Tai Lieu on Tap Thi Tot Nghiep 2012 Mon Toan Thpt

7/31/2019 Tai Lieu on Tap Thi Tot Nghiep 2012 Mon Toan Thpt

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TR NG THPT CHU V N ANTR NG THPT CHU V N ANTR NG THPT CHU V N ANTR NG THPT CHU V N ANT TONT TONT TONT TON

GV: Dng Phc GV: Dng Phc GV: Dng Phc GV: Dng Phc

On tap Tot nghiep

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Dng Phc Sang - 1 - THPT Chu Vn An

Ph nPh nPh nPh n IIII. KH O ST. KH O ST. KH O ST. KH O ST HM S HM S HM S HM S V BI TON LIN QUAN V BI TON LIN QUAN V BI TON LIN QUAN V BI TON LIN QUAN1. 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 nghim 0x (nu c).4 Tnh hai gii hn: lim ; lim

x 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 phng

trnh 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 duynht

th hm s trng phng lun i xng qua trc tung b) Vit phng trnh tip tuyn (dng 1 bit to tip imM 0 )

1 Ch r 0x v 0y (honh & tung ca imM 0)2 Tnh 0( ) 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 gck ) 1 Lp lun c c 0( ) f x k = (*)2 Thay 0( )y x vo (*) tm 0x 3 C 0x , tm 0y v dng cng thc

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

Lu : Tip tuyn song song viy ax b= + c h s gck = a Tip tuyn vung gc vi ( 0)y ax b a = + c h s gc 1

a 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 cam 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 thngd : 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 d bng vi s nghim ca (*)3 m s nghim ca (*) suy ra s giao im ca( )C v d

V D MINH HO Bi 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 s m phng trnh sau y c

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

Cu 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 ; lim

x 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 : doa > 0) x 1 3 + y + 0 0 +

y 5 +

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) 9 f = Phng trnh tip tuyn ca( )C tiA 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 41 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 theoa s 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 = Gii hn: lim ; lim

x x y y

+= + =

Hm s ng bin trn khong (0;1)

Bng bin thin:(ch : doa < 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 1

2 26 12 . 0y x y x y = = = = . im un 1 1

2 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 = = 32

0x x

= =

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

( ;0)B

Ti (0;0)O : (0) 0 f = , phng trnh tip tuyn l: 0y = Ti 3

2( ;0)B : 3 9

2 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 = = 32

a = (*) S nghim phng trnh (*) bng vi s giao im ca th( )C v ng thng 3

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

a 32 a S giao imca ( )C v d

S nghim caphng trnh (*)

23a <

32 1a > 1 1

23

a = 32

1a = 2 223

0a < < 32

0 1a < < 3 3

0a = 32 0a = 2 2

0a > 32 0a < 1 1

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8/74



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

2x x x

y + +=

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

song vi ng thng32:

y x =

c) Tm to cc giao im ca( )C vi ng thng 32 2y x = + Bi gii

Cu a:3 23 3

2x x x

y + += Tp xc nh:D =

o hm23 6 3 0,

2x x

y x + + = do hm s lun ng

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

x x y y

+= = +

Bng bin thin:

12

3 3 0 1y x x y = + = = = im un 1

2( 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 hs gc 30 2( )k f x = =

20 03 6 3

2x x + +

= 32

2 00 0

0

03 6 0

2x

x 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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Vi 0 2x = th 0 ( 2) 1y y = = , tip tuyn tng ng l3 32 2

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

Vy, tip tuyn tho l 32

2y x = +

Cu c: Honh giao im (nu c) ca( )C v 32 2y x = + l nghim

phng trnh3 23 3

2x x x + + = 3 23 2 3 3 3 4

2x x x x x + + + = +

3 2 2 13 4 0 ( 1)( 4 4) 02

x x x x x x

x

= + = + + = =

721x y = = v 2 1x y = = Vy, ( )C v 3

2: 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 x l nghim ca phng trnh ( ) 20 f x = c) Tm cc gi tr ca tham s m 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 = = = = Gii hn: lim ; lim

x x y y

+= + = +

Bng bin thin:x 1 0 1 + y 0 + 0 0 +y + 3 +

4 4Hm s ng bin trn cc khong trn (1;0), (1;+ ) v nghchbin 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 0

0 13 4 1

m m m

m m

< > <

Bi 5 :a) Kho st v v th( )C ca hm s: 4 24 3y x x = + b) Dng th( )C bin lun s nghim pt sau:4 24 0x x m + =

Hng dn gii v p s Cu a:HS t gii c c th:Cu b:Bin i phng trnh ta c: 4 2 4 24 0 4 3 3x x m x x m + = + = Bng kt qu s nghim ca phng trnh cho

m m 3

S giaoim

ca ( )C v d

S nghimca

phngtrnh (*)m > 4 m 3 > 1 0 0m = 4 m 3 = 1 2 2

0 < m < 4 3 < m 3 < 1 4 4m = 0 m 3 = 3 3 3m < 0 m 3 < 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 cam phng trnh sau c 3 nghim phn bit:

3 3 1 2 0x x m + + = .Bi 7 : Cho hm s 3 21 3

2 22y x x = +

a) Kho st s bin thin v v th( )C ca hm s.b) Vit pttt vi( )C song song vi ng thngd : 9

