De Cuong Toan 2014-2015

UBND TNH BC NINH5 GIO DC V O TO

CNGN THI THPT QUC GIA MN TON

0m hc 2014 - 2015

%c Ninh, thng 11 nm 2014

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GD&0Ninh0ngnthiTHPTqugia2014-2015

CHUYN 1. KHO ST HM S V CC BI TON LIN QUANBin son v su tm: Ng Vn Khnh GV trng THPT Nguyn Vn C

1. Ch 1: Bi ton v tip tuyn1.1. Dng 1: Tip tuyn ca th hm s ti mt Lm

0 0M( , ) ( ) : ( )x y C y f x =

* Tnh ' '( )y f x= ; tnh '0

( )k f x= (h s gc ca tip tuyn)

* Tip tuyn ca th hm s ( )y f x= ti Km ( )0 0;M x y c phng trnh( )'0 0 0( )y y f x x x- = - vi 0 0( )y f x=

V d 1: Cho hm s 3 3 5y x x= - + (C). Vit phng trnh tip tuyn ca th (C):a) 6i Km A (-1; 7).b) 6i Km c honh x = 2.c) 6i Km c tung y =5.

Gii:a) Phng trnh tip tuyn ca (C) ti Km

0 0 0( ; )M x y c dng:

0 0 0'( )( )y y f x x x- = -

Ta c 2' 3 3y x= - '( 1) 0y - = .

Do phng trnh tip tuyn ca (C) ti Km A(-1; 7) l: 7 0y - = hay y = 7.b) T 2 7x y= = .

y(2) = 9. Do phng trnh tip tuyn ca (C) ti Km c honh x = 2 l:7 9( 2) 7 9 18 9 11y x y x y x- = - - = - = -

c) Ta c: 3 30

5 3 5 5 3 0 3

3

x

y x x x x x

x

=

= - + = - = = - =

+) Phng trnh tip tuyn ti ca (C) ti Km (0; 5).Ta c y(0) = -3.Do phng trnh tip tuyn l: 5 3( 0)y x- = - - hay y = -3x +5.

+) Phng trnh tip tuyn ti ca (C) ti Km ( 3;5)- .2'( 3) 3( 3) 3 6y - = - - =

Do phng trnh tip tuyn l: 5 6( 3)y x- = + hay 6 6 3 5y x= + + .

+) Tng t phng trnh tip tuyn ca (C) ti ( 3;5)- l: 6 6 3 5y x= - + .

V d 2: Cho th (C) ca hm s 3 22 2 4y x x x= - + - .a) Vit phng trnh tip tuyn vi (C) ti giao Km ca (C) vi trc honh.

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b) Vit phng trnh tip tuyn vi (C) ti giao Km ca (C) vi trc tung. c) Vit phng trnh tip tuyn vi (C) ti Km x0 tha mn y(x0) = 0.Gii:

Ta c 2' 3 4 2y x x= - + . Gi ( )0 0;M x y l tip Km th tip tuyn c phng trnh:

0 0 0 0 0 0'( )( ) '( )( ) (1)y y y x x x y y x x x y- = - = - +

a) Khi ( )M C Ox= th y0 = 0 v x0 l nghim phng trnh:3 22 2 4 0 2x x x x- + - = = ; y(2) = 6, thay cc gi tr bit vo (1) ta c phng

trnh tip tuyn: 6( 2)y x= -

b) Khi ( )M C Oy= th x0 = 0 0 (0) 4y y = = - v 0'( ) '(0) 2y x y= = , thay ccgi trbit vo (1) ta c phng trnh tip tuyn: 2 4y x= - .

c) Khi x0 l nghim phng trnh y= 0. Ta c: y = 6x 4.

y = 00 0

2 2 886 4 0

3 3 27x x x y y

- = = = = = - ;

0

2 2'( ) '

3 3y x y

= =

Thay cc gi tr bit vo (1) ta c phng trnh tip tuyn: 2 1003 27

y x= -

V d 3: Cho hm s 3 3 1y x x= - + (C) a) Vit phng trnh tip tuyn d vi (C) tai Km c honh x=2. b)Tip tuyn d ct li th (C) ti Km N, tm ta ca Km N.Gii

a) Tip tuyn d ti Km M ca th (C) c honh 0 0

2 3x y= =

Ta c 20

'( ) 3 3 '( ) '(2) 9y x x y x y= - = =

Phng trnh tip tuyn d ti Km M ca th (C) l

0 0 0'( )( ) 9( 2) 3 9 15y y x x x y y x y x= - + = - + = -

8y phng trnh tip tuyn d ti Km M ca th (C) l 9 15y x= -b) Gi s tip tuyn d ct (C) ti N

Xt phng trnh

( )( )3 3 2 23 1 9 15 12 16 0 2 2 8 0 4x

x x x x x x x xx

=- + = - - + = - + - = = -8y ( )4; 51N - - l Km cn tmV d 4: Cho hm s 3 3 1 ( )y x x C= - + v Km 0 0( , )A x y (C), tip tuyn ca th

(C) ti Km A ct (C) ti Km B khc Km A. tm honh Km B theo0

x

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/i gii: V Km

0 0( , )A x y (C) 3

0 0 03 1y x x = - + , ' 2 ' 2

0 03 3 ( ) 3 3y x y x x= - = -

Tip tuyn ca th hm c dng:' 2 3

0 0 0 0 0 0 02 30 0 0

( )( ) (3 3)( ) 3 1

(3 3)( ) 2 1 ( )

y y x x x y y x x x x x

y x x x x d

= - + = - - + - +

= - - - +

Phng trnh honh giao Km ca (d) v (C):3 2 3 3 2 3 2

0 0 0 0 0 0 0

200

000

3 1 (3 3)( ) 2 1 3 2 0 ( ) ( 2 ) 0

( ) 0( 0)

22 0

x x x x x x x x x x x x x x

x xx xx

x xx x

- + = - - - + - + = - + = =- = = -+ =

8y Km B c honh 0

2B

x x= -

V d 5: Cho hm s 3 21 2 33

y x x x= - + (C). Vit phng trnh tip tuyn d ca th

(C) ti i m c honh 0

x tha mn ''0

( ) 0y x = v chng minh d l tip tuyn ca (C) cJ s gc nh nht.Gii

Ta c ' 2 ''4 3 2 4y x x y x= - + = -

0 0 0

2''( ) 0 2 4 0 2 (2; )

3y x x x M= - = =

Khi tip tuyn ti M c h s gc0

k = ' '0

( ) (2) 1y x y= = -

8y tip tuyn d ca th (C) ti Km 22;3

M

c phng trnh ( )'0 0 0( )y y f x x x- = -

suy ra ( )2 1 23

y x- = - - hay 83

y x= - +

Tip tuyn d c h s gc0

k = -1/t khc tip tuyn ca thi (C) ti Km by k trn (C) c h s gc

( )2' 2 0( ) 4 3 2 1 1k y x x x x k= = - + = - - - =

Du = xy ra 1x = nn ta tip im trng vi 22;3

M

8y tip tuyn d ca (C) ti Km 22;3

M

c h s gc nh nht.

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V d 6: Vit phng trnh tip tuyn vi th (C): 21

xy

x+

=-

ti cc giao Km ca (C)

Xi ng thng (d): 3 2y x= - .+ Phng trnh honh giao Km ca (d) v (C):

23 2 2 (3 2)( 1)

1x

x x x xx

+= - + = - -

- (x = 1 khng phi l nghim phng trnh)

23 6 0 0 ( 2) 2 ( 4)x x x y x y - = = = - = =

8y c hai giao Km l: M1(0; -2) v M2(2; 4)

+ Ta c:2

3'

( 1)y

x

-=

-.

+ Ti tip Km M1(0; -2) th y(0) = -3 nn tip tuyn c phng trnh: 3 2y x= - -+ Ti tip Km M2(2; 4) th y(2) = -3 nn tip tuyn c phng trnh: 3 10y x= - +Tm li c hai tip tuyn tha mn yu cu bi ton l: 3 2y x= - - v 3 10y x= - + .

V d 7: Cho hm s 3 21 13 2 3

my x x= - + (Cm).)i M l Km thuc th (Cm) c honh

bng -1. Tm m tip tuyn vi (Cm) ti M song song vi ng thng d: 5x-y=0Gii

Ta c ' 2y x mx= -ng thng d: 5x-y=0 c h s gc bng 5, nn tip tuyn ti M song song vi ngthng d trc ht ta cn c '( 1) 5 1 5 4y m m- = + = =

Khi 4m = ta c hm s 3 21 123 3

y x x= - + ta c0

1x = - th0

2y = -

Phng trnh tip tuyn c dng '0 0 0

( )( ) 5( 1) 2 5 3y y x x x y y x y x= - + = + - = +

R rng tip tuyn song song vi ng thng d8y 4m = l gi tr cn tm.V d 8: Cho hm s 3 23y x x m= - + (1).Tm m tip tuyn ca th (1) ti Km c honh bng 1 ct cc trc Ox, Oy ln

Nt ti cc Km A v B sao cho din tch tam gic OAB bng 32

.

Gii8i

0 01 2x y m= = - M(1 ; m 2)

- Tip tuyn ti M l d: 20 0 0

(3 6 )( ) 2y x x x x m= - - + -

d: y = -3x + m + 2.

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- d ct trc Ox ti A: 2 20 3 2 ; 03 3A A

m mx m x A

+ + = - + + =

- d ct trc Oy ti B: 2 (0 ; 2)B

y m B m= + +

- 23 1 3 2| || | | || | 3 2 3 ( 2) 92 2 2 3OAB

mS OA OB OA OB m m

+= = = + = + =

2 3 1

2 3 5

m m

m m

+ = = + = - = - 8y m = 1 v m = - 5

1.2. Dng 2: Vit tip tuyn ca thi hm s ( )y f x= (C) khi bit trc h s gc ca n

+ Gi0 0

( , )M x y l tip Km, gii phng trnh '0 0

( )f x k x x= = ,0 0

( )y f x=

+ n y tr v Fng 1,ta d dng lp c tip tuyn ca th:

0 0( )y k x x y= - +

Cc dng biu din h s gc k:

*) Cho trc tip: 35; 1; 3; ...7

k k k k= = = =

*) Tip tuyn to vi chiu dng ca trc Ox mt gc a , vi 0 0 0 215 ;30 ;45 ; ; .... .3 3p p

a

Khi h s gc k = tan a .*) Tip tuyn song song vi ng thng (d): y = ax + b. Khi h s gc k = a.

*) Tip tuyn vung gc vi ng thng (d): y = ax + b 11ka ka

- = - = .

*) Tip tuyn to vi ng thng (d): y = ax + b mt gc a . Khi : tan1k a

kaa

-=

+.

V d 9: Cho hm s 3 23y x x= - (C). Vit phng trnh tip tuyn ca th (C) bit hU gc ca tip tuyn k = -3.Gii:

Ta c: 2' 3 6y x x= -

)i0 0

( ; )M x y l tip im Tip tuyn ti M c h s gc ' 20 0 0

( ) 3 6k f x x x= = -

Theo gi thit, h s gc ca tip tuyn k = - 3 nn:2 20 0 0 0 0

3 6 3 2 1 0 1x x x x x- = - - + = =

V0 0

1 2 (1; 2)x y M= = - - .

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Phng trinh tip tuyn cn tm l 3( 1) 2 3 1y x y x= - - - = - +

V d 10: Vit phng trnh tip tuyn ca th hm s 3 23 1y x x= - + (C). Bit tiptuyn song song vi ng thng y = 9x + 6.Gii:

Ta c: 2' 3 6y x x= -

)i0 0

( ; )M x y l tip Km Tip tuyn ti M c h s gc ' 20 0 0

( ) 3 6k f x x x= = -

Theo gi thit, tip tuyn song song vi ng thng y = 9x + +6 tip tuyn c h

U gc k = 9 02 20 0 0 0

0

1 ( 1; 3)3 6 9 2 3 0

3 (3;1)

x Mx x x x

x M

= - - -- = - - = = Phng trinh tip tuyn ca (C) ti M(-1;-3) l: 9( 1) 3 9 6y x y x= + - = + (loi)Phng trinh tip tuyn ca (C) ti M(3;1) l: 9( 3) 1 9 26y x y x= - + = -

V d 11: Cho hm s 3 3 2y x x= - + (C). Vit phng trnh tip tuyn ca (C) bit tip

tuyn vung gc vi ng thng 19

y x-

= .

Gii:Ta c 2' 3 3y x= - . Do tip tuyn ca (C) bit tip tuyn vung gc vi ng

thng 19

y x-

= nn h s gc ca tip tuyn k = 9.

Do 2 2' 3 3 9 4 2.y k x x x= - = = = +) Vi x = 2 4y = . Pttt ti Km c honh x = 2 l:

9( 2) 4 9 14.y x y x= - + = -

+) Vi 2 0x y= - = . Pttt ti Km c honh x = - 2 l:9( 2) 0 9 18y x y x= + + = + .

Vy c hai tip tuyn c (C) vung gc vi ng thng 19

y x-

= l:

y =9x - 14 v y = 9x + 18.

V d 12: Lp phng trnh tip tuyn vi th (C) ca hm s: 4 21 24

y x x= + ,

bit tip tuyn vung gc vi ng thng (d): 5 2010 0x y+ - = .

Gii:

(d) c phng trnh: 1 4025

y x= - + nn (d) c h s gc l - 15

.

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)i D l tip tuyn cn tm c h s gc k th 1 . 1 5 ( ( ))5

k k do d- = - = D ^ .

Ta c: 3' 4y x x= + nn honh tip Km l nghim phng trnh: 3 4 5x x+ =

3 2 94 5 0 ( 1)( 5) 0 1 0 14

x x x x x x x y + - = - + + = - = = =

8y tip Km M c ta l 91;4

M

Tip tuyn c phng trnh: 9 115( 1) 54 4

y x y x- = - = -

8y tip tuyn cn tm c phng trnh: 1154

y x= - .

