 2 21 2 3

a a a a = + +

AB AB =

.

cos( , ).

a ba b

a b=

. 0a b a b =

, vi

0

0

a

b

5. Tch c hng ca hai vcta) nh ngha

Cho1 2 3

1 2 3

( ; ; )

( ; ; )

a a a a

b b b b

= =

. Khi , vct [ ]2 3 1 3 1 2

2 3 1 3 1 2, ; ;

a a a a a a

a b b b b b b b

=

c gi l tch c hng ca hai vct a

v b

.

b) Lu :Nu [ , ]n a b=

th n a

v n b

(gi s 0, 0, 0a b n

)

c) ng dng 1: Cho ba vct khc 0

ln lt l , ,a b c

. Khi ,

a

v b

cng phng vi nhau [ , ] 0a b =

,a b

v cng phng vi nhau [ , ]. 0a b c =

A,B,C thng hng [ , ] 0AB BC =

A,B,C,D ng phng [ , ]. 0AB AC AD =

d) ng dng 2: (tnh din tch)

Din tch hnh bnh hnh ABCD

[ , ]ABCD

S AB AD =

Din tch tam gic ABC:

ABCS

12

[ , ]AB AC=

e) ng dng 3: (tnh th tch) Th tch khi hnh hp .ABCD A B C D

[ , ].hh

V AB AD AA=

Th tch khi t din ABCD:

ABCDV = 16

[ , ].AB AC AD

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V D MINH HO

Bi 1: Trong h to ( , , , )O i j k cho 2 3OA i j k = + ,4 3 2 , (2; 7;1)OB i j k BC = + =

v (4;1; 7)A

a) Chng minh rng A,B,Cl 3 nh ca mt tam gic vung.b) Chng minh rng ( )AA ABC

c) Tnh th tch khi t din A ABC .d) Xc nh to cc nh cn li ca hnh hp .ABCD A B C D

Bi giiT gi thit ta c (2;1; 3), (4; 3; 2), (6; 4; 1), (4;1; 7)A B C A

Cu a:(2;2;1)

. 8 10 2 0(4; 5;2)

AB

AB AC AB AC AC

= = + = =

Vy, ABCl tam gic vung ti A

Cu b: Ta c, (2; 0; 4)AA =

v (2;2;1), (4; 5;2)AB AC = =

Do ,. 2.2 0.2 4.1 0

. 2.4 0.( 5) 4.2 0

AA AB

AA AC

= + = = + =

( )AA AB

AA ABC AA AC

Cu c:2 2 2

2 2 2

2 2 1 3

4 ( 5) 2 3 5

AB

AC

= + + = = + + =

. 9 5

2 2ABCAB AC

S = =

2 2 22 0 ( 4) 2 5h AA= = + + =

Vy, 9 5.2 51 13 3 3.2

. 15A ABC ABC

V h S AA = = = =B.

Cu d:ABCDl hnh bnh hnh AD BC =

2 2 4

1 7 6. (4; 6; 2)

3 1 2

D D

D D

D D

x x

y y D

z z

= = = = + = =

Tng t, (6;3; 6)B , (6; 6; 6)D , (8; 4; 5)C

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BI TP V TO CA IM, TO CA VCTBi 2: Trong h to Oxyz, cho cc im (2; 0; 1), (3;2; 3), ( 1;1;1)A B C

a) Chng minh rng A,B,Cl ba nh ca mt tam gic.b) Xc nh to nh Dv tm Ica hnh bnh hnh ABCD.

c) Tm to im Msao cho 2AM OB AC

=

Bi 3: Trong h to Oxyz, cho cc im (2;2; 1), (2;1; 0), (1;1; 1)A B C

a) Chng minh rng ABCl tam gic u.b) Cho im (4;0; 3)A . Xc nh to cc im B v C

.ABC A B C l mt hnh lng tr.c) Chng minh rng .ABC A B C l mt lng tr u.

Bi 4: Trong h to ( , , , )O i j k cho 3 2 3OM i j k = + v A,B,Clnlt l hnh chiu vung gc ca Mln cc trc to Ox,Oy,Oz.a) Chng minh rng ABCl tam gic cn.b) Tnh th tch t din OABC, t tnh khong cch t gc to n mt phng ( )ABC

Bi 5: Trong h to ( , , , )O i j k cho 3 2 3ON i j k = + v A,B,Clnlt l hnh chiu vung gc ca im Nln cc mt phng to Oxy, Oyz, Oxz.

a) Tnh din tch tam gic ABCv th tch ca t din NABC.b) Tnh khong cch t im Nn mt phng ( )ABC Bi 6: Trong khng gian vi h to Oxyz, chng minh rng (0;0;0)O ,

A(0;1;2),B(2;3;1),C(2;2;1) l bn nh ca mt hnh ch nht.