22y x = +

c) Tm cc gi tr cak phng trnh sau y c nghim duynht: 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 theom s 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 a) Kho st s bin thin v v th( )C ca hm s.b) Vit pttt vi( )C ti im trn( )C c honh x tho 1y = c) Tnh din tch hnh phng gii hn bi( )C v : 2 0d y = .d) Tm cc gi tr cam phng trnh sau c nghim duy nht

3 22 9 6 0x x e e m + = Bi 10 : Cho hm s 3 21

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 0.c) Vit pttt ca ( )C song song vi ng thng 8 3y x = d) Tm cc gi tr caa 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 thngd : 1y x =

c) Bin lun theom s 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 thngd : 1 1

3 3y x =

c) Tm cc gi tr caa 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 imA(0; 2)c) Vit pttt ca ( )C bit tip tuyn song song vi9 4 4 0x y = d) Bin lun theom s 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 1

72: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 s m 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 cam phng trnh sau y c ng 2 nghim:4 22 3 2 0x x m + + + = Bi 18 : Cho hm s 1

2y = 4 23x x + 3

2c 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 theom s 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 + a) Kho st s bin thin v v th( )C ca hm s.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 1

4y = 4 22x x

a) Kho st s bin thin v v th( )C ca hm s.b) Vit pttt ca ( )C song song vi 1 : 15 2012d y x = + .c) Vit pttt ca ( )C vung gc vi 2 :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: { }\ d c D = 2 Tnh

2( )ad cb

y cx d

=+

v khng nhy dng hay m, d c

x

3 Suy ra hm s ng bin hay nghch bin trn mi khong xcnh ( ; ),( ; )d d

c c + v khng t cc tr.

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

x

a y

c = v lim

x

a y

c += , suy ra a y

c = l TCN

Tnh( )

limd c

x y

v

( )lim

d c

x y

+ , suy ra d x

c = l TC

5 V bng bin thin ca hm s.6 Lp bng gi tr. 7 V 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 bit lun 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 imM 0 )1 Ch r 0x v 0y (honh & tung ca imM 0)2 Tnh 0( ) 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 gck )

1 Lp lun c c 0( ) f x k = (*)2 Thay 0( )y x vo (*) tm 0x 3 C 0x , tm 0y v dng cng thc 0 0 0( )( )y y f x x x =

Lu : Tip tuyn song song viy ax b= + c h s gck = a Tip tuyn vung gc vi ( 0)y ax b a = + c h s gc 1

a k =

d) S tng giao gia th (C ):y = f (x ) v ng thngd : 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 d bng vi s nghim ca (*)3 m s nghim ca (*) suy ra s giao im ca( )C v d

V D MINH HO Bi 23 : Cho hm s 2 1

1x

y 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

bng52

c) Chng minh rng ng thng: 2d y x m = + lun ct th

( )C ti 2 im phn bit. Bi gii

Cu a:Hm s 2 11

x y

x +=+

Tp xc nh: \{ 1}D =

o hm:2

1 0, 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 2

x x y y

+= = y = 2 l tim cn ngang.

( 1) ( 1)lim ; lim

x 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 = th 2 1 5 2(2 1) 5( 1) 31 2

x x 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 d l nghim phng trnh

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

x x 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 d

lun c 2 im chung phn bit.

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

y x

=

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

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

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17/74


Cu a:Hm s 3 32 2x x

y x x

= = +

Tp xc nh: \ {2}D =

o hm:2

1 0, 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 2lim ; lim

x x y y

+ = = + 2x = l tim cn ng.

Bng bin thin:x 2 + y y 1

+ 1

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

22 0 1

2

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

Cu b: V tip tuyn song song vi ng thngy x = nn c h s gc 0( ) 1k f x = =

20

1 1(2 )x

=

2

0(2 ) 1x =0 0

0 0

2 1 12 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

2x

x m x

= +

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

( )C v d ct nhau ti 2 im phn bit khi v ch khi phngtrnh (*) 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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18/74



BI TP V HM S NHT BIN Bi 25 : Cho hm s 2 1

1x

y x

+=

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 3.c) Vit pttt vi( )C ti im trn( )C c tung bng7

2

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

Bi 26 : Cho hm s 2 11

x y

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.

Bi 27 : Cho hm s 2 12

x y

x =

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 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 321

y x

= +

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 bitBi 29 : Cho hm s 2

3x

y 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 bng54

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

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19/74


Bi 30 : Cho hm s 21

y x

=+

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 3

2 2y x = +

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

Bi 31 : Cho hm s 1

1

x y

x

=+

c th( )C .