V d 13: Cho hm s 22 3x

yx

+=

+ (C). Vit phng trnh tip tuyn vi (C) bit rng tip

tuyn ct trc honh ti A, trc tung ti B sao cho tam gic OAB vung cn ti O, y Ol gc ta .Gii

Ta c: '2

1

(2 3)y

x

-=

+

V tip tuyn to vi hai trc ta mt tam gic vung cn nn h s gc ca tip tuynl: 1k = Khi gi ( )0 0;M x y l tip Km ca tip tuyn vi th (C) ta c ' 0( ) 1y x =

0

200

211

1(2 3)

x

xx

= -- = = -+ 8i

01x = - th

01y = lc tip tuyn c dng y x= - (trng hp ny loi v tip

tuyn i qua gc ta , nn khng to thnh tam gic OAB)8i

02x = - th

04y = - lc tip tuyn c dng 2y x= - -

8y tip tuyn cn tm l 2y x= - -

V d 14: Cho hm s y = 2 11

xx

--

c th (C).

Lp phng trnh tip tuyn ca th (C) sao cho tip tuyn ny ct cc trc Ox, Oy lnNt ti cc Km A v B tha mn OA = 4OB.Gii

Gi s tip tuyn d ca (C) ti0 0

( ; ) ( )M x y C ct Ox ti A, Oy ti B sao cho

4OOA B= .

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Do 'OAB vung ti O nn 1tan4

OBA

OA= = H s gc ca d bng 1

4 hoc 1

4- .

* s gc ca d l0 2 2

0 0

1 1 1( ) 0

4( 1) ( 1)y x

x x = - < - = -

- - 0 0

0 0

31 ( )

25

3 ( )2

x y

x y

= - = = =

Khi c 2 tip tuyn tha mn l:

1 3 1 5( 1)

4 2 4 41 5 1 13

( 3)4 2 4 4

y x y x

y x y x

= - + + = - + = - - + = - +

.

1.3. Dng 3: Tip tuyn i qua imCho th (C): y = f(x). Vit phng trnh tip tuyn vi (C) bit tip tuyn i qua

Km ( ; )A a b .Cch gii

+ Tip tuyn c phng trnh dng:0 0 0

( ) '( )( )y f x f x x x- = - , (vi x0 l honh tip Km).

+ Tip tuyn qua ( ; )A a b nn0 0 0

( ) '( )( ) (*)f x f x xb a- = -

+ Gii phng trnh (*) tm x0 ri suy ra phng trnh tip tuyn.V d 15: Cho th (C): 3 3 1y x x= - + , vit phng trnh tip tuyn vi (C) bittip tuyn i qua Km A(-2; -1).Gii:

Ta c: 2' 3 3y x= -

)i M( )30 0 0; 3 1x x x- + l tip Km. H s gc ca tip tuyn l 20 0'( ) 3 3y x x= - .Phng trnh tip tuyn vi (C) ti M l D : ( )3 20 0 0 03 1 (3 3)( )y x x x x x- - + = - -D qua A(-2;-1) nn ta c: ( )3 20 0 0 01 3 1 (3 3)( 2 )x x x x- - - + = - - - 3 20 03 4 0x x + - =

0 020 0 0

0 0

1 1( 1)( 4 4) 0

2 1

x yx x x

x y

= = - - + + = = - = -8y c hai tip tuyn cn tm c phng trnh l: : 1 ; : 9 17y y xD = - D = +

1.4. Dng 4. Mt s bi ton tip tuyn nng cao.V d 16: Tm hai Km A, B thuc th (C) ca hm s: 3 3 2y x x= - + sao cho

tip tuyn ca (C) ti A v B song song vi nhau v di Qn AB = 4 2 .Gii:

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)i 3 3( ; 3 2) , ( ; 3 2) ,A a a a B b b b a b- + - + l hai Km phn bit trn (C).

Ta c: 2' 3 3y x= - nn cc tip tuyn vi (C) ti A v B c h s gc ln lt l:2 2'( ) 3 3 '( ) 3 3y a a v y b b= - = - .

Tip tuyn ti A v B song song vi nhau khi:2 2'( ) '( ) 3 3 3 3 ( )( ) 0 ( 0)y a y b a b a b a b a b v a b a b= - = - - + = = - -

22 2 3 34 2 32 ( ) ( 3 2) ( 3 2) 32AB AB a b a a b b = = - + - + - - + =

2 22 3 3 2 2 2( ) ( ) 3( ) 32 ( ) ( )( ) 3( ) 32a b a b a b a b a b a ab b a b - + - - - = - + - + + - - =

22 2 2 2( ) ( ) ( ) 3 32a b a b a ab b - + - + + - = , thay a = -b ta c:

( ) ( )2 22 2 2 2 2 2 6 4 24 4 3 32 3 8 0 6 10 8 0b b b b b b b b b+ - = + - - = - + - =

2 4 2 22 2

( 4)( 2 2) 0 4 02 2

b ab b b b

b a

= = - - - + = - = = - =- 8i 2 2a v b= - = ( 2;0) , (2;4)A B-

- 8i 2 2a v b= = - (2;4) , ( 2;0)A B -

Tm li cp Km A, B cn tm c ta l: ( 2; 0) (2; 4)v-

V d 17: Tm hai Km A, B thuc th (C) ca hm s: 2 11

xy

x-

=+

sao cho tip

tuyn ca (C) ti A v B song song vi nhau v di Qn AB = 2 10 .Gii:

Hm sc vit li: 321

yx

= -+

)i 3 3;2 , ;21 1

A a B ba b

- - + + l cp Km trn th (C) tha mn yu cu bi ton.

8i Ku kin: , 1, 1a b a b - - .

Ta c:2

3'

( 1)y

x=

+ nn h s gc ca cc tip tuyn vi (C) ti A v B l:

2 2

3 3'( ) '( )

( 1) ( 1)y a v y b

a b= =

+ +

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10

Tip tuyn ti A v B song song khi:2 2

3 3'( ) '( )

( 1) ( 1)y a y b

a b= =

+ +

1 12

1 1 2

a b a ba b

a b a b

+ = + = = - - + = - - = - - (1) (do a b )

2

2 2 3 32 10 40 ( ) 401 1

AB AB a bb a

= = - + - = + + 2 2

2 23 3 6( 2 2) 40 4( 1) 401 1 1

b bb b b

- - + - = + + = + - - + ( do thay a (1) )

24 2

2

( 1) 1 1 1 1 1( 1) 10( 1) 9 0

1 3 1 3( 1) 9

b b bb b

b bb

+ = + = + = - + - + + = + = + = -+ =

0 2

2 0

2 4

4 2

b a

b a

b a

b a

= = - = - = = = - = - =

%p Km A v B cn tm c ta l: ( 2;5) (0; 1) ; (2;1) ( 4;3)v v- - -

V d 18: Cho hm s: y = x3 + 3x2 + mx + 1 c (Cm); (m l tham s). Xc nh m (Cm) ct ng thng y = 1 ti 3 Km phn bit C(0, 1), D, E sao cho cc tip tuyn ca(Cm) ti D v E vung gc vi nhau.Gii

Phng trnh honh giao Km ca (Cm) v ng thng y = 1 l:

x3 + 3x2 + mx + 1 = 1 x(x2 + 3x + m) = 0 20

3 0 (2)

x

x x m

= + + =

* (Cm) ct ng thng y = 1 ti C(0, 1), D, E phn bit: Phng trnh (2) c 2 nghim xD, xE z 0.

209 4 040 3 0 09

mm

m m

D = - > + +


11

9m + 6m (3) + 4m2 = 1; (v xD + xE = 3; xDxE = m theo nh l Vi-t).

4m2 9m + 1 = 0 m = ( )B1 9 658S: m = ( ) ( )B1 19 65 9 658 8hay m- =

V d 19: .p phng trnh tip tuyn vi th (C) ca hm s: 2 21

xy

x-

=+

, bit

Tng khong cch tKm I(-1; 2) n tip tuyn l ln nht.Gii:

)i D l tip tuyn ca th (C) ti tip Km M ( )2 2; , ( )1

aa M C

a

- + .

Ta c: ( )2 24 4

' '( ) , 1( 1) ( 1)

y y a ax a

= = -+ +

8y 2 22

2 2 4: ( ) 4 ( 1) 2 4 2 0 (*)

1 ( 1)

ay x a x a y a a

a a

-D - = - - + + - - =

+ +

( )2 2

4 4

4( 1) ( 1) .2 2 4 2 8 1;

4 ( 1) 4 ( 1)

a a a ad I

a a

- - + + - - +D = =

+ + + +.

Ta c:2

4 2 2 2 4 24 ( 1) 2 ( 1) 2.2( 1) 4 ( 1) 2.2( 1) 2 1a a a a a a + + = + + + + + + = +

( )8 1

; 42 1

ad I

a

+ D =

+. Vy ( );d I D ln nht khi ( );d I D = 4

2 21 2 1

2 ( 1)1 2 3

a aa

a a

+ = = = + + = - = - . C hai gi tru tha mn 1a

+ Vi a = 1 thay vo (*) ta c phng trnh tip tuyn l:4 4 4 0 1 0x y x y- - = - - =+ Vi a = -3 thay vo (*) ta c phng trnh tip tuyn l:4 4 28 0 7 0x y x y- + = - + =Tm li: C hai tip tuyn cn tm c phng trnh l: 1 0 ; 7 0x y x y- - = - - =

V d 20: Cho (C) l th hm s 12 1x

yx

+=

+. Vit phng trnh tip tuyn vi (C),

bit tip tuyn ct trc honh, trc tung tng ng ti cc Km A, B tha mnD OAB vung cn ti gc ta O.Gii:

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12

)i ( )0 0;M x y l tip Km. Tip tuyn vi (C) ti M phi tha mn song song vi ccng thng y = x hoc y = -x.

Ta c:2

1'

(2 1)y

x= -

+ nn tip tuyn vi (C) ti M c h s gc l:

0 20

1'( ) 0

(2 1)y x

x= - nn hm st cc tiu ti x = 2 v gi tr cc tiu yCT ( ) 423

y-

= = .

Vi d 2: Tm cc tr ca cc hm s sau:

a) 1cos os2 12

y x c x= + - b) 2 13 s inx cos2

xy x += + +

(?) Ta thy hm s ny rt kh xt du ca y, do hy s dng quy tc 2 tm cc tr?

2 1' 2; ' 02

xy x x y

x

ff

2 1' 2; ' 02

xy x x y

x

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19

Giia) TX: D=R

* ' s inx sin2y x= - -

s inx 0' 0 s inx(1 2cos ) 0 1 2cos 2

2 3

x ky x

x x n

pp p

= == + =

= - = +

* " cos 2 os2y x c x= - -

Ta c "( ) os( ) 2 os( 2 ) 1 2 0y k c k c kp p p= - - = - < Hm st cc tiu ti: ( )x k kp= Z

1 31 0

2 22 2 4" 2 os - 2cos3 3 3

y n cp p pp = + = > + = -

Hm st cc tiu ti: 2 2 ( )3

x n n Zp p= +

b) TX: D=R.

* ' 3 cos s inx 1y x= - +

' 0 3 cos s inx 1y x= - = -

3 1 1cos s inx2 2 2

x - = - 1s in x sin3 2 6p p

- = =2

27 26

x k

x k

p p

p p

= +

= +

* " 3 s inx cosy x= - -Ta c:

+ " 2 3 s in cos 3 02 2 2

y kp p pp

= + = - - = -

x Khi ba Km cc tr ca th hm s l:( ) ( ) ( )2 20; 1 , ; 1 , ; 1A m B m m m C m m m- - - + - - + -x

+

21 .2ABC B A C B

S y y x x m m= - - = ; 4 , 2AB AC m m BC m= = + =

( )+

4

3

2

12. .1 1 2 1 0 5 14 4

2ABC

mm m mAB AC BCR m m

S mm m

=+ = = = - + = - =

Bi tp t luynBi 1. Cho hm s ( )3 22 3 1 6y x m x mx= - + + .

a) Tm m hm s c cc tr.

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25

b) Tm m hm s c hai cc tr trn ( )0;+ .c) Tm m th hm s c hai Km cc tr nm v hai pha trc tung.d) Tm m th hm s c hai Km cc tr nm v hai pha trc honh

Bi 2. Cho hm s ( ) ( )3 21 11 3 23 3

y mx m x m x= - - + - + . Tm m hm st cc i

Vi 0x = .

Bi 3. Tm m hm s ( )3 2 22 22 3 13 3

y x mx m x= - - - + c hai Km cc tr1

x v2

x sao

cho: ( )1 2 1 22 1x x x x+ + =

Bi 4. Tm tt c cc gi tr ca m hm s ( )3 2 21 1 33 2

y x mx m x= - + - c cc i ti

x& cc tiu ti CTx sao cho x&, CTx l di cc cnh gc vung ti mt tam gic vung

c di cnh huyn bng 52

.

Bi 5. Xc nh m hm s ( )3 22 1 1y mx m x x= - - - + t cc tr ti 1 2,x x sao cho

1 2

169

x x- = .

Bi 6. Xc nh m hm s ( )3 23 1 9y x m x x m= - + + - t cc tr ti 1 2,x x sao cho

1 2 2x x- = .

Bi 7. Tm m th hm s ( )3 22 3 1 6y x m x mx= - + + c hai Km cc tr A v B saocho ng thng AB vung gc vi ng.

Bi 8. Tm m th hm s 3 2 33 3y x mx m= - + c hai Km cc tr A v B sao chotam gic OAB c din tch bng 48.

Bi 9. Cho hm s 3 2 33 4y x mx m= - + (1), vi m l tham s thc. Tm m th hmU (1) c hai Km cc tr A v B sao cho 2 2 20OA OB+ = .

Bi 10. Cho hm s 3 21 2 33

y x x x= - + (1).

1. Kho st s bin thin v v th ca hm s (1).2. Gi A, B ln lt l cc Km cc i, cc tiu ca th hm s (1). Tm Km M thuctrc honh sao cho tam gic MAB c din tch bng 2.

Bi 11. Cho hm s 3 2 33 12 2

y x mx m= - + Tm m th hm s c hai Km cc i,

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26

Ec tiu i xng qua ng thng y = x.

Bi 12. Cho hm s: 3 2y = x 3mx + 2- (1), m l tham sTm m ng thng quahai Km cc tr ca th hm s (1) to vi cc trc ta mt tam gic c din tchDng 4.

