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Tham gia kha LTH v Luyn gii dnh 9 im Ton tr ln trong k TSH 2014 www.moon.vn 137

MC LC 1. CHUN K NNG I S ...........................................................................................................................01

2. CC TR CA HM S ...............................................................................................................................08

3. TIP TUYN CA TH HM S.........................................................................................................36

4. S TNG GIAO CA HAI TH HM S........................................................................................53

5. CC BI TON V KHONG CCH TRONG HM S......................................................................101

6. BI TON TM IM TRN TH ......................................................................................................112

7. BIN LUN S NGHIM PHNG TRNH BNG TH ...............................................................123

8. TNH N IU CA HM S ................................................................................................................128

9. TNG HP CC BI TON HM S TRONG THI I HC.....................................................136

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NG VIT HNG

BI GI NG TRNG TM V

HM S Chng trnh Luy n thi i hc SAP

NM HC 2013 - 2014

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1. K NNG XT DU CA BIU THC

Nguyn tc:

+ Phn tch biu thc cn xt du hay bt phng trnh v dng tch, ri loi b nhng hng t l ly tha bc chn. + Sp xp cc nghim ca cc hng t sau khi thanh lc cc hng t chn theo th t t b n ln trong bng xt du. + Tin hnh xt du theo quy tc an du khi bit du ca mt khong no . + Vic xt du biu thc chng ta ch c quy ng mu s m khng c nhn cho.

Cc v d in hnh:

V d 1: Xt du cc biu thc sau

a) 2

( ) 3.3 4

+= +

xf x

x b)

1 3( ) 2 .

1=

f x

x x

c) ( 3)(3 2 )

( ) .1

+ =

x xf x

x d)

4 2 2 1 5( ) .

3 2 4

+= x xf x

e) 2 3 2

( ) .1

+=

x xf x x

x f)

2 2( ) .

3 1 2 1

+ = +

x xf x

x x

g) 2

1 1 2( ) .

1

= + + +

xf x

x x x x h)

1 2 3( ) .

2 2= +

+f x

x x x

V d 2: Gii cc bt phng trnh sau

a) 1 2 3

.3 2

+

x x x

x x

e) 4 2

2

4 30.

8 15

+ +

x x

x x f) ( )

3 23 30.

2

+ >

x x x

x x

2. K NNG S DNG LC HOOCNER CHIA A THC

Nguyn tc:

+ f(x) chia cho g(x) c h(x) v d l k th ta c th vit ( ) ( ) ( ) ( )( ) ( ) ( ).= + = +f x k

f x g x h x k h xg x g x

+ chia a thc bng lc Hoocner ta phi sp xp a thc chia theo ly tha gim dn, s hng no khuyt ta cho h s bng 0.

+ Thc hin chia theo quy tc: u ri - nhn ngang - cng cho.

Cc v d in hnh:

V d: Thc hin cc php chia sau

a) 4 3 23 2

3

+ + =+

x x x x

x........................... b)

3 23 2 10

1

+ + =

x x x

x.................................

c) 22

1

+ + =

x mx m

x................................... d)

( )2 22 2 22 1

+ +=

+x m x

x................................

3. K NNG NHM NGHIM CA PHNG TRNH A THC

Bi m u:

CHUN K NNG I S

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Xt phng trnh: ( ) ( )4 3 2 0, 1 .= + + + + =f x ax bx cx dx e Nu x = xo l mt nghim ca phng trnh (1) th ( ) ( ) ( )( )3 21 0 = + + + =of x x x ax b x c x d

( ) 3 2 = + + + o

f xax b x c x d

x x

Nguyn tc:

+ Nu tng cc h s ca phng trnh bng 0 th phng trnh c mt nghim x = 1.

+ Nu tng cc h s bc chn ca x bng tng h s bc l ca x th phng trnh c mt nghim x = 1. + Nu phng trnh khng tun theo hai quy tc trn th chng ta nhm nghim bt u t cc nghim n gin nh 0; 1; 2

+ Vi cc phng trnh c cha tham s, nhm nghim ca phng trnh ta cho phn h s ca tham s m bng 0, c nghim x ta thay vo phng trnh kim tra li.