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

M song song vi ng thngd : y = 2x

Bi 32 : Cho hm s 21

x y

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 thng 1

22012y x = +

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

Bi 33 : Cho hm s 2 31x

y 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 thng

3y x = + ng thi tip xc vi th( )C

Bi 34 : Cho hm s 3 41

x y

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 sy = f (x ) trn on [a ;b ]1 Hm s ( )y f x = lin tc trn on [a ;b].2 Tnh ( )y f x = .3

Cho 0y = tm cc nghim [ ; ]i x a b (nu c) v cc s [ ; ] j x 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 tnh f ca cc x l b loi )5 Chn kt qu ln nht v kt qu nh nht t bc 4 kt lun

v 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 i ti 0x

Nu 00

( ) 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 0

0,0

y y x a

>

Hm s 3 2y ax bx cx d = + + + nghch bin trn 0

0,0

y y x a

> (khng c du =)Hm s ax by

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 HO Bi 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)x y 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: ( ) ; ;4 133 27 (1) 0 (3) 6 f f f = = = Do 13

276 0 < < nn

[1;3]min (3) 6x

y f

= = v[1;3]

maxx

y

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

24 2 2 42 1 1

x x y x x x

+ + = + =

Cho(nhan)(loai)

2 1 [ 3;0]0 2 2 4 02 [ 3;0]

x y x x

x

= = + + = =

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

161 4 ln 2 ln 0e = < v

29 8 ln 2 1 8 ln 0e = + > nn

[ 3;0]min ( 1) 1 4 ln 2

x y f

= = v

[ 3;0]max ( 3) 9 8 ln 2

x 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]

t g 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 s m 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

a m

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 phnbit 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 cam hm s 3 2 23 ( 1) 2y x mx m x = + +

t cc i ti 0 2x = 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 khi

2(2) 0 {1;11}12 11 011

(2) 0 212 6 0 f m m m

m f m m

= + = = < > <

Vy vi 11m = th hm s (*) t cc i ti0 2x =

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23/74


Bi 38 : Chng minh rng nu sinx

x y

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

Bi gii

Hm s sin

.sinx

x

x

y e x e

= = c tp xc nhD = ( ) . sin .(sin ) (cos sin )x x x y e x e x e x x = + = ( ) (cos sin ) (cos sin ) 2 cosx x x y e x x e x x e x = + = 2 2 2 cos 2 (cos sin ) 2 sin 0x x x y y y e x e x x e x + + = + + = Vy, vi .sinx y e x = th 2 2 0y y y + + =

BI TP V CC VN KHC LIN QUAN HM S Bi 39 : Tm gi tr ln nht, gi tr nh nht ca cc hm s sau ya) 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[ ]3

20;

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 2x

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 x x

= trn on[ ]31;e

l) ln( ) x f x x

= trn on 12

[ ;e 2 ]e

Bi 41 : Tm cc gi tr ca tham s m hm s sau y lun ng bin a) 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 s a hm s sau y lun nghch bina) 3 2( 1) (2 1) 3y x a x a x = + + + b) 7

5 3ax a

y x a

+ = +

Bi 43 : Tm cc gi tr cam hm s sau y c cc i v cc tiu

a)3 2 2

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

2 2 42

x mx m y

x + =

+

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

a) 3 2 22 ( 1) ( 4) 1y x m x m x m = + + + + t cc i ti 0 0x = b) 2 3 2(2 1) (2 3) 2y m x mx m x = + + t cc tiu ti

01x =

c) 2 63

m y = 3 1x mx + + t cc tiu ti 0 2x =

d) 12

y = 4 2x mx n + t cc tiu bng 2 ti 0 1x = Bi 45 : Chng minh rng

a) Nu (cos2 sin2 )x y e x x = + th 2 5 0y y y + = b) Nu 4 2x x y e e = + th 13 12y y y =

c) Nu ln x y x

= th 23 0y xy x y + + =

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25/74


Ph nPh nPh nPh n II.II.II.II. PH NG TRNHPH NG TRNHPH NG TRNHPH NG TRNH ---- B T PH NG TRNH M!B T PH NG TRNH M!B T PH NG TRNH M!B T PH NG TRNH M! &&&& LGARITLGARITLGARITLGARIT

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

n m n m n m mn

m m n m n m n

n

n n n n

a a a a a

a a a a

a

a a a a

+

= =

= =

= =

i i

i i

i i

( )( ) ( )

( ) .n

n

n n n

n a a b b

n n a bb 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 a a 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:

0Bin i phng trnh theo

( ) f x

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

( )( ) 1. . 0

f x f x

a m a n p+ + =

1 t ( ) f x t a = (km iu kin chot ) v thay vo phng trnh2 Gii phng trnh mi theot tm nghim0t (nu c)3 i chiu nghim0t tm c vi iu kin bc 1 ri tmx .

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

( ) 1 f x

f x

a a =

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 x b

d) Phng php lgarit ho: vi0 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 tmx (nu c)i chiux tm c vi iu kin kt lun

Cc cng thc v quy tc tnh lgarit: vi0 1a < v b > 0, 0 : log 1 0a = log ( ) logn m m

a a n b b= ( 0n )

log ( )a a = .log ( ) log loga a a m n m n = + ( , 0m n > )

loga ba b= ( )log log logm a a a n m n = ( , 0m n > ) log ( ) . loga a b b

= loglog

log c c

ba a

b = ( 0 1c < )

1log loga a b b = 1loglog ba a b = ( 1b )

a) Phng trnh lgarit c bn: vi 0a > v 1a , ta c log ba x 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 ( ) 0 f x > ) log ( ) ( ) b

a f x b f x a = =

Lu : Nu c ( ) 0 f x > th 2log ( ) 2 log ( )n a a f x n f x =

Nu ch c ( ) 0 f x th 2log ( ) 2 log ( )n a a f x n f x = Bin i sau yrt d sai 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 ( )( )

log f 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( )( )

log f x a g x

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27/74


c) Phng php t n s ph:0 Bin i phng trnh theolog ( )a f x , chng hn:

2. log ( ) . log ( ) 0a a m f x n f x p+ + =

1 t log ( )a t f x = v thay vo phng trnh.2 Gii phng trnh mi theot tm nghim0t (nu c)3 T 0t t = ta gii phng trnh lgarit c bn tmx .

d) Phng php m ho: vi0 1a < v 0 1b< , ta clog ( ) 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 lgarit

cn ch so snh c s a vi 1 s dng tnh ng bin, nghch binca hm s m v hm s lgarit.