Bi 13. Cho hm s ( ) ( )3 2 2 23 3 1 3 1 1y x x m x m= - + + - - - Tm m hm s (1) cEc i, cc tiu, ng thi cc Km cc i v cc tiu cng vi gc ta O to thnhOt tam gic vung ti O.Bi 14. Cho hm s y = 2x3 + 9mx2 + 12m2x + 1, trong m l tham s.Tm tt c ccgi tr ca m hm s c cc i ti x%, cc tiu ti xCT tha mn: x2%= xCT.

Bi 15. Cho hm s ( ) ( )3 23 3 1 1 3 my x x m x m C= - + - + + Tm m hm s c cci, cc tiu, ng thi cc Km cc i v cc tiu cng vi gc ta O to thnh mttam gic c din tch bng 4.

Bi 16. Cho hm s 3 2 2 23 3(1 ) 2 2 1y x x m x m m= - + - + - - (m l tham s)Tm tt ccc gi tr ca tham s thc m hm s cho c cc i, cc tiu; ng thi hai Km cctr ca th hm si xng nhau qua ng thng : 4 5 0.d x y- - =

Bi 17. Cho hm s 3 23 ( 2) 3( 1) 12

y x m x m x= - - - - + (1), m l tham s.

a) Kho st s bin thin v v th ca hm s (1) khi 2m = - .b) Tm 0m > th hm s (1) c gi tr cc i, gi tr cc tiu ln lt l

,

C CTy y

tha mn

2 4C CT

y y+ = .

Bi 18. Cho hm s 3 21 5 4 4 ( )3 2

y x mx mx C= - - - . Tm m hm st cc tr ti1 2,x x

sao cho biu thc22

2 12 2

1 2

5 12

5 12

x mx mmA

x mx m m

+ += +

+ +t gi tr nh nht.

Bi 19. Tm m hm s ( ) ( )3 21 11 3 23 3

y mx m x m x= - - + - + t cc tr ti x1, x2 tha

mn x1 + 2x2 = 1.Bi 20. Tm m hm s ( )4 2 29 10y mx m x= + - + c 3 Km cc tr.Bi 21. Tm m th hm s y = -x4 +2(m+2)x2 2m 3 ch c cc i, khng c cctiu.

Bi 22. Tm m (C): ( ) ( )4 21 3 1 2 14

y x m x m= - + + + c 3 Km cc tr lp thnh mt

tam gic c trng tm l gc ta .

Bi 23. Cho hm s 4 22( 1)y x m x m= - + + (1), m l tham s.

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27

a) Kho st s bin thin v v th hm s (1) khi m = 1.b) Tm m th hm s (1) c ba Km cc tr A, B, C sao cho OA = BC, O l gc

Va , A l cc tr thuc trc tung, B v C l hai Km cc tr cn li.Bi 24. Cho hm s 4 22 4y x mx= - + - c th ( )mC . (m l tham s thc)Tm tt c cc gi tr ca m cc Km cc tr ca th ( )mC Pm trn cc trc ta .Bi 25. Cho hm s ( )4 2 2 42 1y x m x m m= - + + , m l tham s thc.Tm m th hm s ( )1 c ba Km cc tr lp thnh mt tam gic c din tch bng 1Bi 26. Cho hm s 4 22 2 1y x mx m= - + - + (1), m l tham s.

a) Kho st s bin thin v v th hm s (1) khi 2m = . b) Tm mTHS (1) c ba Km cc tr nm trn mt ng trn c bn knh bng 1.

Bi 27. Cho hm s ( )4 24 1 2 1y x m x m= - - + - c th ( )mC

a) Kho st s bin thin v v th ( )C ca hm s khi 32

m = .

b) Xc nh tham s m hm s c 3 cc tr to thnh 3 nh ca mt tam gic uBi 28. Cho hm s 4 2 2 42 1(1).y x m x m= - + + Tm m th ca hm s (1) c ba

Km cc tr , ,A B C sao cho cc Km , ,A B C v Km O nm trn mt ng trn, trong

O l gc ta .

Bi 29. Cho hm s ( )4 2 2 42 1y x m x m m= - + + , m l tham s thc. a) Kho st s bin thin v v th ca hm s khi 1m = - . b) Tm m th hm s ( )1 c ba Km cc tr lp thnh mt tam gic c din tchDng 32.

Bi 30. Cho hm s 4 22 1y x mx m= - + - c th ( )mC .Tm cc gi tr thc ca thamU m th ( )mC c ba Km cc tr nm trn ng trn c bn knh bng 1.

Bi 31. Cho hm s 4 21 2 2 (1)3

y x mx= - + , vi m l tham s. Tm m th ca hm

U (1) c ba Km cc tr to thnh mt tam gic c tm ng trn ngoi tip trng vi gcVa .

Bi 32. Cho hm s ( ) ( )4 2 22 2 5 5y f x x m x m m= = + - + - +a) Kho st s bin thin v v th (C ) hm s vi m = 1.b) Tm cc gi tr ca m th hm s c cc Km cc i, cc tiu to thnh 1

tam gic vung cn.

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28

3. Ch 3: Bi ton tng giao3.1. Kin thc c bn3.1.1. Bi ton tng giao tng qut:Cho hai th hm s: y = f(x, m) v y = g(x,m). Honh giao Km ca hai th lnghim ca phng trnh

f(x, m) = g(x,m) (1).) Nhn xt: S nghim ca (1) chnh l s giao Km ca hai th hm s.Sau lp phng trnh tng giao ca d v (C).3.1.2. Bi ton c bn:Cho hai th hm s: y = f(x, m) v d: y =ax+bHonh giao Km ca hai th l nghim ca phng trnh f(x,m) = ax+b. (1)Ch :

+ 0u ng thng d i qua Km M(x0; y0) v c h s gc k th phng trnh d c&ng: y y0 = k(x x0).

+ Khai thc ta giao i m ( ( ; )M M

M x y Ea (C) v d, ta cn ch :M

x l nghim

Ea (1);M thuc d nnM M

y ax b= +

+ Nu (1) d n n mt phng trinh b c hai, ta c th s dng nh l Viet) Phng php nhm nghim hu tCho phng trnh: 1

1 1 0( ) ... 0n n

n nf x a x a x a x a-

-= + + + + = .

0u phng trnh c nghim hu t pxq

= (p, q)=1 th \n

q a v0

\p a .

) Phng php hm sChuyn phng trnh honh tng giao v: g(x) = m.Khi s nghim chnh l s giao Km ca th y = g(x) v ng thng y = m.3.2. V d v bi tp

Vi d 1. Cho hm s 3 23 1y x x= - + -

a) Kho st v v th hm s trn. b) Da vo th bin lun theo m s nghim ca phng trnh 3 23 0x x m- + = .

Giia)

x TX: D = R.x 2' 3x 6xy = - +

20

' 0 3x 6x=02

xy

x

== - + =

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29

x Gii hn: lim , limx x

y y- +

= + = -

x $ng bin thin:

x Hm sng bin trn (0 ; 2); hm s nghch bin trn ( ;0)- v (2; )+ .x Hm st cc i ti x = 2, y% = 3; hm st cc tiu ti x = 0, yCT = -1.x th: Km c bit: (0;-1), (-1; 3), (3; -1), (1; 1)

b)x 3 2 3 23 0 3 1 1x x m x x m- + = - + - = -x 5 nghim ca phng trnh l s giao Km ca th hm s 3 23 1y x x= - + -Xi ng thng y = m 1.

8y1 3 4m m- > > : Phng trnh c 1 nghim.1 3 4m m- = = : Phng trnh c 2 nghim.

3 1 1 4 0m m> - > - > > : Phng trnh c 3 nghim.1 1 0m m- = - = : Phng trnh c 2 nghim.1 1 0m m- < - < : Phng trnh c 1 nghim.

Vi d 2.Cho hm s 4 23x 1y x= - + + c th (C)a) Kho st s bin thin v v th hm s (C).b) &a vo th (C) tm m phng trnh 4 2x 3x 0m- + = c 4 nghim phn bit.

Giia)

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30

Thc hin cc bc tng t nh bi tp 2, ta c th hm s sau:

b)x 4 2 4 2x 3x 0 3 1 1m x x m- + = - + + = +x 5 nghim ca phng trnh l s giao Km ca th (C) vi ng thng y=m+1.x &a vo th, phng trnh c 4 nghim phn bit

13 91 1 0

4 4m m < + < < "

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31

x 8y vi mi m th ng thng y = x m ct th (C) ti hai Km phn bit.V d 4.Cho hm s ( )3 23 4y x x C= - + .Gi d l ng thng i qua Km A(- 1; 0) viJ s gc l k ( k thuc R). Tm k ng thng d ct (C) ti ba Km phn bit v haigiao Km B, C (B, C khc A ) cng vi gc ta O to thnh mt tam gic c din tchDng 1.Gii

ng thng d i qua A(-1; 0) vi h s gc l k, c phng trnh l: y = k(x+1) = kx+ k.0u d ct (C) ti ba Km phn bit th phng trnh: x3 3x2 + 4 = kx + k x3 3x2 kx + 4 k = 0 (x + 1)( x2 4x + 4 k ) = 0

21

( ) 4 4 0

x

g x x x k

= - = - + - =

c ba nghim phn bit g(x) = x2 4x + 4 k = 0 c hai

nghim phn bit khc - 1' 0 0

0 9 (*)( 1) 0 9 0

kk

g k

D > > < - - 8i Ku kin: (*) th d ct (C) ti ba Km phn bit A, B, C.Vi A(-1;0), do B,C chonh l hai nghim ca phng trnh g(x) = 0.)i ( ) ( )1 1 2 2; ; ;B x y C x y Xi 1 2;x x l hai nghim ca phng trnh: 2 4 4 0x x k- + - = . Cn

1 1 2 2;y kx k y kx k= + = + .

Ta c: ( )( ) ( ) ( ) ( )JJJG 2 2 2

2 1 2 1 2 1 2 1; 1 1BC x x k x x BC x x k x x k= - - = - + = - +

Khong cch t O n ng thng d:21

kh

k=

+8y theo gi thit:

2 3 3 3

32

1 1 1 1 1. . 2 1 2 1

2 2 2 4 41

kS h BC k k k k k k

k= = + = = = = =

+

V d 5. Cho hm s ( )2 11

xy C

x+

=+

Tm tham s m ng thng d: y = - 2x + m ct

th ti hai Km phn bit A, B sao cho din tch tam gic OAB bng 3 .Gii

Xt phng trinh honh giao im ca d v (C):

22 1 2 ( 1) ( ) 2 ( 4) 1 0 (1)1

xx m x g x x m x m

x+

= - + - = - - + - =+

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32

D ct (C) ti 2 i m phn bit (1) c hai nghim phn bit khc -1.2 2( 4) 8(1 ) 0 8 0

( 1) 0 ( 1) 1 0

m m m

g g

D = - - - > + > - - = -

2 8 0m m R + > .

Chng t vi mi m d lun ct (C) ti hai Km phn bit A, B)i ( ) ( )1 1 2 2; 2 ; ; 2A x x m B x x m- + - + . Vi: 1 2,x x l hai nghim ca phng trnh (1)

Ta c ( )( ) ( ) ( )JJJG

1

2 2

2 1 2 2 1 2 1 2 1;2 4 5AB x x x x AB x x x x x x= - - = - + - = - .

)i H l hnh chiu vung gc ca O trn d, th khong cch t O n d l h:

2 52 1

m mh = =

+

Theo gi thit: 2 1 21 1 1 1. 5 . . 8 32 2 2 2 45

x xS AB h m

- D= = = = + =

8y: 2 2 2 2 28 4 .3 8 4 .3 40 2 10 (*)m m m m+ = + = = =

8i m tha mn Ku kin (*) th d ct (C) ti A, B tha mn yu cu bi ton.

V d 6. Cho hm s ( )3 22 3 4y x mx m x= + + + + (1). Tm m ng thng d: y = x + 4Et th hm s (1) ti ba Km phn bit A, B, C sao cho tam gic MBC c din tchDng 4. (Km B, C c honh khc khng ; M(1;3) ).Gii

th (1) ct d ti ba Km A, B, C c honh l nghim ca phng trnh:

( )3 2 2 20

2 3 4 4; 2 2 02 2 0

xx mx m x x x x mx m

x mx m

= + + + + = + + + + = + + + =2' 2 0 1 2 (*)m m m m D = - - > < - >

8i m tha mn (*) th d ct (1) ti ba Km A(0; 4), cn hai Km B,C c honh l hainghim ca phng trnh:

22 ' 2 02 2 0 1 2; 2

2 0

m mx mx m m m m

m

D = - - > + + + = < - > - +

- Ta c ( ) ( ) ( )1 1 2 2 2 1 2 1; 4 ; ; 4 ;B x x C x x BC x x x x+ + = - -JJJG

( ) ( )2 22 1 2 1 2 1 2BC x x x x x x = - + - = --Gi H l hnh chiu vung gc ca M trn d. h l khong cch t M n d th:

2 1 2 1

1 3 4 1 12 . 2. 2

2 22h S BC h x x x x

- + = = = = - = -

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33

- Theo gi thit: S = 4 2 22 1

4; 2 ' 4; 2 4 6 0x x m m m m - = D = - - = - - =

-t lun: vi m tha mn: 2 3 3m m m= - = = (chn ).

V d 7. Cho hm s ( ) ( )4 21 my x m x m C= - + + . Xc nh 1m > th ( )mC cttrc Ox ti 4 Km phn bit sao cho hnh phng gii hn bi ( )mC v trc Ox c din tchphn pha trn trc Ox bng din tch phn pha di trc Ox.Gii th hm s ct Ox ti 4 Km phn bit ( )4 21 0x m x m - + + = (1) c 4 nghimphn bit

( )2 1 0t m t m - + + = (2) c 2 nghim dng phn bit

( )21 4 01 0 0& 1

0

m m

m m m

m

D = + - > + > > >Hai nghim ca (2) l 1,t t m= = , do 1m > nn 4 nghim phn bit ca (1) theo th t

Vng l: , 1,1,m m- -

Hm s l chn nn hnh phng trong bi ton nhn Oy lmtrc i xng. Khi th c dng nh hnh bn.Bi ton tha mn

( ) ( )

( )( ) ( )( )

( )( )

1 2

14 2 4 2

0 11

4 2 4 2

0 1

4 2

0

1 1

1 1

1 0

H H

m

m

m

S S

x m x m dx x m x m dx

x m x m dx x m x m dx

x m x m dx

=

- + + = - + +

- + + = - - + +

- + + =

( )5 3

0

11 0 1 0 5

5 3 5 3

m

x x m mm mx m

+ - + + = - + = = .