Cc v d in hnh:

V d 1: Phn tch cc a thc sau thnh nhn t

a) ( ) 4 3 22 4 3 2 1= + f x x x x x b) ( ) 3 24 2 7 1= f x x x x c) ( ) ( ) ( )3 21 1 2 1= + + f x x m x m x m

Hng dn gii :

a) ( ) 4 3 22 4 3 2 1= + f x x x x x Xt phng trnh ( ) 4 3 20 2 4 3 2 1 0= + =f x x x x x Ta nhn thy phng trnh c tng cc h s bng 0 nn c mt nghim l x = 1.

Khi ( ) ( ) ( ) ( )4 3 2

4 3 2 2 4 3 2 10 1 . 2 4 3 2 11

+ = = + =

x x x xf x x g x x x x x g x

x

Dng lc Hoocner ta c

( )( )4 3 2

3 2 4 3 2 3 22 4 3 2 1 2 6 3 1 2 4 3 2 1 1 2 6 3 11

+ = + + + + = + + +

x x x xx x x x x x x x x x x

x

b) ( ) 3 24 2 7 1= f x x x x Xt phng trnh ( ) 3 20 4 2 7 1 0= =f x x x x Tng h s bc chn l 2 1 = 3, tng h s bc l ca phng trnh l 4 7 = 3 T ta thy phng trnh c mt nghim x = 1.

Khi ( ) ( ) ( ) ( ) ( ) ( )3 2

3 2 4 2 7 11 . 4 2 7 1 1 .1

= + = + =+

x x xf x x g x x x x x g x g x

x

Dng lc Hoocner ta c

( ) ( ) ( )( )3 2

2 3 2 24 2 7 1 4 6 1 4 2 7 1 1 4 6 11

= = = = + +

x x xg x x x f x x x x x x x

x

c) ( ) ( ) ( )3 21 1 2 1= + + f x x m x m x m Tng cc h s a thc l ( ) ( )1 1 1 2 1 0 + + =m m m nn f(x) = 0 c mt nghim x = 1. Tin hnh chia a thc ta c ( ) ( ) ( ) ( )( )3 2 21 1 2 1 1 2 1= + + = +f x x m x m x m x x mx m V d 2: Phn tch cc a thc sau thnh nhn t

a) ( ) 4 23 2 6= + +f x x x x = ... b) ( ) 3 24 6 1= + +f x x x x = c) ( ) 3 2= + f x x mx x m= . d) ( ) ( )3 22 1= + +f x x x m x m= . e) ( ) 3 2 6 8= + f x x x x = .

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f) ( ) 3 22 4 4= + f x x x x = . 4. K NNG X L VI TAM THC BC HAI V PHNG TRNH BC HAI

Xt phng trnh bc hai: ( ) ( )2 0, 1= + + =f x ax bx c a) Gii v bin lun phng trnh (1):

Nu a = 0 th ( ) ( )1 0, * + =bx c + nu b = 0 v c = 0 th (*) nghim ng vi mi x. + nu b = 0 v c 0 th (*) v nghim.

+ nu b 0 th ( )* = cxb

Nu a 0 th (1) l phng trnh bc hai c bit thc ( )

2

2

4

; 2

=

= =

b ac

b ac b b

+ nu > 0 th (1) c hai nghim phn bit 2

1;2

4.

2 2

= =b b b acxa a

+ nu = 0 th (1) c nghim kp .2

= bxa

+ nu = 0 th (1) v nghim.

b) H thc Vi-t:

Khi (1) c hai nghim phn bit x1 v x2 th ta c h thc Vi-t: 1 2

1 2

= + = = =

bS x x

ac

P x xa

Mt s cc kt qu cn lu :

( )22 2 21 2 1 2 1 22 2+ = + = x x x x x x S P ( ) ( )33 3 31 2 1 2 1 2 1 23 3+ = + + = x x x x x x x x S SP ( ) ( )2 24 4 2 2 2 2 2 21 2 1 2 1 22 2 2x x x x x x S P P+ = + = ( ) ( )2 2 21 2 1 2 1 24 4 = + = x x x x x x S P c) Tnh cht nghim ca phng trnh bc hai:

Phng trnh c hai nghim dng phn bit khi

2

1 21 2

1 2

4 00

0; 0

0

> > = + = > > = = >

b ac

bS x x

x x ac

P x xa

Phng trnh c hai nghim m phn bit khi

2

1 21 2

1 2

4 00

0; 0

0

> > = + =



Phng trnh c hai nghim phn bit v u ln hn khi

( )( ) ( )