Hm s m x y a = ng bin khia > 1, nghch bin khi0 1a < < Hm s lgarit loga y x = cng ng bin khia > 1 v nghch bin

khi 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 giiCu 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 1

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

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

t t t t t t = > + = = >

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

Vy, phng trnh cho c 2 nghim:x = 1 v 3log 2x =

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

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

5x x x

x + = + =

Hng dn: t 5 ( 0)x t t = > . p s: 1x = Cu d: 6.9 13.6 6.4 0x x x + = . Chia 2 v ca phng trnh cho4x ta

c: ( ) ( ) ( ) ( )29 6 3 34 4 2 26 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 82log 2 log log 13x x x + + = d)

233

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

Hng dn gii v p s Cu a: 2 2log 4 log 1 1x x + = (1)

iu kin:4 0 4

41 0 1

x x x

x x

> > > > >

. Khi ,

(1) 2log ( 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 kinx > 4 ta ch nhn nghimx = 5Vy, phng trnh cho c nghim duy nht lx = 5

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29/74


Cu b: 5 25 0,2log log logx x + = 13 (2).

Vi iu kinx > 0, ( )2 11

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

+ =

p s: 3 3x = Cu c: 24 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: 233log ( 2) log ( 4) 0x x + = (4).

iu kin: 22 0 2

4( 4) 0x x

x x

> > (I). Khi ,

23 3(4) 2 log ( 2) log ( 4) 0x x + =

22 23 3 3log ( 2) log ( 4) 0 log ( 2)( 4) 0x x x x + = =

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

( 2)( 4) 1x x

x x x x

= = =

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

a) 22 2log 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:22 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 = = (thox > 0)Vi 2t = th 22log 2 2x x

= = (thox > 0)Vy, tp nghim ca phng trnh (5) l: 1

4{ ;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: t 2logt x = . p s: 12x = 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 2

5 11 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 giiCu 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 + 32[ ;1]x (gii bng bng xt du )

Vy, tp nghim ca bt phng trnh(8) l S = 32

[ ;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 0x x

+ < (10)

t 2x t = (t > 0), (10) tr thnh: 2 3 2 0t t + < vit > 0Bng xt du: cho2 3 2 0 1; 2t t t t + = = =

t 0 1 2 +2 3 2t t + + 0 0 +

Nh vy,12

t

t

>< hay

2 1 00 1

12 2

x

x

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 02 0 :

x x

x

+ >+ >

hoac12

2x x < > (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: 1

2[0; ) (2;5]S =

Cu c: 1 13 3

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

iu kin:hoac2 2 36 0

322 4 0x x x x

x x x

< > > > > + >

Vi iu kinx > 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 kinx > 3 ta nhn cc gi tr3 5x < Vy, tp nghim ca bt phng trnh l:

(3;5]S

=

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BI TP V PHNG TRNH M Bi 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 x e 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 x e 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, 5 x x + = d) ( )

52 22 16

9(0, 75) 0x x 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 0x x x +

= v)1

4 2 .4 3 0x x x

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

a) 2log( 6 5) log(1 )x x x + = b) 4 22ln . log ( 2 ) 3 lnx x x x = c) 1

7

27log ( 2) log (8 ) 0x x + + = d) 1

3

23log ( 10) log (3 ) 0x x + =

e) ln(4 4) ln( 1) lnx x x = f) 22log ( 1) log (7 )x x =

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

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

i) 2 2log ( 1) log (2 11) 1x x = j) 2 4 0,5log (2 ) log logx x x + = k) 2 0,5log ( 3) log ( 1) 3x x + = l) 5 0,25log log log 2x x x + =

m) 3 9 27log 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 0,2log log 12 0x x + = d)2ln ln( ) 1 0x ex =

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

3 log log log (2 )x x x =

g) ( )22 4 8log 6 log 7x x = h) 20,2 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 3x x + = l) 3log 27 3 log 8x x = m)

22 log 2 log 5x x + = n) ( )6 62 log 5 log 6x x x =

Bi 12 : Gii cc phng trnh sau ya) 23 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)23 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 21log log ( 1)( 4) 24

x x x

x + + = +

BI TP V BT PHNG TRNH M - LGARIT Bi 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 ya) 2 2log ( 5) log (3 2 ) 4x x + b) 1

3

52

log log 3 x x >

c) 28 8 32 log ( 2) log ( 3)x x > d) 13

3 1log 12

x x

>+

e) 4 4log ( 7) log (1 )x x + > f)22 2log log 0x +

Bi 16 : Gii cc bt phng trnh sau ya) 1 1

2 2

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

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

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

d) 0,2 0,2log (3 5) log ( 1)x x > + e) 3 3log ( 3) log ( 5) 1x x + <

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Ph nPh nPh nPh n III. NGUYN HMIII. NGUYN HMIII. NGUYN HMIII. NGUYN HM ---- TCH PHN TCH PHN TCH PHN TCH PHN V (NG D*NG V (NG D*NG V (NG D*NG V (NG D*NGI. TM TT CNG THC V PHNG PHP GII1. Bng cng thc nguyn hm v nguyn hm m rng

11

2 2

1. .( )1. ( ) .