KL: 5m = tha mn yu cu

V d 8. Gi ( )mC l th ca hm s ( )4 22 1 2 2y x m x m= - + + + . Tm mngthng 1y = Et ( )mC Vi bn Km phn bit A, B, C, D sao cho

4 2 2OA OB OC OD+ + + = +

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34

GiiXt phng trnh honh giao Km

( ) ( )4 2 4 22 1 2 2 1 2 1 2 1 0x m x m x m x m- + + + = - + + + =t ( )2 0t x t= , ta c phng trnh ( ) ( )2 2 1 2 1 0, *t m t m- + + + = c 4 giao Km phn bit th phng trnh (*) phi c 2 nghim dng phn bit

( ) ( )

( )

20 1 2 1 0 00 2 1 0 10 2 1 0 2

m m mP m

mS m

D > + - + > > + > > - > + > 8i Ku kin trn phng trnh (*) c hai nghim dng

1 2,t t . Theo Vi-et ta c,

( )1 2 1 22 1 , 2 1t t m t t m+ = + = +6 2 2

1 1 2 2;t x x t t x x t= = = =

t1 1 2 2, , ,

A B C Dx t x t x t x t= = - = = -

( ) ( ) ( ) ( )1 1 1 2;1 , ;1 , ;1 , ;1A t B t C t D t - -1 1

2 1 2 1OA OB OC OD t t + + + = + + +

Theo 1 22 1 2 1 4 2 2t t + + + = +

( )1 2

2

1 2

1 1 2 2

1 1 6 4 2

t t

t t

+ + + = +

+ + + = +

1 2 1 2 1 22 1 4 4 2t t t t t t + + + + + = +

( ) ( )2 1 2 2 1 2 1 1 4 4 24 4 1 2 2

m m m

m m

+ + + + + + = +

+ = + -

( )2

1 2 2 0

4 4 1 2 2

m

m m

+ - + = + -( )2

1 2 2

2 3 2 2 5 4 2 0

m

m m

+ - + + + =

1m =

8y Ku kin phi tm l 1m =V d 9. Cho hm s ( )4 22 1 2 1y x m x m= - + + + c th l ( )mC . nh m th( )mC ct trc honh ti 4 Km phn bit c honh lp thnh cp s cng.

Gii Xt phng trnh honh giao Km: ( )4 22 1 2 1 0x m x m- + + + = (1)

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35

t 2, 0t x t= th (1) tr thnh: ( )2( ) 2 1 2 1 0f t t m t m= - + + + = . (Cm) ct Ox ti 4 Km phn bit th ( ) 0f t = phi c 2 nghim dng phn bit

( )2' 0 1

2 1 0 202 1 0

mm

S mmP m

D = > > - = + > = + >

(*)

8i (*), gi1 2t t< l 2 nghim ca ( ) 0f t = , khi honh giao Km ca (Cm) vi Ox ln

Nt l: 1 2 2 1 3 1 4 2; ; ;x t x t x t x t= - = - = =

1 2 3 4, , ,x x x x lp thnh cp s cng

2 1 3 2 4 3 2 19x x x x x x t t - = - = - =

( ) ( )45 4 4

1 9 1 5 4 1 45 4 49

mm mm m m m m m

m m m

= = + + + = + - = + - = + = -

8y 44;9

m = -

V d 10.Tm mng thng 1y = - ct th (Cm) ti 4 Km phn bit u c honh nh hn 2.Gii Phng trnh honh giao Km ca (Cm) v ng thng 1y = - :

4 2n (3 2) 3 1x m x m+ + = - 4 2n (3 2) 3 1 0x m x m+ + + = 2 1

3 1 (*)

x

x m

= = +

ng thng 1y = - Et (Cm) ti 4 Km phn bit c honh nh hn 2 khi v ch khiphng trnh (*) c hai nghim phn bit khc r1 v nh hn 2

0 3 1 4

3 1 1

m

m

< +


36

Bi 2. Cho hm s 3 23 3 3 2y x mx x m= - - + + v ng thng : 5 1.d y x= - Tm mng thng d ct th (C) ti ba Km phn bit

a) c honh ln hn 1

b) c honh 1 2 3; ;x x x tha mn 2 2 2

1 2 315x x x+ + >

Bi 3. Cho hm s 3 23 ( 1) 1y x mx m x m= - + - + + v ng thng : 2 1.d y x m= - -Tm mng thng d ct th (C) ti ba Km phn bit c honh ln hn hocDng 1.

Bi 4. Cho hm s 3 2 2 23 3( 1) ( 1)y x mx m x m= - + - - -

Tm m th (C) ct Ox ti ba Km phn bit c honh dng.Bi 5. Cho hm s y = 2x3 3x2 1, c th l (C). Gi (dk) l ng thng i qua A(0; 1)v c h s gc bng k. Tm kng thng dk ct (C) ti

a) 3 Km phn bit.b) 3 Km phn bit, trong hai Km c honh dng.

Bi 6. Cho hm s y = x3 + 2mx2 + (m + 3)x + 4, c th (Cm).a) Kho st v v th ca hm s khi m = 1.b) Cho d l ng thng c phng trnh y = x + 4 v Km K(1 ; 3). Tm m d ct

(Cm) ti ba Km phn bit A(0 ; 4), B, Csao cho tam gic KBC c din tch bng 8 2 .

Bi 7. Cho hm s 3 22 3( 1) 2y x mx m x= + + - + (1), m l tham s thca) Kho st s bin thin v v th hm s khi 0m = .b) Tm m th hm s ct ng thng : 2y xD = - + ti 3 Km phn bit

(0;2)A ; B; C sao cho tam gic MBC c din tch 2 2 , vi (3;1).M

Bi 8. Cho hm s 3 26 9 3y x x x= + + + c th l (C) v hai Km A( 1;3), B(1; 1)- -

a) Kho st s bin thin v v th (C) ca hm s.b) Tm cc Km M thuc (C) sao cho tam gic ABM cn ti M

Bi 9. Cho hm s: 3 3 1y x x= - -a) Kho st s bin thin v v th (C) ca hm s.b) Vit phng trnh ng thng d Et (C) ti 3 Km phn bit A, M, N sao cho2

Ax = v 2 2MN =

Bi 10. Cho hm s 3 3 2.y x x= - +a)Kho st s bin thin v v th (C) ca hm s cho.b)Gi A, B l hai Km cc tr ca th (C). Tm ta cc Km M thuc (C) sao

cho tam gic MAB cn ti M.

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37

Bi 11. Cho hm s: 3 21 12 33 3

y x x x= - + -

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

b) Tm m ng thng 1:3

y mxD = - ct (C) ti ba Km phn bit A, B, C sao

cho A cnh v din tch tam gic OBC gp hai ln din tch tam gic OAB

Bi 12. Cho hm s 3 22 ( 3) 4y x mx m x= - + + + c th l (Cm).Tm m ng

thng (d): y = x + 4 ct (Cm) ti ba Km phn bit A(0; 4), B, C sao cho 2 2BCDSD =Xi D(1; 3).

Bi 13. Cho hm s ( ) ( )3 23 1 1 1y x x m x= - + + + c th ( )mC vi m l tham sa) Kho st s bin thin v v th (C) ca hm s (1) khi 1m = -b) Tm m ng thng ( ) : 1d y x= + ct th ( )mC ti 3 Km phn bit

( )0,1 , ,P M N sao cho bn knh ng trn ngoi tip tam gic OMN bng 5 22

vi ( )0;0O

Bi 14. Cho hm s: 3 23 (3 1) 6y x mx m x m= - + - + (C)

a) Kho st v v th (C) khi m=1b) Tm m th hm s (C) ct trc honh ti ba Km phn bit c honh

1 2 3, ,x x x tha mn Ku kin 2 2 2

1 2 3 1 2 320x x x x x x+ + + =

Bi 15. Cho hm s y = x3 + 3x2 + mx + 1 c th l (Cm); (m l tham s) Xc nh m (Cm) ct ng thng y = 1 ti ba Km phn bit C(0;1), D, E sao cho cc tip tuyn ca(Cm) ti D v E vung gc vi nhauBi 16. Cho hm s y = x3- (m+1)x2 + (m - 1)x + 1Chng t rng vi mi gi tr khc 0Ea m, th ca hm s ct trc honh ti ba Km phn bit A, B, C trong B, C chonh ph thuc tham s m. Tm gi tr ca m cc tip tuyn ti B, C song songXi nhau.

Bi 17. Cho hm s ( )4 22 1 2 1y x m x m= - + + + c th l (Cm), m l tham s. a) Kho st s bin thin v v th ca hm s khi m = 0. b) Tm m th (Cm) ct trc honh ti 3 Km phn bit u c honh nh hn 3.Bi 18. Cho y =x4 -2(m+1)x2 +2m+1Tm m th hm s ct trc honh ti 4 Km

phn bit c honh lp thnh cp s cng

Bi 19. Cho hm s: 2 32

xy

x+

=-

c th ( C ).

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

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38

b)Xc nh m ng thng (d): y x m= + ct th (C) ti hai Km phn bit

A, B sao cho tam gic OAB c din tch bng 2 3 (vi O l gc ta ).

Bi 20. (KB-2010) Cho hm s: y = 2 11

xx

++

a) Kho st s bin thin v v th ( C ).b) Tm m ng thng y = -2x + m ct th (C) ti hai Km phn bit A, B

sao cho tam gic OAB c din tch bng 3 (O l gc ta ).

Bi 21. Cho hm s ( )2 41x

y Cx

+=

-.

a) Kho st hm s b) Gi (d) l ng thng qua A( 1; 1 ) v c h s gc k. Tm k sao cho (d) ct ( C )

Vi hai Km M, N v 3 10MN = .

Bi 22. Cho hm s 2 11

xy

x-

=-

c th (C).

a) Kho st s bin thin v v th (C) ca hm s.b) Tm m ng thng y x m= + Et (C) ti hai Km A, B sao cho 4AB = .


xy

x+

=+

c th l (C). Chng minh ng thng d: y = -x + m

lun lun ct th (C) ti hai Km phn bit A, B. Tm m Qn AB c di nh nht.


xy

x+

=-

c th l (C).

a) Kho st s bin thin v v th hm s b) Tm m ng thng (d): y = mx+3 ct (C) ti hai Km phn bit A, B saocho tam gic OAB vung ti O.

Bi 25. Cho hm s 22 1x

yx-

=-

( C )

a) Kho st v v th ( C ) ca hm s.b) Tm m (dm) ct (C) ti hai Km phn bit thuc cng mt nhnh ca (C).

Bi 26. Cho hm s y = 2 42

xx

+-

(1). Tm m ng thng dm: y = mx + 2 - 2m ct

th ca hm s (1) ti hai Km phn bit thuc hai nhnh khc nhau.

Bi 27. Cho hm s:2

2

xy

x

+=

-.

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

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39

b) Chng minh rng vi mi gi tr m th trn (C) lun c cp Km A, B nm v hai

nhnh ca (C) tha mn0

0A A

B B

x y m

x y m

- + = - + =

Bi 28. Cho hm s y = 21

xx

+-

(C) v ng thng d: y = x+m ct th ( )C ti cc Km

A vB sao cho tam gic IAB nhn Km ( )4; 2H - lm trc tm. Vi I l giao Km cahai ng tim cn

Bi 29. Cho hm s2

x my

x- +

=+

(C). Tm s thc dng m ng thng

( ) : 2 2 1 0d x y+ - = ct (C) ti hai Km A v B sao cho tam gic OAB c din tch bng1 trong O l gc ta .


xy

x- +

=-

. Tm nhng Km trn (C) sao cho tip tuyn vi (C) ti

Km to vi hai trc ta mt tam gic c trng tm cch trc honh mt khong

Dng 53

.


xy

x+

=-

(1).Gi I l giao Km hai tim cn ca (C), ng thng

( ) : 2 5 0d x y- + = Et ( )C ti hai Km A, B vi A c honh dng. Vit phng trnhcc tip tuyn ca( )C vung gc vi IA.


xy

x-

=-

(C).

a) Kho st s bin thin v v th (C) ca hm s.b) Tm mng thng d: y x m= + ct (C) ti hai Km phn bit A, B sao cho

'OAB vung ti O.

Bi 33. Cho hm s 11

xy

x-

=+

.Tm a v b ng thng (d): y ax b= + ct (C) ti hai

Km phn bit i xng nhau qua ng thng (D ): 2 3 0x y- + = .

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40

4. Php bin i th4.1. Kin thc lin quan th cha du tr tuyt i

y = f(x) c th (C) ( )y f x= c th (C) ( )y f x= c th (C)( ) 0,y f x x D= " .

Ta c: y = f( x ) =

( ) 0.

( ) 0.

f x khi x

f x khi x

-


41

1

x

y

o 2

-4

1

.. .

.

..

.

.-1-2

1

x

y

o 2

-4

1

.. .

.

.

xy

y

0-f +f1 2

0+ +-

-f

+f1

0

%(1; 1) ; CT(2; 0) *) Bng bin thin

2) Da vo th (C) v th cc hm s:a) y = ~2x3 9x2 + 12x 4~ (t f(x) = 2x3 9x2 + 12x 4) Ta c y = ~2x3 9x2 + 12x 4~= ~f(x)~Do th hm s: y = ~2x3 9x2 + 12x 4~ gm:+) Phn t trc honh tr ln ca th hm s y = f(x).+) i xng phn th pha di trc honh ca thhm s y = f(x) qua trc honhb) y = 2~x~3 9x2 + 12~x~ 4

t f(x) = 2x3 9x2 + 12x 4)

Ta c: y = 2~x~3 9x2 + 12~x~ 4

= f( x ) =( ) 0.