2 2

1 2 1 2 1 21 2

21 221 2 1 2

4 0 4 00

02 2 2

0

0 0

> >

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

b ac b ac

b bx x S x x S x x

x ,x a ax x c bx x x x .

a a

Phng trnh c hai nghim phn bit v u nh hn khi

( )( ) ( )

2 2

1 2 1 2 1 21 2

21 221 2 1 2

4 0 4 00

02 2 2

0

0 0

> >

> > + < = + = < = + = + + >

b ac b ac

b bx x S x x S x x

x ,x a ax x c bx x x x .

a a

Phng trnh c hai nghim phn bit v u khc khi ( ) 21 20 00

; 0 0

> > > + + x x g a b c

Phng trnh c mt nghim v nghim ny ln hn khi

( )( ) ( )

1 21 21 21 2

2 21 2 1 2 1 2 1 2

0000

22220000

0 0 0

= = = = = = > = = >= = >= = > > > > > < < < + + < + + < < < + + < + + > > <

th (1) c 2 nghim phn bit

2

1;22 2 5 3

.1

m m mx

m

+=+

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b) Phng trnh (1) c hai nghim phn bit khi ( )23

0 2 5 3 0 *1

2

mm m

m

> > > <

Gi hai nghim phn bit l x1 ; x2 vi x2 > x1.

Theo nh l Vi-t ta c 1 2

1 2

4

12 3

1

b mx x

a mc m

x xa m

+ = = + + = = +

Hai nghim u dng khi 1 2

1 2

1 0400 11 .

0 2 3 301 2

o

mmx x mm vnx x m

mm

< + > > + > + > < +

c) Hai nghim u nh hn 1 khi

( )( ) ( )1 2 1 2 1 21 2 1 2

2 3 4 4 1 41 0 01 1 0 1 0 1 1 1 1 4.14 42 2 2 2 0 1

1 1

m m m mx x x x x x m m m mm

m mx x x xm

m m

+ + < > + + > + + + > + + + < + < + < < > < + +

i chiu vi iu kin (*) v tn ti hai nghim phn bit ta c 3 < m < 4 l gi tr cn tm.

V d 2: Cho phng trnh ( )( ) ( )+ + + =22 2 1 0, 1x x mx m . a) Tm m phng trnh c ba nghim phn bit. b) Tm m phng trnh c ba nghim phn bit, trong c hai nghim m.

c) Tm m phng trnh c ba nghim phn bit x1; x2; x3 tha mn + + + > > + > <

+

Vy vi

4 2 5

4 2 5

5

4

m

m

m

> + <

th phng trnh cho c 3 nghim phn bit.

b) Do nghim x = 2 < 0 nn (1) c 3 nghim trong 2 nghim m th (2) phi c hai nghim tri du.

T ta c 1

0 1 2 0 .2

P m m< < >

Gi tr ny tha mn iu kin (*) nn l gi tr cn tm. c) Khng mt tnh tng qut, gi s x1 = 2. Khi x2 ; x3 l hai nghim phn bit ca (2).

Theo nh l Vi-t ta c 2 3

2 3 1 2

x x m

x x m

+ = =

Khi ( ) ( )22 2 2 2 21 2 3 2 3 2 37 4 2 7 2 1 2 3 0 4 5 0 5 1.x x x x x x x m m m m m+ + < + + < < + < < < Kt hp vi iu kin (*) ta c 4 2 5 1m + < < l gi tr cn tm.

BI TP LUYN TP:

Bi 1: Cho phng trnh ( ) 21 2 1 0. + + =m x mx m

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a) Chng minh rng phng trnh lun c hai nghim phn bit vi mi gi tr ca m 1. b) Xc nh gi tr ca m phng trnh c tch hai nghim bng 5, t hy tnh tng hai nghim ca phng trnh.

c) Tm m phng trnh c nghim x1, x2 tho mn h thc 1 22 1

50.