Bi 7: Trong h to ( , , , )O i j k cho t din ABCD sao cho(2; 4; 1), 4 , (2; 4; 3), (0; 2; 0)A OB i j k C AD = + =

a) Chng minh rng AB, ACv ADi mt vung gc vi nhau.

b) Tnh din tch tam gic ABCv th tch t din ABCD.Bi 8: Trong h to Oxyzcho (2;1; 3), (4; 3; 2), (6; 4; 1)A B C

a) Chng minh rng A,B,Cl ba nh ca mt tam gic vung.b) Tm to im D A,B,C,Dl 4 nh ca mt hnh ch nht

Bi 9: Tm to cc nh cn li ca hnh hp .ABCD A B C D bitrng (2;4; 1), (1;4; 1), (2;4;3), (2;2; 1)A B C OA

=

Bi 10:Tm im Ntrn Oycch u hai im (3;1;0)A v ( 2; 4;1)B Bi 11:Tm im M trn mt phng ( )Oxz cch u ba im (1;1;1)A ,

( 1;1; 0)B v (3;1; 1)C

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6. Phng trnh mt cua) Dng 1: mt cu ( )S tm I(a;b;c), bn knh R c phng trnh:

( ) ( )2 2 2 2( ) x a y b z c R+ + =

b) Dng 2: vi iu kin2 2 2

0a b c d + + > th2 2 2 2 2 2 0x y z ax by cz d + + + = l phng trnh mt cu Tm I(a;b;c)

Bn knh 2 2 2R a b c d = + + c) Lu : mt cu ( , )S I R tip xc vi mt phng ( ) ( , )d I R = 7. Phng trnh tng qut ca mt phnga) Cng thc: Nu mt phng ( )P i qua im

0 0 0 0

( ; ; )M x y z v c vct

php tuyn ( ; ; ) 0n = A B C th ( )P c phng trnh tng qut l:

0 0 0( ) ( ) ( ) 0A x x B y y C z z + + =

b) Lu v cch xc nh vct php tuyn (vtpt) cho mt phng:

Nu ( )P AB th ( )P nhn n AB=

lm vct php tuyn.

Nu a

v b

l hai vct khng cng phng, c gi song song hoc

cha trong ( )P th ( )P nhn [ , ]n a b

=

lm vct php tuyn.

Cho trc ( ) : 0Q Ax By Cz D + + + = . Nu ( )( )P Q th ( )P c

phng trnh dng 0Ax By Cz D + + + = (vi D D )

Mt phng ( ) : 0P Ax By Cz D + + + = c vtpt ( ; ; )n A B C = c) Phng trnh mt phng theo on chn

Mt phng ( )P i qua ba im phn bit

( ;0;0)A a , (0; ;0), (0; 0; )B b C c c phng trnh

1x y z

a b c+ + =

d) Khong cch t im Mo n mt phng (P)

0 0 0

2 2 20( ,( )) Ax By Cz D

A B C

d M P + + ++ +

=

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8. Phng trnh ca ng thngCho ng thng di qua im

0 0 0 0( ; ; )M x y z v c vtcp ( ; ; )u a b c =

a) Phng trnh tham s ca d:

0

0

0

( )

x x at

y y bt t z z ct

= += + = +

b) Phng trnh chnh tc ca d: 0 0 0x x y y z z

a b c

= =

(gi s a,b,cu khc 0)c) Cch xc nh vct ch phng (vtcp) cho ng thng d

di qua 2 im A v Bphn bit th dc vtcp u AB=

Cho ng thng c vtcp u . Nu d th dc vtcp u u=

Cho mt phng ( )P c vtptP

n

. Nu d(P) th dc vtcpP

u n=

Cho hai vct khng cng phng a

v b

. Nu dvung gc vi gi

ca 2 vct a

v b

th dc vtcp [ , ]u a b

=

Cho ng thng c vtcp u

v mt phng ( )P c vtptP

n

. Nu

dsong song vi ( )P v vung gc vi th dc vtcp [ ],P

u n u=

Cho hai mt phng ( )P v ( )Q ln lt c vtptP

n

vQ

n

.

Nu dl giao tuyn ca ( )P v ( )Q th dc vtcp [ ],P Q

u n n=

Cho hai ng thng1

d v2

d ln lt c vtcp1

u

v2

u

khng

cng phng. Nu dvung gc vi c hai ng thng1

d v2

d th

dc vtcp1 2

[ , ]u u u=

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V D MINH HOBi 12: Cho A(1;3;1), B(2;1;2), C(0;2; 6) v ( ) : 2 2 1 0P x y z + + =

a) Vit phng trnh mt cu tm B, i qua Ab) Vit phng trnh mt cu ng knh BC.

c) Vit phng trnh mt cu tm C, tip xc vi mt phng ( )P d) Vit phng trnh mt cu ngoi tip t din OABC.

Bi giiCu a:Gi

1( )S l mt cu tm B(2;1;2)v i qua im A. Khi

1( )S

c bn knh1

R AB=

Ta c 2 2 2(1; 2;1) 1 ( 2) 1 6AB AB = = + + =

1

( )S c phng trnh 2 2 2( 2) ( 1) ( 2) 6x y z + + =

Cu b: Gi2

( )S l mt cu ng knh BCth2

( )S c tm (1 2)32

; ;I l

trung im ca on thng BCv bn knh2

BCR =

v 2 2 2( 2;1; 8) ( 2) 1 ( 8) 69BC BC = = + +

=

nn 692 2

BCR = =

Phng trnh mt cu 2( )S l2 2 23 69

2 4( 1) ( ) ( 2)x y z + + + = Cu c: Gi

3( )S l mt cu tm C(0;2;6), tip xc vi ( )P . Khi

3( )S

c bn knh3

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