( ) 0.

f x khi x

f x khi x

-


42

Do th hm s: y = f( x ) = 2~x~3 9x2 + 12~x~ 4 gm:

+) Phn bn phi Oy ca th hm s y = f(x).+) i xng phn th trn qua Oy

Vi du 2. Cho hm s 11

xy

x+

=- +

c th (C)

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

2. Bin lun theo m s nghim ca phng trnh1

.1

xm

x

+=

- +

Gii* 7p xc nh: D=R\{1}* 6 bin thin:

( )( ) ( )2

2' 0, ;1 1;

1y x

x= > " - +

-

Hm sng bin trn cc khong ( ) ( );1 v 1;+- .&c tr: Hm s khng c cc tr.Gii hn, tim cn:

1 1 1 1

1 1lim lim ; lim lim

1 1x x x xx x

y yx x- - + + + +

= = + = = -- + - +

Do ng thng x = 1 l tim cn ng.1 1

lim lim 1; lim lim 11 1x x x x

x xy y

x x- - + ++ +

= = - = = -- + - +

Do ng thng y = - 1 l tim cn ngang.%ng bin thin:

* th: th ct trc Oy ti Km (0; 1) v ct trc honh ti Km (-1; 0).

++-1

-1

1

-f

+f

+f-f

y

y'

x

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43

th c tm i xng l giao Km I(1; -1) ca hai tim cn.

b)Bin lun theo m s nghim ca phng trnh ( )1

. 11

xm

x

+=

- +

Np lun suy t th (C) sang th ( )1

' .1

xy C

x

+=

- +

5 nghim ca pt (1) bng s giao Km ca th1

1

xy

x

+=

- + v g thng y = m.

Suy ra p s: 1; 1 :m m< - > phng trnh c 2 nghim phn bit.1 :m = phng trnh c 1 nghim.

1 1 :m- < phng trnh v nghim.

Bi tp t luynBi 1 a) Kho st s bin thin v v th (C) ca hm s: y = x3 3x2 + 2

b) Bin lun theo m s nghim ca phng trnh: 2 2 21

mx x

x- - =

-

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Bi 2 Cho hm s: 3 23y x x= -a) Kho st s bin thin v v th (C) ca.

b) Bin lun theo m s nghim ca phng trnh x =2 3

m

x x-

Bi 3 Cho hm s y = (x+1)2(x-2). (C) a) Kho st v v th (C) ca hm s. b) Bin lun theo m s nghim ca phng trnh 22 ( 1)x x m- + =

Bi 4. a) Kho st v v th hm s 34 3y x x= -

b) Bin lun theo m s nghim ca phng trnh3

4 3x x m- =

Bi 5. a) V th hm s 3 23 6y x x= - - (C)

b) Bin lun theo m s nghim ca phng trnh 3 23 6x x m- - =

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45

CHUYN 2. PHNG TRNH, BT PHNG TRNH M LOGARITBin son v su tm: Nguyn Quang Tun GV Trng THPT Hn Thuyn

1. Kin thc cn nh1.1. Cng thc ly thaCho 0, 0a b> > v \,m n . Khi :

.m n m na a a += .( )m n m na a= ( ) .n n nab a b=

mm n

n

aa

a-=

mn m na a=

m m

m

a ab b

=

1 nn

aa

-=1n

na

a-=

n na bb a

- =

1.2. Cng thc lgarit8i cc Ku kin thch hp ta c:

logab a baa= = log 1 0

a=

log 1aa = log

aaa a=

loga ba b= log loga ab ba a=

1log log

aab ba a

= log logmn

aa

nb b

m=

log ( . ) log loga a a

m n m n= + log log loga a a

mm n

n= -

loglog

logc

ac

bb

a=

1log

logab

ba

=

2. Phng trnh v bt phng trnh m2.1. Phng php a v cng c s:Cho a l mt s dng khc 1. Ta c:

a) ( ) ( )( ) ( )f x g xa a f x g x= =

b) ( ) ( )0

logxf

a

ba b

f x b

>= =* Lu : Vi 0b < th phng trnh v nghim.

c) ( ) ( )(*)xf g xa a>

- Vi 1a > th ( ) ( ) ( )( )xf g x fa g xa x> >

- Vi 0 1a< < th ( ) ( ) ( )( )xf g x fa g xa x> - Vi 0,b bt phng trnh nghim ng vi mi ,x D D l tp xc nh ca

( )f x .- Vi 0 :b >

+ 1a > : ( ) ( ) logf x aa b f x b> >

+ 0 1a< < : ( ) ( ) logf x aa b f x b> < .Bi 1 (TN). Gii cc phng trnh sau:

2 3x)5 625xa + =2 3 6) 2 16x xb - - = 1) 2 .5 200x xc + =

/i gii2 23 3 4 2

2

)5 625 5 5 3 4

13 4 0

4

x x x xa x x

xx x

x

+ += = + = = + - = = -

8y phng trnh c nghim x = 1 v x = -4.2 23 6 3 6 4 2

2

) 2 16 2 2 3 6 4

53 10 0

2

x x x xb x x

xx x

x

- - - -= = - - = = - - = = -

8y phng trnh c nghim x = 5 v x = -2.1) 2 .5 200 2.2 .5 200

10 100 2

x x x x

x

c

x

+ = =

= =

8y phng trnh c nghim x = 2.Bi 2. (TN) Gii cc bt phng trnh sau:

26 3 7) 7 49x xa + -

2 7 23 9

)5 25

x x

b- + + >

/i gii2 26 3 7 6 3 7 2 2 2) 7 49 7 7 6 3 7 2 6 3 9 0x x x xa x x x x+ - + - + - + -

21

0 6 3 9 03

xVT x x

x

== + - = = -Xt du VT ta c tp nghim ca bt phng trnh S = [-3; 1].

2 27 2 7 2 2

2 23 9 3 3) 7 2 2 7 05 25 5 5

x x x x

b x x x x- + + - + + > > - + + < - +


47

6p nghim ca bt phng trnh ( ) ( ); 0 7;S = - +Bi 3 (H). Gii phng trnh:

2 2 23 10 4 22 4 2 16 0x x x x x x- - - - + ++ - - = ./i gii

Phng trnh tng ng:2 2 2 2 2 23 10 2 2 8 2 3 14 2 2 12 22 2 2 16 0 2 2 2 1 0x x x x x x x x x x x x- - - - + + - - - - + -+ - - = + - - =

2 2 22 2 12 2 2 2 12(2 1)(2 1) 0 2 1 0x x x x x x- - + - - - - + = - =

22 2 12 0 22

2 2 2 2 12 03

x xx

x xx

- - = - = - - = =

8y phng trnh c 2 nghim 2, 3.x x= - =BI TP T LUYNGii phng trnh, bt phng trnh:

1) \3 32 2 2 2 4 44 2 4 2 ( )x x x x x x x+ + + + + -+ = + 3) 1 2 1 22 2 2 3 3 3x x x x x x- - - -+ + = - +

2)5 177 332 0,25.128

x xx x

+ +- -= 4)

2 2 23 2 6 5 2 3 74 4 4 1x x x x x x- + + + + ++ = +

5)2

1

2

12

2

x

x x

-

- 6)

22 2( 10 3) ( 10 3)

xx x

-- ++ > -

7)2 25 6 1 6 52 2 2.2 1x x x x- + - -+ = + 8)

1 12 12 24 3 3 2

x xx x- + -- = -

9)1 2 1 29 9 9 4 4 4x x x x x x+ + + ++ + < + + 10) ( ) ( )

11

15 2 5 2x

xx

--

++ -

11)( )

15

2 14 102.0, 5 16 0

xx

++ - = 12)

13 3x2 4 .0,125 4 2x x =

2.2. Phng php t n ph:9 t , 0xt a t= > .9 Thay vo phng trnh hoc bt phng trnh bin i phng trnh theo t.9 Gii phng trnh, bt phng trnh tm t, i chiu Ku kin.9 0u c nghim tha th thay xt a= tm x v kt lun.

Bi 1. (TN) Gii cc phng trnh sau:) 9 10.3 9 0x xa - + = ) 25 3.5 10 0x xb + - =

3) 2 2 2 0x xc -- - = ) 6.9 13.6 6.4 0x x xd - + =

/i gii2) 9 10.3 9 0 3 10.3 9 0x x x xa - + = - + =

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48

t 3 , 0xt t= > .

Phng trnh tr thnh: 21 ( )

10 9 09 ( )

t nhant t

t nhan

=- + = =1 3 1 0

9 9 2

x

x

t x

t x x

= = =

= = =8y phng trnh c hai nghim x = 0 v x = 2.

2) 25 3.5 10 0 5 3.5 10 0x x x xb + - = + - =

t 5 , 0xt t= >

Phng trnh tr thnh: 22( )

3 10 05( )

t nhant t

t loai

=+ - = = -

52 5 2 log 2xt x= = =

8y phng trnh cho c nghim5

log 2x = .

3 28) 2 2 2 0 2 2 0 2 2.2 8 02

x x x x xx

c -- - = - - = - - =

t 2 , 0xt t= >

Phng trnh tr thnh: 24 ( )

2. 8 02 ( )

t nhant t

t loai

=- - = = -

4 2 4 2xt x= = =8y phng trnh c nghim x = 2.

29 6 3 3

) 6.9 13.6 6.4 0 6 13 6 0 6 13 6 04 4 2 2

x x x x

x x xd - + = - + = - + =

t 3 , 02

x

t t = >

. Phng trnh tr thnh 23

( )26 13 6 02

( )3

t nhant t

t nhan

=- + = =

3 3 31

2 2 2

2 3 21

3 2 3

x

x

t x

t x

= = = = = = -

8y phng trnh c nghim x = -1 v x = 1.

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49

Bi 2 (TN). Gii cc bt phng trnh:4 3.2 2 0x x- + gii$t phng trnh 24 3.2 2 0 2 3.2 2 0x x x x- + < - + /ppt 2 , 0xt t= /pp$t phng trnh tr thnh: 2 3 2 0t t- + < 1 2 1 2 2 0 1xt x< < < < < 8y bt phng trnh c nghim S = (0; 1)./ppBi 3 (H). Gii phng trnh: ( ) ( )3 3log log 2/pp10 1 10 13/ppx x x+ - - = ./pp/i giiKu kin: x 0/pp Ta c phng trinhg tng ng vi: ( ) ( )3 3 3log log/pplog210 1 10 1 .33/ppx xx+ - - =/pp3 3log log/pp10 1 10 1 23 3 3/ppx x + - - = . t/pp3log/pp10 13/ppx/ppt + = /pp(t > 0).

Phng trnh tr thnh: 21 2 3 2 3 03

t t tt

- = - - =

1 103

1 103

t

t

+ = - =

(loi)

8i t = 1 103

+ ta gii c x = 3

Vy phng trnh cho c nghim duy nht x =3.Bi 4 (H). Gii cc bt phng trnh

a)2

22log 2 log2 20 0

x xx+ -

b)2 22 1 2 1 4(2 3) (2 3)

2 3x x x x- + - -+ + -

-.i gii

Ku kin: x> 0 ; BPT 22 24 log 2 log2 20 0x xx+ - t

2logt x= . Khi 2tx = .

BPT tr thnh2 22 24 2 20 0t t+ - . t

222 ty = ; y t 1.BPT tr thnh y2 + y - 20 d 0 - 5 d y d 4.i chiu Ku kin ta c:

22 2 22 4 2 2 1t t t - 1 d t d 1.

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50

Do - 1 d2

log x d 1 1 22

x .

b) Bpt ( ) ( )2 22 2

2 3 2 3 4x x x x- -

+ + -

t ( )2 2

2 3 ( 0)x x

t t-

= + >

BPTTT: 21 4 4 1 0 2 3 2 3t t t tt

+ - + - + (tm)

( )2 2

2 3 2 3 2 3x x-

- + + 21 2 1x x - -

2 2 1 0 1 2 1 2x x x- - - +BI TP T LUYNGii phng trnh v bt phng trnh:

1)2 25 1 54 12.2 8 0x x x x- - - - -- + = 2) ( ) ( )7 4 3 3 2 3 2 0

x x

+ - - + =

3) 3(3 5) 16.(3 5) 2x x x++ + - = 4)2 2 22 1 2 1 225 9 34.15x x x x x x- + - + -+ =

5) 1 1 14 5.2 16 0x x x x+ - + - +- + 6) 1 2 1 23 2 12 0x

x x+ +- - 1 th log ( ) log ( ) 0 ( ) ( )a af x g x f x g x< < < .

* Nu 0 < a < 1 th log ( ) log ( ) ( ) ( ) 0a af x g x f x g x< > >

Bi 1.(TN) Gii cc phng trnh sau

2 4 8) log log log 11a x x x+ + =

5 25 0,2

1) log log log

3b x x+ =

22 2) log log 6 0c x x- - =

22 2

) 4 log log 2d x x+ =

23 3) 3 log 10 log 3e x x= -

2) ln( 6 7) ln( 3)f x x x- + = -

/i gii

2 4 8) log log log 11a x x x+ + = (1)

Ku kin: x > 0.

2 32 2 2(1) log log log 11x x x + + =

2 2 2

2

62

1 1log log log 11

2 311

log 116log 6 2 64 ( )

x x x

x

x x nhan

+ + =

=

= = =

8y phng trnh c nghim x = 64.

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52

5 25 0,2

1) log log log

3b x x+ = (2)

Ku kin: x > 0.

( )2 11

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

-

+ =5 5 5

1log log log 3

2x x + =

( )5 5 5 5

23 3

5 5 5 5

3

3 2log log 3 log log 3

2 3

log log 3 log log 3

3

x x

x x

x

= =

= =

=

8y phng trnh c nghim 3 3x = .22 2) log log 6 0c x x- - = (3)

Ku kin: x > 0.

t2

logt x= . PT (3) tr thnh 23

6 02

tt t

t

=- - = =9 3

23 log 3 2 8t x x= = = = (tha mn)

9 22

2 log 2 2 4t x x= = = = (tha mn)

8y phng trnh c nghim x = 4 v x = 8.22 2

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

Ku kin x > 0.