1 11 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 bx

x dx C ax b dx C a ax b

dx x C dx C x ax b a

ax bdx x C dx C

a x ax b

dx C dx C x a ax bx ax b

e e 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

a ax b

x dx x C ax b dx C a

ax bdx x C dx C

a x ax bax b

dx x C dx C a x ax b

++= + + = +

+= + + = ++= + = +

++= + = +

+

i i

i ii 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

a a f x dx F x F b F a = =

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

I 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 bt a

I f t dt = v tnh tch phn mi ny (bint )

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

( )( )

t x dx

t x

( )t t x = mu

( ) . ( )t x e 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 dx f x

x 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 cosx

2(tan )

cos

dx f x

x tant x = 2cos

dx

x i km biu thc theo tan x

2(cot )

sin

dx f x

x

cott x = 2sin

dx

x

i km biu thc theo cot x

( ).ax ax f e e dx ax t e = ax e dx i km biu thc theo ax e 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

a a a u 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

==

( ). .ax P x e dx , ta t ( ).ax u P x dv e dx ==

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. sin .ax e bx dx , ta t sin .ax u e

dv bx dx

==

(khong co )

( ). ln . ,n f x x dx dx x

ta t ln( ).n

u x dv 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 , H l 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 phngH l: ( ) ( )

b

a S f x g x dx =

Lu 1: nu 2( )C l trc honh th ( ) 0g x = v ( )b

a S f x dx =

Lu 2: Khi tnh tch phn ( )b

a s x dx ta cn lu nh sau:

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

b b

a a s x dx s x dx =

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

b b

a a s 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 n c c c < <


38/74



V D MINH HO

Bi 1 : Tnh3

0 2

3

1

x A dx

x =

+ 2

3

1 cossin (1 cos )

x C dx

x x

=+

2213 . .x B x e dx

= 42 ln 1. lnx D dx x x += Bi gii

Cu a:3

0 2

3

1

x A 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 13. 3. 3 6 3 3

tdt A dt t t

= = = = = Cu b: 2

2

13 . .x B 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

t t e dt B e e e = = =

Cu c: 2 23 3

21 cos sin

sin (1 cos ) (1 cos )x x

C dx dx x x x

= =+ +

t 1 cos sin .t x dt x dx = + = sin .x dx dt = i cn:

2x = 1t =

3x = 3

2t =

Vy, ( )33

2232

112 21 11 . t

dt C dt

t t = = = ( )2 1 13 1 3= = Cu d:

4

2

ln 1. lnx

D 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 lnt D 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 . x F x e dx

=

2 21

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

Cu e: 20

( 1) sinE x xdx =

t1

sin cosu 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: 213 . x F x e dx

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

Nh vy, ( ) ( )2 22 2 1

11 13 . 3 6 3 3x x x F 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 21

(3 1) ln .G x x dx = t 2 31ln

(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

1x H x e dx x

=

2 20 ( 1).I x x xdx = + + 3

21

2 1e t t J 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 x H x e dx xe dx xe dx dx x

= = =

Xt 21 1 :x H xe dx = t x x u x du dx dv e dx v e = = = =

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( ) ( )2 22 2 2

1 11 1. 2x x x H 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 22 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 2

2 01.I 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 11 1 2 12 ln

e e e t t t

t t t t J dt t dt t + = = + =

( ) ( )2 21 1 1 1 32 2 1 2 22 ln 2 ln 1e e e e e = = 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 0cosa a 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 1d) 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 2

3 3 2x x + + 0 +

Vy, ( )2 31 3 2S x x dx = + ( )4 2 23 21

4 2 412x 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 22

3 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 31

3 2S x x dx

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

x 1 23 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 l

1 3 22

2S x x x dx

= + HD: xt du 3 2 2x x x + v a n cng thc

0 13 2 3 22 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 Ox bit (H ) gii hn bi: siny x = ,Ox , 0x = v 3

2x =

Bi gii Ta c, ( ) sin f x x = . Xt on[ ]3

20;

Th tch cn tm l:32 2

0(sin )V x dx

= 3 3 32 2 22

0 0 01 cos 2 1 cos 2sin

2 2 2x x V 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)

1 2

0

.(2 1)x x dx

b)

ln2

0

(3. 5)x x e e dx

c)

1 3

1

(2 3 )x dx

d)2

1

1 t te t dt

t + e)

2

1

(1 ) x x

x e x dx

xe

+ f)23

1

3 2t t dt

t +

g) ( )22 1

1 t t dt h) ( )

21 22 x

x x dx

+ i)

1 30

(1 )x x dx j) 4

6

cos4 .cos3x xdx

k) 64

sin 3 . sin .t t dt

l) 4 20 tan xdx

m) 120

1 .cos

x x e e dx x

+ n)2 1ln2

01x

x e dx

e + + o) 20 1 x dx

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p)32

1

2 5t t dt

t q)

22

0

3 11

x x dx

x + r) 12

1 3 1( 1)x

dx x x

++

m) 36

2 2

2

tan cos

sin

x x dx

x

n) 3

20

2 cos 2 1

cos

x dx

x

o) 4 2

0sin .x dx

Bi 7 : Tnh cc tch phn sau ya) 2

0

sin1 3 cos

x dx

x

+ b)2

21

12 3

x dx

x x

c)

21 10

. x x e dx

d)1/2

21

x e dx

x e)2

62

cos(1 sin )

xdx

x

+ f)20

41 (1 )x

dx x

g) 20sin .