2+ + =x x

x x

Bi 2: Cho hm s y = (x 1)(x2 + mx + m). a) Vi m = 2, tnh y v gii phng trnh y = 0.

b) Tm m tip tuyn ti im c honh x = 1 song song vi ng thng d: y = 2x 3 c) Tm m phng trnh y = 0 c ba nghim phn bit x1; x2; x3 tha mn 2 2 21 2 3 4.+ +



DNG 1. CC TR CA HM A THC BC BA

Xt hm s bc ba : 3 3 23 3= + + + = + +y ax bx cx d y ax bx c

DNG 1. TM IU KIN V S CC TR CA HM S

Nu a = 0 th 3 03

= + = = cy bx c y xb

Trong trng hp ny hm s c 1 cc tr. Nu a 0 :

+ Hm s khng c cc tr khi y khng i du, tc l phng trnh y = 0 v nghim hoc c nghim kp, tc l 0. + Hm s c 2 im cc tr khi y i du hai ln, tc l phng trnh y = 0 c hai nghim phn bit. T ta c iu kin hm s c hai cc tr l > 0.

Vy, vi hm bc ba th hm s ch c hai cc tr hoc khng c cc tr.

V d 1: Bin lun s cc tr ca hm s ( )= + 3 21 1 13

y x m x mx ty theo gi tr ca tham s m.

Hng dn gii: Ta c ( )2 2 1 . = + +y x m x m Hm s khng c cc tr khi y khng i du trn min xc nh (hay hm s lun ng bin hoc nghch bin trn min xc nh), iu xy ra khi y = 0 v nghim hoc c nghim kp.

T ta c iu kin ( )2 2 3 5 3 50 1 0 3 1 0 .2 2

+ + m m m m m

Hm s c hai cc tr khi y i du trn min xc nh, iu xy ra khi y = 0 c hai nghim phn bit.

2

3 5

20 3 1 03 5

2

+> > + >



Hm s c hai cc tr khi y = 0 c hai nghim phn bit

2

2 2 6 2 2 6005 5

0 5 4 4 00

+ < +



c) Tm m hm s t cc tiu ti x = 2. Hng dn gii :

Ta c ( )2 2( 2) 2 2 2 . = + = +y x m x m y x m

a) Hm s c cc tr khi phng trnh y = 0 c hai nghim phn bit 21

0 5 4 04

> > + + > <

mm m

m

b) Tm m hm s t cc i ti ti x = 0. Cch 1: + Hm s t cc i ti x = 0 th ( )0 0 0. = =y m

+ Vi m = 0 th ta c 20

4 04

= = = =

xy x x

x

Ta c bng bin thin:

x 0 4 +

y + 0 0 +

y C + CT

T bng bin thin ta thy hm s cho t cc i ti x = 0. Vy m = 0 l gi tr cn tm. Cch 2:

Hm s t cc i ti ( )( )0 0 0

0 02( 2) 00 0

= == = + >



Tnh cht 2: Cc im cc tr c honh cng dng, cng m, cng ln hn hoc nh hn cho trc.

Hai im phn bit cc tr cng c honh dng.

Khi ta c 1 22 11 2

000

00

BS x x Ax xP x x C

A

>= + > > > = > >

Hai im cc tr cng c honh m.

Khi ta c 1 21 21 2

000

00

BS x x Ax xP x x C

A

Hai im cc tr c honh tri du.

Khi ta c 1 2 1 20 0 0C

x x P x xA

< < = < <

Hai im cc tr cng c honh ln hn .

Khi ta c ( )( ) ( )

221 2 1 2

1 22 1

1 2

0 0 0

2 2

2

C Bx x x x

x x A Ax x B

x x BA

A

+ > + + > > > > + > > >

Hai im cc tr cng c honh nh hn .

Khi ta c ( )( ) ( )

221 2 1 2

1 21 2

1 2

0 0 0

2 2

2

C Bx x x x

x x A Ax x B

x x BA

A

+ > + + > > < < + < + + >



Vy vi

7 3 5

2

7 3 5

2

+> > + > > > + > > >

x x x x mx x m

x x mx x m

880

8 5.35

5

+ > > < < < <

mm

mm

m

i chiu vi iu kin (*) ta c 7 3 5

82

m < < l gi tr cn tm.

d) Ta c 1

21 2

2

1

60 3 2( 1) 3 01

6

= = = + = < + = =

mx x

y x m x x xm

x x

Bng bin thin

x x1 x2 +

y + 0 0 +

y C + CT

Ta thy hm s t cc i ti im c honh 11

6

= mx

Theo bi ta c ( )1 25 01

1 1 6 56 5

= > > < <

mmx m m

m

2 2

5 58 5.

3 247 1 10 25

< > + + < + +

m mm

mm m m m


82

m < < l gi tr cn tm.