1

2

2 22 2 2

2

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

t2

logt x= . PT (4) tr thnh 21

4 2 2 0 12

tt t

t

= -+ - = =

9 12

11 log 1 2 ( / )

2t x x t m-= - = - = =

91

2

2

1 1log 2 2 ( / )

2 2t x x t m= = = =

8y phng trnh c nghim 12

x = v 2x =

23 3) 3 log 10 log 3e x x= - (5)

Ku kin x > 0

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53

t3

logt x= ta c 2 23

3 10 3 3 10 3 0 13

tt t t t

t

== - - + = =

9 333 log 3 3 27 ( )t x x nhan= = = =

91

333

1 1log 3 3 ( )

3 3t x x nhan= = = =

8y phng trnh c hai nghim x = 27 v 3 3x = .2) ln( 6 7) ln( 3)f x x x- + = - (6)

Ku kin2 6 7 0

3 0

x x

x

- + > - >

2 22 ( )

(6) 6 7 3 7 10 05 ( )

x loaix x x x x

x nhan

= - + = - - + = =8y phng trnh c nghim x = 5.Bi 2. (TN) Gii cc bt phng trnh sau:

3) log (4 3) 2a x - < 2

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

21 13 3

) log (2 4) log ( 6)c x x x+ - - 2) lg(7 1) lg(10 11 1)d x x x+ - +

/i gii

3) log (4 3) 2a x - >

23

log (4 3) 2 4 3 3 4 12 3x x x x- < - < <

( ) 12 2 20,5log ( 5 6) 1 5 6 0,5 5 4 0 1 4x x x x x x x-

- + - - + - +

-t hp Ku kin bt phng trnh c nghim ) (1;2 3;4S = 2

1 13 3

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

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Ku kin: 2

22 4 0

2 36 0

3

xx

x xx x

x

> - + > < - > - - > >2 2

1 13 3

2

log (2 4) log ( 6) 2 4 6

3 10 0 2 5

x x x x x x

x x x

+ - - + - -

- - -

-t hp vi Ku kin, bt phng trnh c nghim (3;5S = 2) lg(7 1) lg(10 11 1)d x x x+ - +

Ku kin: ( )2

177 1 0 1 1

1 ; 1;10 11 1 0 7 10

101

xx

xxx x

x

> - + > - + < - + > >2 2

2

lg(7 1) lg(10 11 1) 7 1 10 11 1

910 18 0 0

5

x x x x x x

x x x

+ - + + - +

-

-t hp Ku kin, bt phng trnh c nghim 1 90; 1;10 5

S =

Bi 3 (H). Gii cc phng trnh:

a) ( ) ( )23 3log 1 log 2 1 2x x- + - =

b) ( ) ( )2 34 82log 1 2 log 4 log 4x x x+ + = - + + (2)/i gii

Ku kin: 1 12

x<

( )3 32 log 1 2 log 2 1 2x x - + - = ( )3 3log 1 log 2 1 1x x - + - =

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

2

11

22 3 4 0( )

1

2 3 2 0

x

x x vn

x

x x

< - - =

2x =

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b) Ku kin:1 0

4 44 0

14 0

xx

xx

x

+ - < - + >

( ) ( ) ( )( )

22 2 2 2 2

2 22 2

(2) log 1 2 log 4 log 4 log 1 2 log 16

log 4 1 log 16 4 1 16

x x x x x

x x x x

+ + = - + + + + = -

+ = - + = -

+ Vi 1 4x- < < ta c phng trnh 2 4 12 0 (3)x x+ - = ;

( )loi2

(3)6

x

x

= = -

+ Vi 4 1x- < < - ta c phng trnh 2 4 20 0x x- - = (4);

( )( )loi

2 244

2 24

x

x

= - = +

8y phng trnh cho c hai nghim l 2x = hoc ( )2 1 6x = - .

Bi 4 (H). Gii cc bt phng trnh:

a) ( )23 1 13 3

1log 5 6 log 2 log 3

2x x x x- + + - > +

b) ( ) ( )2 21 5 3 13 5

log log 1 log log 1x x x x+ + > + -

/i giiKu kin: 3x >

$t phng trnh cho tng ng:

( ) ( ) ( )1 123 3 31 1 1

log 5 6 log 2 log 32 2 2

x x x x- -- + + - > +

( ) ( ) ( )23 3 31 1 1

log 5 6 log 2 log 32 2 2

x x x x - + - - > - +

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

( )( )3 32

log 2 3 log3

xx x

x

- - - > + ( )( ) 22 3

3x

x xx

- - - >

+

210

9 110

xx

x

< - - > >

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Giao vi Ku kin, ta c nghim ca phng trnh cho l 10x > .b) k: 0x >

( ) ( ) ( )( ) ( )

( )

2 23 1 3 5

5

2 23 1 5

5

2 25

1 log log 1 log log 1 0

log log 1 . log 1 0

log 1 1

x x x x

x x x x

x x

+ - + + +

Phng trnh ( )33 3

1 42 log 1

log 9 1 logx

x x - - =

-3

3 3

2 log 41

2 log 1 log

x

x x

- - =

+ -

t:3

logt x= , thnh 22 4 1 3 4 02 1

tt t

t t-

- = - - =+ -

(v t = -2, t = 1 khng l

nghim)1 4t hay t = - =

Do , (1)3

1log 1 4 81

3x hay x x hay x = - = = = .

b) 2 316 42

log 14 log 40. log 0x x xx x x- + = (1)

k: 0, 1/ 4, 1 / 16, 2x x x x> (*)

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Khi , phng trnh tng ng vi16 4

2

2. log 42. log 20.log 0x x xx x x- + = (2)

Nhn thy x =1 lun l nghim ca pt.

8i 0 < x 1, pt (2) 2 42 20 0log 16 log 4

log2

x xx

x x x- + =

t t = logx2, phng trnh trn tr thnh2 42 20

01 1 4 1 2t t t

- + =- + +

(3)

(3) 2t2 + 3t 2 = 0 t = 1/2 hoc t = -2(tmk)

* Vi t = -2 th logx2 = -2 1

2x =

* Vi t = 1/2 th logx2 = 1/2 x = 4.

-t hp k ta c nghim ca phng trnh l x = 4, x = 12

Bi 3 (H). Gii bt phng trnh3

4 2 22 1 2 12

2 2

32log log 9 log 4 log

8x

x xx

- + 0.

$t phng trnh2

4 3 2 22 2 2 2 2 2

log ( ) log log 8 9 log 32 log 4 log ( )x x x x - - + -


59

7)5

2 log log 125 1x

x - < 8)2 1

2

1 21

5 log 1 logx x+ > suy ra 2n 3 2.v u x x= - +

PT cho tr thnh3 3 33 3

lolo g log log log gu vu

v u v u u u vv

v- = - += =- +

(1). Xt hm c trng: ( ) 3log , 0f t t t t= + > .

Ta c ' 1( ) 1 0, 0. ln 3

f t tt

= + > " > nn hm sng bin khi t > 0.

6 (1) ta c f(u) = f(v), suy ra u = v hay v-u=0, tc l x2-3x+2=0.Phng trnh c nghim 1, 2x x= = .

.u : Vi phng trnh dng loga

uv u

v= - Xi u > 0, v > 0 v 1 < a, ta thng bin i

logau - logav = v u logau + u = logav. V hm s f(t) = logat + t ng bin khi t > 0, suy ra u = v.

Bi 2 (H). Gii bt phng trnh 22 1

2

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

2x x x x x

- + - > - + -

/i gii

K: ( )2 2

11 10 12 *2 2 1 24 4 1 0 (2 1) 0

2

xx xx

x x x x

8i Ku kin (*) bt phng trnh tng ng vi:

2 22 log (1 2 ) 2 2 ( 2) log (1 2 ) 1x x x x - - > + + - - 2log (1 2 ) 1 0x x

- + > > - + < - < - < > < < - >

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-t hp vi Ku kin (*) ta c: 1 14 2

x< < hoc x < 0.

Bi 3 (H). Gii bt phng trnh2 2

(3 log 2) 9 log 2x x x- > -

/i giiKu kin: 0x > $t phng trnh 23( 3)log 2( 1)x x x- > -

Nhn thy x=3 khng l nghim ca bt phng trnh.

TH1 Nu 3x > BPT 23 1

log2 3

xx

x-

>-

Xt hm s:2

3( ) log

2f x x= ng bin trn khong ( )0;+ 1( )

3x

g xx

-=

- nghch bin

trn khong ( )3;+ *Vi 4x > :Ta c( ) (4) 3

( ) (4) 3

f x f

g x g

> = < = Bpt c nghim 4x > * Vi

4x < :Ta c( ) (4) 3

( ) (4) 3

f x f

g x g

< = > = Bpt v nghim

TH 2:Nu 0 3x< < BPT 23 1

log2 3

xx

x-

:Ta c

( ) (1) 0

( ) (1) 0

f x f

g x g

> = < = Bpt v

nghim

* Vi 1x < :Ta c( ) (1) 0

( ) (1) 0

f x f

g x g

< = > = Bpt c nghim 0 1x< < Vy Bpt c ngh

4

0 1

x

x

> <


61

CHUYN 3. PHNG TRNH LNG GICBin son v su tm: Nguyn Minh Nhin S GD&T

1. Phng trnh a v dng tch1.1. Phng trnh s dng cc cng thc bin i lng gic: cng thc bin tch thnh tng, tngthnh tch, cng thc h bc,V d 1. Gii phng trnh: sinx+sin2x+sin3x+sin4x+sin5x+sin6x=0 (1)Gii

( ) ( ) ( ) ( )

( )

1 sin 6 sin sin 5 sin 2 sin 4 sin 3 0

7 5 3 7 3 2 sin cos cos cos 0 4 sin cos 2 cos 1 0

2 2 2 2 2 2

27sin 0

723 2

cos 0 ;2 3 3

2 cos 1 0 22

3

x x x x x x

x x x x x xx

kxx

x kx k Z

xx k

p

p p

pp

+ + + + + = + + = + =

== = = + + =

= +

*.u : Khi ghp cp ra tng ( hoc hiu ) sin ( hoc cos ) cn n gc sao cho tng hochiu cc gc bng nhau

V d 2. Gii phng trnh: 3 32 3 2

cos 3 cos sin 3 sin8

x x x x-

- = (2)

Gii

( ) ( ) ( )2 21 1 2 3 22 cos cos 4 cos2 sin cos2 cos 42 2 8

x x x x x x-

+ - - =

( ) ( )

( ) ( )

2 2 2 2 22 3 2 2 3 2cos 4 cos sin cos 2 cos sin cos 4 cos 24 4

24 cos 4 2 1 cos 4 2 3 2 cos 4

2 16 2

x x x x x x x x

kx x x x k Z

p p

- - + + - = + =

+ + = - = = +

*.u : Vic kho lo s dng cng thc bin tch thnh tng c th gip ta trnh c vic sFng cng thc nhn 3

V d 3. Gii phng trnh : 2 22 cos 2 3 cos 4 4 cos 14

x x xp - + = -

(3)

Gii

( ) ( )2 23 1 cos 4 3 cos 4 4 cos 1 sin 4 3 cos 4 2 2 cos 12

x x x x x xp + - + = - + = -

1 3sin 4 cos 4 cos2 cos 4 cos2

2 2 6x x x x x

p + = - =

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62

] ,12 36 3

kx k x k

p p pp = + = +

1.2. Phng trnh s dng mt s bin i khcVic a phng trnh v dng tch Ku quan trng nht vn l lm sao pht hin ra

nhn t chung nhanh nht, sau y l mt s bin i c th gip ta lm c Ku

( )( ) ( )( )( )( )

( )( )

2 2

2

2

sin 1 cos 1 cos , cos 1 sin 1 sin

cos2 cos sin cos sin1 cos 2 sin 2 2 cos (sin cos )

1 sin 2 sin cos1 cos 2 sin 2 2 sin (sin cos )

1 sin 2 sin cos

sin cos1 tan

cos

2 sin

x x x x x x

x x x x xx x x x x

x x xx x x x x

x x x

x xx

x

x

= - + = - +

= - ++ + = +

+ = +- + = +

- = -

+ + =

sin cos4

x xp + = +

V d 4. Gii phng trnh: 2 sin (1 cos2 ) sin2 1 2 cosx x x x+ + = + (4)

GiiCch 1: ( ) ( )( )24 2 sin .2 cos 2 sin cos 1 2 cos 2 cos 1 2 sin cos 1 0x x x x x x x x + = + + - =

1cos

2sin 2 1

x

x

= - =

phn cn li dnh cho bn c

Cch 2: ( )4 2 sin cos 2 (1 sin2 ) 2(cos sin ) 0x x x x x - - - - =

( )( ) ( ) ( )22sin x cos sin cos sin cos sin 2 cos sin 0x x x x x x x x - + - - - - =( )( )2cos sin 2 sin x cos 2 sin cos sin 2 0x x x x x x - + - + - =( )( )2cos sin 2 sin cos 2 cos cos sin 0x x x x x x x - - - + = (phn cn li HS t

lm)V d 5.Gii phng trnh: cos2 3 sin2 5 sin 3 cos 3x x x x+ + - = (5)Gii

( ) 25 (6 sin cos 3 cos ) (2 sin 5 sin 2) 0 3 cos (2 sin 1) (2 sin 1)(sin 2) 0

(2 sin 1)(3 cos sin 2) 0

x x x x x

x x x x

x x x

- - - + = - - - - = - - + =

Phng trnh ny tng ng vi 2 phng trnh c bn (HS t lm)2. Phng trnh cha n mu

8i loi phng trnh ny khi gii rt d dn n tha hoc thiu nghim, Ku quan trngnht ca dng ny l t Ku kin v kim tra Ku kin xc nh.Thng thng ta hay dng

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63

ng trn lng gic loi nghim. Ngoi ra, ta cng gp nhiu phng trnh cha tan, cot. Khi, c th s dng mt s cng thc

( ) ( )

( ) ( )

sin sintan tan cota cotb=

cos cos cos cosos os

tan cot tana-cotb=cos sin cos sin

2tan cot c

sin 2

a b b aa b

a b a bc a b c a b

a ba b a b

a aa

=

- - + + =

+ =

( ) ( )ot tan 2 cot2

os os1 tan tan 1 tana tan

cos cos cos cos

a a a

c a b c a ba b b

a b a b

- =

- - + + = - =

%n lu cc Ku kin xc nh ca tng cng thc

V d 6. Gii phng trnh:2 cos 4

cot tansin 2

xx x

x= + (6)

Gii.