8 cos 1x dx

x

+ h)19

0 3 2

3

8

xdx

x + i)2

1

1 lne x dx x

+

j) 11

(1 ln )e

e dx

x 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)

1 20

1x x dx + o)7 3

0. 1x x dx +

p)2

2

3

sin .cos .x x dx

q) 40 sin2

.cos2x

e xdx r)0

5 4 .x x dx s)

22

sin21 cos

x dx

x

+ t)1

2 20

4(2 1)

x dx

x + u)ln 3

0 1 x dx

e + Bi 8 : Tnh cc tch phn sau y

a)1

0( 1) x x e dx + b)

1

0(2 1) x x e dx c)

1 2 10

. x x e dx

d)ln 5

ln2 2 ( 1)x

x e dx e)ln2

0 ( 1)x

x e dx

f)2

0 2 .cos .x x dx

g) 4

0(2 1)cosx xdx

h)0

(1 ) cosx xdx

i) 20 2 .sinx xdx

j) 4

0( 1) sin 2x xdx

+ k) 40 sin2x xdx

l) 1 ln .e

x dx m)

12 .(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). x x e dx + q) 40 sinx e xdx

r)4

1

x e dx

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Bi 9 : Tnh cc tch phn sau y

a)1

0(3. 5 )x x e e x dx b) ( )0 cosx x x dx

+ c)

2 20

( )x x x e dx +

d)2

1

lnx x dx x

+

e)4

1

x x e dx x

+

f) 211 lne x x

dx x

+

g) ( )1 ln 1

e x x dx + h) 40 ( cos ) sinx x xdx

+ i)2

1( 2 )x x xe dx +

j)1

0

11

x

x

xe x dx

e

+ ++ k)

2

01 sin1 cos

x dx

x

+ l)

2

21

( 1). lnx x dx

x

Bi 10 : Tnh cc tch phn sau y

1) ( )0 2 11 x x e e dx 2) 21 ( 1)dx x x + 3) 60 cos2 sin 1xdx x

+ 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 x dx

x + 8)

4

1

ln .e x dx x 9)

2 22

1

lnx x dx

x +

10)

1

0

2 11

x

dx x

+ 11)

4

1 ( 2)dx

x x + 12)32 2

0 2 1

x dx

x + 13) 4

tan

20 cos

x e dx

x

14) 20

cos sin1 cos

x x dx

x

+ 15)

2ln2

30 ( 4)

x

x

e dx

e +

16)0

ln 63.x x e e dx + 17) 0 ( cos )x x e x dx

+ 18) 0 2 sinx xdx

19) 3

4

30

cos sin

cos

x x dx

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) x x e dx +

25)2

1

ln 1e x x dx

x + 26) 20

sin 2 .1 cos

x dx x

+ 27) 20sin 2 .3 sin 1

x dx

x

+

28)0

(1 cos ) cos .x x dx

29) ( )20 4 1x x dx + 30)1

0( 3)x xe dx +

31) ( cos 2)x x dx

32) 1 ( ln 2)e x x x dx + 33) 21 30 x x e dx

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BI TP V NG DNG HNH HC CA TCH PHN Bi 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 ( )19 1y x = j) ( ) : 1 , 1C xy x x = + = v tip tuyn vi( )C ti im( )322; .k) 3 1 , , 0

1x

y Ox x x

+= =

l) 1ln , ,e

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

m) ln1 x y x x

= + , 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) ,x y e x = trc honh v 1x = ( l trc honh)

e) 2 ,2

y x

=

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)x F x e x = + l mt nguyn hm

ca 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

x G x

= l

nguyn hm ca cng mt hm s vi mix thuc Bi 16 : Tm gi tr ca tham s m 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( ) ( ) x F x ax bx c e = + + l mt nguyn hm ca

hm 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 = bit

rng ( ) 1F =

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

x += tha mn iu

kin ( 1) 3F = . Bi 20 : Tm nguyn hm ( )F x ca hm s

21 ln( ) x f 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 mn

iu kin ( 1) 3F = .

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

= tha mn

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

iu kin (1)F e = .

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

x f x

+= tha mniu kin (1)F e = .

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Ph nPh nPh nPh n IV. S PH(CIV. S PH(CIV. S PH(CIV. S PH(C 1. Cc khi nim v php ton lin quan n s phc

n v oi : 2 1i = i 3i i = i 4 1i =i S phcz a bi = + l s c phn thc la v phn ob Mun ca s phcz a bi = + l: 2 2z a b= + S phc lin hp ca s phcz 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 22 2 2

.