V d 2: Cho hm s = + + 3 23( 1) 9 .y x m x x m Tm m hm s t cc i, cc tiu ti x1 ; x2 tha mn 1 2 2.x x

Hng dn gii:

Ta c 23 6( 1) 9. = + +y x m x

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Hm s t cc i, cc tiu ti x1; x2 khi y = 0 c hai nghim phn bit 1 2; 0 >x x

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

> + + >

<

mm

m

Theo nh l Vi-et ta c 1 2

1 2

2( 1)

3

+ = + =

x x m

x x

Khi : ( ) ( )2 2 21 2 1 2 1 22 4 4 4 1 12 4 ( 1) 4 3 1 + + + x x x x x x m m m


1 3 1

< + <

m

m l cc gi tr cn tm.

V d 3: Cho hm s = + + + +3 2(1 2 ) (2 .) 2y x m x m x m

Tm m hm s t cc i, cc tiu ti x1 ; x2 tha mn >1 21

.3

x x

Hng dn gii:

Ta c 23 (1 2 .)2 2= + + x m x my Hm s t cc i, cc tiu ti x1; x2 khi y = 0 c hai nghim phn bit x1; x2

( )2 25

(1 2 ) 3(2 ) 4 5 0 *41

> = = >

<

mm m m m

m

Theo nh l Vi-et ta c 1 2

1 2

(1 2 )

32

2

.

3

+ = =

mx x

mx x

Khi ( ) ( )2 2 21 21 2 2 1 211

4 4(1 2 ) 4(2 11

93) = + > >> x x x x mx mx x x

2

3 29

816 12 5 03 29

8

+> >

<

m

m

l cc gi tr cn tm.

V d 4: Cho hm s = + +3 21 1( 1) 3( 2) .3 3

y x m x m x

Tm m hm s t cc i, cc tiu ti x1 ; x2 tha mn + =1 22 1.x x Hng dn gii:

Ta c 2 2( 1) 3( 2). = + y x m x m Hm s t cc i, cc tiu ti x1; x2 khi y = 0 c hai nghim phn bit x1; x2 2 5 7 0, = + > m m m

Khi ta c ( )( )

1 2 2 2 2

1 2 1 2 1

1 2 1 2 1 2

2( 1) 1 2 2( 1) 3 23( 2) 3( 2) 1 2(2 2 ) 4 3

2 1 2 1 3( 2) 3 2 4 3 3 6

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

x x m x x m x mx x m x x m x m mx x x x x x m m m m

2 4 348 16 9 0 .4

+ = =m m m Vy 4 344

=m l cc gi tr cn tm.

V d 5: Cho hm s = + + + +3 2(1 2 ) (2 ) 2.y x m x m x m Tm m hm s c cc i, cc tiu ng thi honh im cc tiu nh hn 1.

Hng dn gii:

Ta c 23 2(1 2 ) 2 ( ). = + + =y x m x m g x Do h s a = 3 > 0 nn yu cu bi ton tr thnh y = 0 c hai nghim phn bit x1; x2 tha mn

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2

1 2

4 5 05 7

1 (1) 5 7 04 52 1

12 3

= >< < = + > < < = y x mx m Khi

1 2

1 2

1 2

4

9.

6 21

4

= + = = =

x x

mx x m

x x

BI TP LUYN TP: Bi 1: Cho hm s 3 2( 2) ( 1) 2.= + + +y x m x m x a) Tm m hm s c cc i, cc tiu. b) Tm m hm s t cc tiu ti ti x = 3. c) Tm m hm s t cc i, cc tiu ti x1 ; x2 tha mn

2 21 2 10.+ 0.

V d 1: Cho hm s = + + +3 23 2y x x mx m , vi m l tham s. Tm m hm s c cc i, cc tiu v cc im cc i, cc tiu nm v hai pha ca trc honh.

Hng dn gii:

Ta c 23 6 = + +y x x m, hm s c cc i, cc tiu khi phng trnh y = 0 c hai nghim phn bit. Tc l 9 3 0 3. = >



Phng trnh honh giao im ca (Cm) v Ox: ( )3 2

2

13 2 0

( ) 2 2 0, 1= + + + = = + + =

xx x mx m

g x x x m

Hm s c 2 im cc tr nm v 2 pha i vi trc Ox khi (1) c 2 nghim phn bit khc 1

Ta c iu kin 3 0 3( 1) 3 0

= > + + >m m m

2

13 3 21

213 5 013 3 21

2

+> + > m m m Hm s c cc im cc i, cc tiu nm v cng mt pha ca trc tung khi phng trnh y = 0 c hai nghim

cng du 1

0 2 1 0 .2

> > >ac m m

Kt hp iu kin ta c 1

12

< m tha mn yu cu bi ton.