K:sin 0

cos 0 sin2 0 ,2

sin 2 0

xk

x x x k Z

x

p

( ) 2 os4 2 cos2 2 os46 cot tan os4 os2 ,sin 2 sin2 sin 2

3

x lc x x c x

x x c x c x l Zlx x x x

pp

= - = = = =

Kim tra Ku kin ta c ,3

x l l Zp

p= +

V d 7. Gii phng trnh:( ) ( )3 2

2

4 os 2 os 2 sin 1 sin 2 2 sin cos0

2 sin 1

c x c x x x x x

x

+ - - - +=

- (7)

Gii.

K: 22 sin 1 0 os2 0 ,4 2

kx c x x k Z

p p- +

( ) ( ) ( ) ( )

( )( )( )

27 4 os sin cos 2 cos sin cos 2 sin cos 0

42 sin cos cos 1 2 cos 1 0 2 ,

22

3

c x x x x x x x x

x m

x x x x x m m Z

x m

pp

pp

p

+ - + - + = = - +

+ - + = = = +

Kim tra Ku kin ta c nghim2

,3

mx m Z

p=

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64

V d 8. Gii phng trnh:2

3 tan 3 cot2 2 tansin 4

x x xx

+ = + (8)

Gii

K:

os3x 0

sin2x 0 6 3 ,cos 0

4sin 4 0

ck

xk Z

x kx

x

p p

p

+

(*)

( ) ( ) ( )

( )

2 2 sin2 cos 28 2 tan 3 tan tan 3 cot2

sin 4 os3 cos os3 sin 2 sin 44 sin 4 sin 2 os2 cos 2 os3 4 sin 4 sin os3 cos 2 os3

4 sin 4 sin os3 cos 8 sin 2 os2 sin 2 sin 2 sin (*)

os2

x xx x x x

x c x x c x x xx x c x x c x x x c x x c x

x x c x x xc x x x x do

c x

- + + = + =

+ = + + =

= - = -

=1 1 1

arccos ,4 2 4

x m m Zp - - = +

Nghim ny tha mn KBI TP T LUYN

2

1) os3 os2 cos 1 0 2) 2 2 sin cos 112

1 13)(1 tan )(1 sin 2 ) 1 tan 4)sin 2 sin 2 cot2

sin 2 2 sin5)sin 2 os2 3 sin cos 2 0

6) tan cos cos sin

c x c x x x x

x x x x x xx x

x c x x x

x x x x

p + - - = - =

- + = + + - - =

+ + - - =

+ - =

( )3

3 3

1 tan tan2

2 cos sin17)2 2 os 3 cos sin 0 8)

4 tan cot2 cot 11

9)cos cos2 os3 sin sin 2 sin 32

10)sin os os2 tan tan4 4

11) tan

xx

x xc x x x

x x x

x xc x x x x

x c x c x x x

p

p p

+ - - - - = = + -

+ =

- = + -

( )

2 2

2 2

22 3 3 2

tan2 sin 3 cos2

712)sin cos 4 sin 2 4 sin

4 2 2

13)sin sin cos sin 1 2 cos2 2 4 2

14)2 sin cot 2 sin 2 1

sin 315)sin cos 3 sin sin 3 cos sin sin 3

3 sin 4

x x x x

xx x x

x x xx x

x x x

xx x x x x x x

x

p

p

+ = - - = - -

- + = - + = +

+ + =

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65

CHUYN 4: S PHCBin son v su tm: Ng Vn Khnh GV trng THPT Nguyn Vn C

1. Kin thc c bn.1.1. Cc khi nim

1.2. Cc php ton trn s phc.* Php cng v php tr, nhn hai s phc.

Cho hai s phc z = a + bi v z = a + bi. Ta nh ngha:' ( ') ( ')

' ( ') ( ')

z z a a b b i

z z a a b b i

+ = + + + - = - + -' ' ' ( ' ' )zz aa bb ab a b i= - + -

* Php chia s phc khc 0.Cho s phc z = a + bi 0 (tc l a2+b2 > 0 )Ta nh ngha s nghch o z-1 ca s phc z 0 l s

z-1=2 2 2

1 1z z

a b z=

+

Thng 'zzEa php chia s phc z cho s phc z 0 c xc nh nh sau:

1

2

' ' ..

z z zz z

z z

-= =

2. Cc dng bi tp.2.1. Dng 1: Cc php ton trn s phc.

V d 1: Cho s phc z = . Tnh cc s phc sau: ; z2; ( )3; 1 + z + z23 12 2

i z z

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66

Gii:

*V z = =

*Ta c z2 = = =

( )2 =

( )3 =( )2. =

Ta c: 1 + z + z2 =

V d 2: Tm s phc lin hp ca:

Gii:

Ta c .

Suy ra s phc lin hp ca z l:

V d 3: Tm phn o ca s phc z bit ( ) ( )2

2 1 2z i i= + -

Gii:

( )( )1 2 2 1 2 5 2z i i i= + - = + . Suy ra, 5 2z i= - Phn o ca s phc 2z = -

V d 4: Tm m un ca s phc

Gii: Ta c:

8y m un ca z bng:

V d 5: Cho s phc z tha mn( )

3

1 3

1

iz

i

-=

-. Tm mun ca s phc .z iz+

Gii:

3 12 2

i z 3 12 2

i2

3 12 2

i

23 1 34 4 2

i i 1 32 2

i

z2

23 1 3 1 3 1 32 2 4 4 2 2 2

i i i i

z z z1 3 3 1 3 1 3 32 2 2 2 4 2 4 4

i i i i i

3 1 1 3 3 3 1 312 2 2 2 2 2

i i i

1(1 )(3 2 )3

z i ii

3 35 5(3 )(3 ) 10

i iz i ii i

53 910 10

z i

(1 )(2 )1 2i iz

i

5 115 5

iz i

21 2615 5

z

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67

Ta c: ( )3

1 3 8i- = - Do 8 4 4 4 41

z i z ii

-= = - - = - +

-

( )4 4 4 4 8 8z iz i i i i + = - - + - + = - - Vy 8 2.z iz+ =

V d 6: Tm cc s thc ,x y tha mn ng thc:

a) 3x + y + 5xi = 2y 1 +(x y)ib) (2x + 3y + 1) + ( x + 2y)i = (3x 2y + 2) + (4x y 3) i.

c) ( ) ( )33 5 1 2 35 23x i y i i+ + - = - +Gii: a) Theo gi thit: 3x + y + 5xi = 2y 1 +(x y)i (3x + y) + (5x)i = (2y 1) +(x y)i

b) Theo gi thit ta c:

92 3 1 3 2 2 5 1 11

2 4 3 5 3 3 411

xx y x y x y

x y x y x yy

= + + = - + - + = - + = - - - + = - =

c) Ta c ( ) ( ) ( ) ( )( )3 21 2 1 2 1 2 3 4 1 2 2 11i i i i i i- = - - = - - - = - .

Suy ra ( ) ( )33 5 1 2 35 23x i y i i+ + - = - + ( ) ( )3 5 2 11 35 23x i y i i + + - = - +

( ) ( )3 11 35 3

3 11 5 2 35 235 2 23 4

x y xx y x y i i

x y y

- = - = - + + = - + + = = Bi tp t luyn

Bi 1. Tm cc s thc x, y bit:a) (3x 2) + (2y +1)i = (x + 1) (y 5)i;b) (2x + y) + (2y x)i = (x 2y + 3) + (y + 2x +1)i;

Bi 2. Chng minh z = (1+2i)(2 - 3i)(2+i) (3-2i ) l mt s thc

Bi 3. Tm cc s thc x, y tha mn ng thc: 3(3 2 ) (1 2 ) 11 42 3

x iy i i

i-

+ - = ++

Bi 4. Cho hai s phc:1 2

z 2 5 ; z 3 4i i= + = - . Xc nh phn thc, phn o ca s

phc1 2.z z

Bi 5. Tm phn thc, phn o v m un ca s phc:a) z (2 3 )(1 ) 4i i i= + - - b) 3(2 2 )(3 2 )(5 4 ) (2 3 )z i i i i= - + - - +

3 2 15

x y yx x y

17

47

x

y

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68

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

d) z = 2 5(1 3 )( 2 )(1 )

ii i i

- ++ - - +

e) z = (1 2 )( 4 )(1 )(4 3 )

i ii i

+ - +- +

Bi 6. Tm cc s phc: 2z z+ v 25iz

, bit z 3 4i= - .

Bi 7. Cho s phc z = 2 + 3i.Tm phn thc v phn o ca s phc1

z iw

iz+

=-

Bi 8. Cho s phc 1 7 (3 2 )( 1 3 )1 2

iz i i

i+

= + - - ++

Tnh m un ca z v tm ta

Km biu din hnh hc ca z trong h ta Oxy.

Bi 9. Cho z tha mn (2 + i)z + 2(1 2 ) 7 81

ii

i+

= ++

. Tm mun ca s phc w = z + 1

+ iBi 10. 5 phc z tha mn (1+i)2(2i)z=8+i+(1+2i)z. Tm phn thc, phn o ca z.

Bi 11. Cho s phc z tha mn ( ) ( )21 2 31

ii z i z

i-

- - = -+

.Tm ta Km biu din

Ea z trong mt phng ta Oxy.Bi 12. Tim s phc z bit z3 = 18 + 26i, trong z = x + yi (x,y Z)

2.2. Dng 2: Tinh ni v p dng

Ch : i4n = 1; i4n+1 = i; i4n+2 = -1; i4n+3 = -i; n N*8y in {-1;1;-i;i}, n N*

2(1 ) 2i i+ = ;( )21 2i i- = - D ;

3D

V d 1: Tnh: i105 + i23 + i20 i34

Gii:Ta c i105 + i23 + i20 i34 = i4.26+1 + i4.5+3 + i4.5 i4.8+2 = i i + 1 + 1 = 2

V d 2: Tnh s phc sau:

a) z = (1+i)15 b) z =16 8

1 11 1

i ii i

+ - + - +

Gii:

a) Ta c: (1 + i)2 = 1 + 2i 1 = 2i (1 + i)14 = (2i)7 = 128.i7 = -128.i

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69

nn z = (1+i)15 = (1+i)14(1+i) = -128i (1+i) = -128 (-1 + i) = 128 128i.

b) Ta c: 1 (1 )(1 ) 21 2 2

i i i ii

i+ + +

= = =-

11

ii

i-

= -+

. Vy16 8

1 11 1

i ii i

+ - + - + =i16 +(-i)8 = 2

V d 3: Tm phn thc, phn o ca s phc sau:

( ) ( ) ( ) ( )2 3 201 1 1 1 ... 1i i i i+ + + + + + + + +Gii:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( ) ( )

21

2 20

2021 2 10 10

10

10 10

1 11 1 1 ... 1

1 1 1 2 1 2 1

2 1 12 2 1

iP i i i

i

i i i i i i

iP i

i

+ -= + + + + + + + =

+ = + + = + = - +

- + - = = - + +

8y phn thc l 102- v phn o l 102 1+ Bi tp t luynBi 1. Tm phn thc, phn o ca cc s phc sau:

z = ( ) ( )( )2

101 11 2 3 2 3

1i

i i ii i

+ + - + + - + -

Bi 2. Tm phn thc v phn o ca s phc z tha mn: ( )( ) 20112 3 1 (1 )z i i i+ - - = + .Bi 3. Tm phn thc, phn o ca s phc z = 19(1 )i+

2.3. Dng 3: Tim s phc da vo Dng i s ca s phc.

0u trong h thc tim s phc z xut hin 2 hay nhiu i lng sau: , , ,...z z z ta se

U dng Dng i s ca z l z x yi= + vi ,x y R

V d 1: Tm s phc z bit ( )2 3 1 9z i z i- + = -Gii:)i z= a+ bi (a,b R ) ta c:

( ) ( )( )2 3 1 9 2 3 1 9z i z i a bi i a bi i- + = - + - + - = -

( )3 1 2

3 3 3 1 93 3 9 1

a b aa b a b i i

a b b

- - = = - - - - = - - = = -

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8y z= 2-i

V d 2: Tnh m un ca s phc z bit rng: ( )( ) ( )( )2 1 1 1 1 2 2z i z i i- + + + - = -Gii:)i z= a+ bi (a, b R )

Ta c

( )( ) ( )( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( )

2 1 1 1 1 2 2

2 1 2 1 1 1 2 2

2 2 1 2 2 1 1 1 2 2

13 3 2 33 3 2 2 2

2 2 13

z i z i i

a bi i a bi i i

a b a b i a b a b i i

aa ba b a b i i

a bb

- + + + - = -

- + + + + - - = - - - + + - + - + - + + = -

= - = - + + - = - + - = - = -

Suy ra m un: 2 2 23

z a b= + =

V d 3: Tm s phc z tha mn:22

2 . 8z z z z+ + = v 2z z+ = .

Gii

)i z = x + iy (x, yR), ta c22 2 2;z x iy z z zz x y= - = = = +

22 2 2 2 22 . 8 4( ) 8 ( ) 2 (1)z z z z x y x y+ + = + = + =

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

6 (1) v (2) tm c x = 1 ; y = 18y cc s phc cn tm l 1 + i v 1 - i

V d 4: Tm s phc z tha mn hai Ku kin: 1 2 3 4z i z i+ - = + + v 2z iz i

-

+l mt s

thun o.Giit z= x+ yi (x,y R )

Theo bi ra ta c

( ) ( )( ) ( ) ( ) ( )2 2 2 21 2 3 4

1 2 3 4 5

x y i x y i

x y x y y x

+ + - = + + -

+ + - = + + - = +

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5 phc( )( )

( )( ) ( )( )

2

22

2 2 1 2 32w

1 1

x y i x y y x y iz i

x y iz i x y

+ - - - - + --= = =

+ -+ + -

w l mt so khi v ch khi

( )( )( )

2

22

2 1 0 1271 0

235

7

x y yx

x yyy x

- - - = = - + - > == +

8y 12 237 7

z i= - +

V d 5: Tm tt c cc s phc z bit22z z z= +

Gii: Gi z= a+ bi (a, b R ) ta c:

( )

( )

2 22 2 2

2 2 2 2

22 2 2 2

2

02 1 1

;2 2 1 0 2 2

1 1;

2 2

z z z a bi a b a bi

a b abi a b a bi

a ba ba b a b a

a bab b b a

a b

+ + + = + + -

- + = + + - = = = -- = + + = - = = - + = - = - =

8y z=0; 1 1 1 1;2 2 2 2

z i z i= - + = - -

V d 6: Tm s phc z tha mn 2z = v z2 l s thun o.