.z z z

z z z = (nhn c t ln mu cho2z )

S phc nghch o caz l: 1.z

z z z =

Mi s thca m c 2 cn bc hai phc l: .a i Ch : s phc ch c phn o (phn thc bng 0) gi l s thun o

2. Gii phng trnh bc hai h s thc ( < 0) trn tp s phcCho 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

v 2z =2

b i

a

+

Lu :Ch c dng cng thc nghim nu trn khi < 0Trng hp 0 ta gii pt bc hai trn tp s thc (nh trc).Khi gii phng trnh trng phng trnC , ta t 2t z = (khng cniu kin chot )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 1

3 2 (3 2 )(3 2 ) 13 133 4 3 4i i i i i i i

i i i i i

+ + + ++ + +

= = = =

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

(1 )(2 )i

i i z ++

=

Bi giiCu 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 21 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 ra

22 4 2 4 2 41 1 1 1

2 4 (2 4 )(2 4 ) 20 10 54 16i i i

z i i i i i + + +

+ = = = = = +

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 14

5 2 (5 2 )(5 2 ) 29 295 4i i i i i i

i i i i z 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

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

Cu b: 4 22 3 0z z + = (2) t 2t z = , phng trnh (2) tr thnh:

2 321

03

t t t

t

=+ = = . T ,

2

2

113.3

z z

z i z

= =

= =

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

1 2 31, 1, 3.z z z i = = = v

43.z i =

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

1 0 ( 1)( 1) 01 0

z z 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 2

i z += ; 1 32 2i z =

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

1 1z = , 312 2 2z i = + v31

3 2 2z i =

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

Bi giiCu 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 23

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 33 4 3 4 55 17 4

a b a z i z a 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 4i i

+ l) 12 i

m)2

11

i i

+

n)5

11

i i

+

o)2

2(2 )i

p) (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 2i i

i i +

+ l)(2 3 )(1 2 )

(2 4 )1i i

i i +

+ + 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 31 2

i i z

i i + += +

f) 2 1 31 2 2

i i z

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 rnga) 2(1 2. )z i = + b)3

4(1 )(1 )

i

i z +

=

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 : Cho 1 22 3 , 1z i z i = + = + . Tnh2

1 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 ( )3

311 2 2

z i = + v ( )3

311 2 2

z i = + . Tnh 1 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 phcz c phn thc v phn o i nhau v 2 2z = Bi 17 : Cho 1 2,z z l hai nghim phc ca phng trnh25 2 1 0z z + =

Chng minh rng tng nghch o ca1z v 2z bng 2.Bi 18 : Cho 1 2,z z l hai nghim phc ca phng trnh

23 2 4 0z z + = Chng minh rng 1 2 1 2. 2z z z z + + =

Bi 19 : Cho 1 2,z z l hai nghim phc ca phng trnh2 4 5 0z z + =

Chng minh rng2 2

1 2 6z z + = Bi 20 : Cho 1 2,z z l hai nghim phc ca phng trnh25 2 2 0z z + =

Chng minh rng 1 2 1 2.z z z z + = Bi 21 : Cho 1 2,z z l hai nghim phc ca phng trnh

23 2 1 0z z + = v 2z c phn o l mt s m. Tnh1 22z z +

Bi 22 : Tm s phcz 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 z bit rng 5z = .Bi 25 : Cho s phc ( 1) ( 1) ,z m m i m = + + . Tm z bit . 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 bng5 .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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Ph n V.Ph n V.Ph n V.Ph n V. PH NG PHP TO+ PH NG PHP TO+ PH NG PHP TO+ PH NG PHP TO+ TRONG KHNG GIAN TRONG KHNG GIAN TRONG KHNG GIAN TRONG KHNG GIAN

1. H to Oxyz Gm 3 trcOx ,Oy ,Oz i mt vung gc nhauc 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 imI ca onAB Trng tmG ca tam gicABC

2

2

2

A B I

A B I

A B I

x x x

y y y

z z z

+=

+=+=

3

3

3

A B C G

A B C G

A B C G

x x x x

y y y y

z z z z

+ +=

+ +=+ +=

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)M M y mp( )Oxz : 13( ; 0; )M M M x z Trc Oz : 3(0;0; )M M z mp( )Oyz : 23(0; ; )M M M y z

3. To ca vct a) nh ngha: 1 2 3 1 2 3( ; ; ) . . .a a a a a a i a j a k = = + +

b) Cng thc to ca vctNu ( ; ; ), ( ; ; )A A A B B B A x y z B x y z th ( ; ; )B A B A B AAB x x y y z z =

Nu 1 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 vib

tn ti s thct 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 31 2 3

( ; ; )( ; ; )

a a a a

b b b b

==

th 1 1 2 2 3 3. . . .a b a b a b a b= + +

b) ng dng: 2 2 21 2 3a a a a = + + AB AB =

.cos( , )

.a b

a ba b

= . 0a b a b =

, vi00

a

b

5. Tch c hng ca hai vct a) 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 vcta 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 khc0

ln lt l , ,a b c

. Khi , a v b

cng phng vi nhau [ , ] 0a b =

,a b

v c ng 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 hnhABCD

[ , ]ABCD S AB AD =

Din tch tam gicABC :ABC S

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 dinABCD :