Tnh cht 5: Cc bi ton cc tr khi y = 0 gii c nghim p

Khi phng trnh y = 0 c ( )2ax b = + th iu kin hm s c cc tr l ( )20 0 .bax b xa

> + >

Khi , 1

2

0x x

yx x

= = = v s dng yu cu ca bi gii ra tham s.

V d 1: Cho hm s = + +3 2 2 33 3( 1) .y x mx m x m m Tm gi tr ca m hm s c cc tr . Khi , tm m khong cch t im cc i n gc ta bng

2 ln khong cch t im cc tiu n gc ta O. Hng dn gii :

Ta c 2 2 2 23 6 3( 1) 0 2 1 0 = + = + =y x mx m y x mx m Hm s c cc i, cc tiu khi y = 0 c hai nghim phn bit 1 0, = > m

Khi ( )( )

1 1;2 20

1 1; 2 2

= =

= + +

x m A m my

x m B m m

Do h s a = 1 > 0 v m + 1 > m 1 nn A l im cc i v B l im cc tiu ca hm s.

Theo bi ta c 23 2 2

2 6 1 03 2 2

= += + + =

=

mOA OB m m

m

Vy 3 2 2= m l cc gi tr cn tm.

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V d 2: Cho hm s ( )

= + + 2

3 3 11 (3 2) 1.3 2

m xy x m x m

Tm gi tr ca m a) hm s c cc i, cc tiu. b) hm s t cc i, cc tiu ti cc im c honh ln hn 2. c) hm s t cc i, cc tiu ti cc im c honh x1 ; x2 tha mn + >3 31 2 28x x

d) hm s t cc i, cc tiu ti cc im c honh x1 ; x2 tha mn + =2 21 22 12x x Hng dn gii :

Ta c ( ) ( )2 23 1 3 2 0 3 1 3 2 0.y x m x m y x m x m = + = + = a) Hm s c cc i, cc tiu khi y = 0 c hai nghim phn bit. Ta c iu kin ( ) ( )2 20 3 1 4. 3 2 0 9 18 9 0 1m m m m m > > + >

b) Vi

( )

( )

3 1 3 11

21 03 1 3 1

3 12

m mx

m ym m

x m

= =

= +

= =

Honh cc im cc i, cc tiu ln hn 2 khi 3 1 2 1.m m > > Vy vi m > 1 th hm s cho c cc i, cc tiu v honh cc i, cc tiu ln hn 2.

c) Ta c ( )33 31 24

28 1 3 1 28 3 1 3 .3

x x m m m+ > + > > >

d) Do vai tr bnh ng ca x1 ; x2 nn ta c hai trng hp xy ra

Vi ( )22 21 2 1 21 10

1; 3 1 2 12 2 3 1 12 3 1 103

x x m x x m m m= = + = + = = =

Kt hp vi iu kin tn ti cc tr ta c 1 10

.3

m=

Vi ( )22 21 2 1 222 2 22

3 1; 1 2 12 2 3 1 1 12 3 12 6

x m x x x m m m= = + = + = = =

Kt hp vi iu kin tn ti cc tr ta c 2 22

.6

m=

V d 3: Cho hm s += +3 23y x x m

Tm m hm s t cc i, cc tiu ti cc im A, B sao cho .= 0120AOB Hng dn gii :

Ta c 20

3 6 02 4

= = = + = = = +

x y my x x y

x y m

Vy hm s c hai im cc tr A(0 ; m) v B(2 ; m + 4).

Ta c (0; ), ( 2; 4).= = +OA m OB m

01

120 cos2

= = AOB AOB

( )( )2 2 2

2 2

4 0( 4) 14 ( 4) 2 ( 4)

3 24 44 024 ( 4)

4 012 2 3 2

412 2 33 3

3

<



Khi d, cc im cc tr ca hm s l 3 3(0; 3 1), (2 ;4 3 1) (2 ;4 ) A m B m m m AB m m

Trung im I ca AB c to 3( ;2 3 1) I m m m ng thng d: ( ) : 8 74 0+ =d x y c mt vc t ch phng (8; 1)=

u .