Gii:

Gi z= a+ bi (a, b R ) Ta c 2 2z a b= + v 2 2 2 2z a b abi= - +

Yu cu bi ton tha mn khi v ch khi2 2 2

2 2 2

2 1 1

10 1

a b a a

ba b b

+ = = = = - = = 8y cc s phc cn tm l 1+i; 1-i; -1+i; -1-i

V d 7: Tm s phc z bit 5 3 1 0izz

+- - =

Gii: Gi z= a+ bi (a, b R ) v 2 2 0a b+ ta c

2 25 3 5 31 0 1 0 5 3 0i i

z a bi a b i a biz a bi

+ +- - = - - - = + - - - - =

+

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( ) ( )2 2

2 25 0

5 3 03 0

a b aa b a b i

b

+ - - = + - - - + = + =2 2 0 1; 3

3 2 2; 3

a a a b

b a b

- - = = - = - = - = = = -

8y 1 3z i= - - hoc 2 3z i= +

V d 8: Tm s phc z tha mn 2z i- = v ( )( )1z z i- + l s thcGii:

Gi s z= x+ yi (x, y R )Khi ,

( ) ( )( )( ) ( ) ( )( ) ( ) ( ) ( )

222 1 2 1

1 1 1 1 1 1

z i x y

z z i x yi x y i x x y y x y i

- = + - =

- + = - + - - = - + - + + -

( )( ) ( )1 1 0 2z z i R x y- + + - =6 (1) v (2) ta c x=1; y=0 hoc x=-1; y=28y z=1; z=-1+ 2i Bi tp t luynBi 1. Tm s phc z tha mn: 2 2z i- + = . Bit phn o nh hn phn thc 3 n v.

Bi 2. Tm s phc z tha mn: | z | - iz = 1 2i

Bi 3. Tm s phc z tha mn: ( )2 10z i- + = v . 25z z = .

Bi 4. Tm s phc z tha mn ( )1 2 26z i- + = v . 25z z = .Bi 5. Tm s phc z tha mn tng trng hp:a) 2z = v z l s thun o. b) 5z = v phn thc ca z bng hai ln phn o ca n.

Bi 6. Tim s phc z tho mn 2z = v z2 l s thun o.

Bi 7. Gii phng trnh:

a) 2 0z z+ = . b) 2z z z+ =

Bi 8. Tm s phc z bit 21( 1)(1 ) | | .1z

z i zi

-+ + + =

-

Bi 9. Tm s phc z bit: 1 1z - = v (1 )( 1)i z+ - c phn o bng 1.

Bi 10. Tm s phc z tha mn: 1 5z - = v_ _

17( ) 5z z z z+ = .

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Bi 11. Tm s phc z tha mn

( ) ( )

5

5 22

42 1

z

zi

i

= - + = +

.

2.4. Dng 4: Biu di n hi nh hc mt s phc. Tim tp hp im biu di n s phc z.Trong dng ny, ta gp cc bi ton biu din hnh hc ca s phc hay cn gi l

tm tp hp Km biu din mt s phc z trong s phc z tha mn mt h thc no (thng l h thc lin quan n mun ca s phc). Khi ta gii bi ton ny nh sau:

Gi s z = x+yi (x, y R). Khi s phc z biu din trn mt phng phc bi KmM(x;y). S dng d kin ca bi tm mi lin h gia x v y t suy ra tp hp KmM.V d 1: Gi s M(z) l Km trn mt phng phc biu din s phc z. Tm tp hp ccKm M(z) tha mn mt trong cc Ku kin sau y: a) =2 b) c)

Gii:t z = x +yi (x, y R) c biu di n bi i m M(x;y)

a) Xt h thc: =2 (1)

t z = x +yi (x, y R) z 1 + i = (x 1) + (y + 1)i.Khi (1)

(x-1)2 + (y + 1)2 = 4. Tp hp cc Km M(z) trn mt phng ta biudin s phc z tha mn (1) l ng trn c tm ti I(1;-1) v bn knh R = 2.b) Xt h thc |(x+2) +yi| = |-x+(1-y)i| (x+2)2 + y2 = x2 + (1-y)2 4x + 2y + 3 = 0.8y tp hp cc Km M l ng thng 4x + 2y + 3 = 0.

Nhn xt:ng thng 4x + 2y + 3 = 0 chnh lng trung trc ca Qn AB.c) Xt h thc:

Xt F1, F2 tng ng biu din cc Km 4i v -4i tc l F1 (0;4) v F2 =(0;-4). Do : MF1 + MF2 = 10

Ta c F1F2 = 8 Tp hp tt c cc Km M nm trn (E) c hai tiu Km l F1 v F2 vc di trc ln bng 10.

Phng trnh ca (E) l:

1z i 2 1z i 4 4 10z i z i

1z i

2 2( 1) ( 1) 2x y

2 z z i

4 4 10z i z i

4 4 10z i z i

2 2

19 16x y

-2 -1 1 2

-2

-1

1

2

x

y

A

B

O

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V d 2: Trong mt phng Oxy, tm tp hp Km biu din cc s phc z tha mn

( )1z i i z- = +Gii:t z= x+ yi (x,y R )Ta c:

( ) ( ) ( ) ( )( ) ( ) ( )2 2 221 1

1

z i i z x y i x y x y i

x y x y x y

- = + + - = - + +

+ - = - + +

( )22 2 22 1 0 1 2x y xy x y + + - = + + =8y tp hp cc Km M biu din cc s phc z l ng trn c phng trnh

( )22 1 2x y+ + =

V d 3: Cho s phc( )

3

1 5

1 3

(1 )

iz

i

+=

+. Tm tp hp Km biu din 2A z iz= + , bit rng

1 0x y- - = .Gii

20

4 04

tt t

t

= - = =

( ) ( )0 0; 1 , 4; 1t B C= - -( ) ( )4 4; 1 , 0; 1t B C= - -

Gi s2

z x yi= + ,x y R biu din bi Km M(x;y). Khi ta c:

( )JJG

2 2 2, , , 0P

n a b c a b c= + +

8y tp hp Km biu din cho s phc2

z l ng trn tm O, bn knh 2

V d 4: Trong cc s phc z tha mn Ku kin 2 4 2z i z i- - = - .Tm s phc z c

mun nh nht.Gi s s phc z cn tm c dng z = x + yi (x,y R) c biu di n bi im M(x;y).

Ta c 2 ( 4) ( 2)x y i x y i- + - = + - (1) 2 2 2 2( 2) ( 4) ( 2)x y x y - + - = + -

4y x = - + . Do tp hp cc Km M biu din cho cc s phc z tha mn (1) l

ng thng x + y = 4. Mt khc 2 2 2 2 28 16 2 8 16z x y x x x x x= + = + - + = - +

Hay ( )22 2 8 2 2z x= - +

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Do min

2 2z x y = = . Vy 2 2z i= +

V d 5: Bit rng s phc z tha mn ( )( )3 1 3u z i z i= + - + + l mt s thc. Tm gi trnh nht ca z .

Giit z= x+ yi (x, y R ) ta c

( ) ( ) ( ) ( ) ( )2 23 1 1 3 4 4 6 2 4u x y i x y i x y x y x y i = + + - + - - = + + - + + - - - Ta c: 4 0u R x y - - =6p hp cc Km biu din ca z l ng thng d: x-y-4=0, M(x;y) l Km biu din caz th m un ca z nh nht khi v ch khi di OM nh nht OM d ^ Tm c M(-2;2) suy ra z=-2+2i.

V d 6: Tm s phc Z c m un ln nht v tha mn Ku kin ( ) 131 3 22

Z i i+ - + =

Gii)i ( , )z x yi x y R z x yi= + = -

2 213 39(1 ) 3 2 5 02 8

z i i x y x y+ - + = + - - + =

)i M (x;y) l Km biu din ca z trong mt phng ta Oxy ( )M C l ng trn

c tm 1 5( ; )2 2

I v bn knh 264

R =

)i d l ng thng i qua O v I : 5d y x =

)i M1, M2 l hai giao Km ca d v (C) 13 15

( ; )4 4

M v2

1 5( ; )4 4

M

Ta thy 1 21

( ( ))

OM OM

OM OI R OM M C

> = +

U phc cn tm ng vi Km biu din M1 hay3 154 4

z i= +

V d 7: Tm tp hp cc Km biu din ca s phc z sao cho 2 3z iuz i

+ +=

- l mt s

thun o.Giit z= x+ yi (x, y R ), khi :

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( ) ( )( )

( ) ( ) ( )( )22

2 3 12 3

1 1

x y i x y ix y iu

x y i x y

+ + + - -+ + + = =+ - + -

( ) ( )( )

2 2

22

2 2 3 2 2 1

1

x y x y x y i

x y

+ + + - + - +=

+ -

u l s thun o khi v ch khi( )

( ) ( )( ) ( )

2 22 2

22

2 2 3 0 1 1 5

1 0 ; 0;1

x y x y x y

x y x y

+ + + - = + + + = + - >

8y tp hp cc Km biu din ca z l ng trn tm I(-1;-1), bn knh 5 trKm(0;1)

Bi tp t luynBi 1. Gi s M(z) l Km trn mt phng ta biu din s phc z. Tm tp hp

nhng Km M(z) tha mn Ku kin sau

a) (1 3 ) 3 2z i z i+ - = + - b) 2 2z i z z i- = - + c) ( )3 4 2z i- - =

Bi 2. Trong cc s phc tha mn 32 32

z i- + = . Tm s phc z c mun nh nht.

Bi 3. Trong mt phng ta . Tm tp hp Km biu din cc s phc z tha mn

Ku kin: 3 2z i z i- = - - . Trong cc s phc tha mn Ku kin trn, tm s

phc c mdun nh nhtBi 4. Trong cc s phc z tha mn Ku kin 2 4 2z i z i- - = - .Tm s phc z c

mun nh nht.

Bi 5. Trong cc s phc z tha mn Ku kin 1 5 3z i z i+ - = + - . Tm s phc z c

mun nh nht.

Bi 6. Trong cc s phc z tha mn 2 52z i- - = , tm s phc z m 4 2z i- + l

nh nht.Bi 7. Tm s phc Z c m un ln nht v tha mn Ku kin Trong tt c cc s

phc z tha mn 2 2 1z i- + = , hy tm s phc c z nh nht

Bi 8. Trong cc s phc z tha mn Ku kin( )1

2 11

i z

i

++ =

-.Tm s phc c m un

nh nht, ln nht.

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2.5. Dng 5. Phng trinh bc hai trn tp s phc2.5.1. Vn 1. Tm cn bc hai ca mt s phc. (c thm)

Cho s phc w = a + bi. Tm cn bc hai ca s phc ny.Phng php:

+) Nu w = 0 w c mt cn bc hai l 0+) Nu w = a > 0 (a R) w c hai cn bc hai l v -+) Nu w = a < 0 (a R) w c hai cn bc hai l v -+) Nu w = a + bi (b z 0)Gi s z = x +yi (x, y thuc R) l mt cn bc hai ca w z2 = w (x+yi)2 = a +

bi

tm cn bc hai ca w ta cn gii h ny tm x, y. Mi cp (x, y) nghim ngphng trnh cho ta mt cn bc hai ca w.

Nhn xt: Mi s phc khc 0 c hai cn bc hai l hai si nhau.V d: Tm cc cn bc hai ca mi s phc sau:

a) 4 + 6 i b) -1-2 iGii:1) Gi s z = x +yi (x, y thuc R) l mt cn bc hai ca w = 4 + 6 i

Khi : z2 = w (x+yi)2 = 4 + 6 i

(2) x4 4x2 45 = 0 x2 = 9 x = 3.x = 3 y =x = -3 y = -

8y s phc w = 4 + 6 i c hai cn bc hai l: z1 = 3 + i v z2 = -3 - i2) Gi s z = x +yi (x, y thuc R) l mt cn bc hai ca w = -1-2 i

Khi : z2 = w (x+yi)2 = -1-2 i

(2) x4 + x2 6 = 0 x2 = 2 x = .x = y = -x = - y =

8y s phc w = 4 + 6 i c hai cn bc hai l: z1 = - i v z2 = - + i

a a

ai ai

2 2

2x y a

xy b

5 6

5

52 2

22

3 5 (1)4

452 6 5 4 (2)

yx y xxy x

x

5

5

5 5 5

6

62 2

22

6 (1)1

62 2 6 1 (2)

yx y xxy x

x

2

2 3

2 3

5 2 3 2 3

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2.5.2. Vn 2: Gii phng trnh bc haiCho phng trnh bc hai: Az2 +Bz +C = 0 (1) (A, B, C C, A z 0)

Phng php:Tnh ' = B2 4AC

*) Nu ' z 0 th phng trnh (1) c hai nghim phn bit z1 = , z2 =

(trong G l mt cn bc hai ca ').*) Nu ' = 0 th phng trnh (1) c nghim kp: z1 = z2 =

V d 1: Gii cc phng trnh sau trn tp s phc2) 1 0a z z- + = 2) 2 5 0b x x+ + = 4 2) 2 3 0c z z+ - =

Gii:2) 1 0a z z- + =

9 21 4 3 3iD = - = - =9 En bc hai ca D l 3i

9 Phng trnh c nghim: 1 21 3 1 3 1 3

,2 2 2 2 2i

z i z i+

= = + = -

2) 2 5 0b x x+ + =

9 24

De Cuong Toan 2014-2015

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