ABCD V = 16 [ , ].AB AC AD

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V D MINH HOBi 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 rngA,B ,C l 3 nh ca mt tam gic vung.b) Chng minh rng ( )AA ABC c) Tnh th tch khi t dinA 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,ABC l tam gic vung tiA 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 52 2ABC

AB 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: ABCD l hnh bnh hnh AD BC =

2 2 41 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 rngA,B ,C l ba nh ca mt tam gic.b) Xc nh to nhD v tm I ca hnh bnh hnhABCD .

c) Tm to imM sao 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 rngABC l tam gic u.b) Cho im (4;0; 3)A . Xc nh to cc imB 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 ,C lnlt l hnh chiu vung gc caM ln cc trc to Ox ,Oy ,Oz .a) Chng minh rngABC l tam gic cn.b) Tnh th tch t dinOABC , 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 ,C lnlt l hnh chiu vung gc ca imN ln cc mt phng to Oxy , Oyz , Oxz .

a) Tnh din tch tam gicABC v th tch ca t dinNABC .b) Tnh khong cch t imN n 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 dinABCD sao cho

(2; 4; 1), 4 , (2; 4; 3), (0; 2;0)A OB i j k C AD = + =

a) Chng minh rngAB , AC v AD i mt vung gc vi nhau.b) Tnh din tch tam gicABC v th tch t dinABCD .Bi 8 : Trong h to Oxyz cho (2;1; 3), (4; 3; 2), (6; 4; 1)A B C a) Chng minh rngA,B ,C l ba nh ca mt tam gic vung.b) Tm to imD A,B ,C ,D l 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 imN trn Oy cch u hai im (3;1;0)A v ( 2;4;1)B

Bi 11 :Tm imM 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 cuTm 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.

Nua v b

l hai vct khng cng phng, c gi song song hoccha 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 imM o 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 thng Cho ng thngd i qua im 0 0 0 0( ; ; )M x y z v c vtcp ( ; ; )u a b c =

a) Phng trnh tham s cad :

0

0

0( )

x x at

y y bt t z z ct

= +

= + = +

b) Phng trnh chnh tc cad : 0 0 0x x y y z z

a b c

= = (gi sa ,b,c u khc 0)

c) Cch xc nh vct ch phng (vtcp) cho ng thngd d i qua 2 imA v B phn bit thd c vtcp u AB =

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

Cho mt phng( )P c vtpt P n . Nud (P ) th d c vtcp P u n =

Cho hai vct khng cng phnga v b

. Nud vung gc vi gi

ca 2 vcta v b

th d c vtcp [ , ]u a b=

Cho ng thng c vtcp u v mt phng( )P c vtpt P n

. Nud song song vi( )P v vung gc vi th d c vtcp [ ],P u n u =

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

.

Nud l giao tuyn ca( )P v ( )Q th d c vtcp [ ],P Q u n n =

Cho hai ng thng1d v 2d ln lt c vtcp 1u v 2u

khngcng phng. Nud vung gc vi c hai ng thng1d v 2d thd c vtcp 1 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 tmB , i qua A b) Vit phng trnh mt cu ng knhBC .c) Vit phng trnh mt cu tmC , tip xc vi mt phng( )P d) Vit phng trnh mt cu ngoi tip t dinOABC .

Bi giiCu a: Gi 1( )S l mt cu tmB (2;1;2) v i qua imA. Khi 1( )S

c bn knh 1R 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:Gi 2( )S l mt cu ng knhBC th 2( )S c tm (1 2)32; ;I ltrung im ca on thngBC v bn knh

2BC R =

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

nn 692 2

BC R = =

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 tmC (0;2;6), tip xc vi( )P . Khi 3( )S

c bn knh 3 ( , ( ) )R d C P = 2 2 20 2.2 2( 6) 1 15

31 ( 2) 25 + +

+ += = =

3( )S c phng trnh:2 2 2( 2) ( 6) 25x y z + + + =

Cu d: Gi s 2 2 24

( ) : 2 2 2 0S x y z ax by cx d + + + = l mt cui qua O (0;0;0),A(1;3;1),B (2;1;2),C (0;2; 6) thd = 0 v

921310

2910

11 2 6 2 0 2 6 2 119 4 2 4 0 4 2 4 940 4 12 0 4 12 40

a a b c a b c

a b c a b c b

b c b c c

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

M ( ) ( ) ( )2 2 22 2 2 9 13 292 10 10 0a b c d + + = + + > nn phng trnh camt cu 4( )S cn tm l

2 2 2 9x y z x + + 13 295 5

0y x + =

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Bi 13 : Vit phng trnh mt phng( ) trong cc trng hp sau y:a) ( ) i qua (1; 2;2)A v vung gc viOM bit (3; 1;2)M b) ( ) i qua ba im (0;1;2), ( 3;1;4), (1; 2; 1)A K D .c) ( ) i qua hai imA, B v song song vi ng thngCD bit

(1;1;1), (2;1;2), ( 1;2;2), (2;1; 1)A B C D d) ( ) l mt trung trc ca onMN vi (2; 3;1), ( 4;1;5)M N

Tai Lieu on Tap Thi Tot Nghiep 2012 Mon Toan Thpt

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