A v B i xng vi nhau qua d ( )38(2 3 1) 74 0

2. 0

+ = = = m m mI d

d mAB d AB u

Vy m = 2 l gi tr cn tm.

V d 5: Cho hm s ( )= + +3 23 3 1 12m

y x x m x

Tm m a) hm s c cc i, cc tiu. b) hm s t cc tiu ti im x = 2. c) hm s t cc i ti x = 0. d) hm s khng c cc i, cc tiu. e) ng thng qua cc im cc i, cc tiu song song vi ng thng d: y = 9x + 1

Hng dn gii :

a) Ta c ( ) ( ) ( )3 2 2 23 3 1 1 3 3 3 1 3 12

my x x m x y x mx m x mx m= + + = + = +

Hm s c cc i, cc tiu khi y = 0 c hai nghim phn bit. Ta c iu kin ( )20 2 0 2.m m > > Vy vi m 2 th hm s cho c cc i, cc tiu.

b) Hm s t cc tiu ti im x = 2 khi h sau c nghim (2) 0

, ( )(2) 0

= >

yI

y

Ta c y = 6x 3m, khi 4 2 1 0 3

( ) 312 3 0 4

m m mI m

m m

+ = = = >

yI

y

Ta c y = 6x 3m, khi h 1 0 1

( ) 13 0 0

m mI m

m m

= = = < >

Gi tr m = 1 tha mn iu kin (*) nn l gi tr cn tm. d) Hm s khng c cc i, cc tiu khi y khng i du y = 0 v nghim 0 (m 2)2 0 Bt phng trnh trn ch c nghim duy nht m =2. Vy vi m = 2 th hm s cho khng c cc tr. e) Xt phng trnh y = 0 ta c x2 mx + m 1 = 0

( )11

2

2

2 2

3 221

2222 3 5 41

2 2

mm m yx mm

m m m mx y

+ == = =

+ + = = =

Gi A(x1, y1) v B(x2, y2) l cc im cc i, cc tiu. Khi 23 8 6

2 ;2

m mAB m

+=

ng thng qua cc im cc i, cc tiu song song vi d : y = 9x + 1 khi

( )2

2 2

3 8 62 2/ / 2 4 9 3 8 6 27 74 58 0

9 1d o

m mm

AB u m m m m m vn

+ = = + + =

BI TP LUYN TP:

Bi 1: Cho hm s ( ) 23 2 11 2(2 1) 3.

3 2

m xy x m x

= + +

Tm gi tr ca m a) hm s c cc i, cc tiu. b) hm s t cc i, cc tiu ti cc im c honh m.

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c) hm s t cc i, cc tiu ti cc im c honh x1 ; x2 tha mn 4 41 2 17x x+ >

d) hm s t cc i, cc tiu ti cc im c honh x1 ; x2 tha mn 2 21 22 12x x+ =

Bi 2: Cho hm s ( ) 23 23 11 (2 ) 2.

3 2

m xy x m m x

+= +

Tm gi tr ca m a) hm s c cc i, cc tiu. b) hm s t cc i ti im x = 3. c) hm s t cc i, cc tiu ti cc im c honh x1 ; x2 tha mn 2 21 2 40x x = d) hm s t cc i, cc tiu ng thi cc im cc i, cc tiu nm hai pha ca trc Oy. Bi 3: Cho hm s 3 3 2= +y x mx a) Tm m ng thng i qua im cc i, cc tiu ca hm s ct ng trn tm I(1; 1) bn knh bng 1 ti hai im phn bit A, B sao cho din tch tam gic IAB t gi tr ln nht. b) Tm m hm s c C, CT v khong cch t O n ng thng i qua C, CT ln nht.

Tnh cht 6: Phng trnh ng thng qua cc i, cc tiu v mt s ng dng in hnh

Ly y chia cho y ta c . ( ) ,= + +y y g x ax b khi ng thng d : y = ax + b chnh l ng thng i qua cc im cc i, cc tiu ca hm s. Tc dng ln nht ca vic tm c phng trnh ng thng qua cc i, cc tiu l ly c tung ca chng. Tht vy, gi M, N l cc im cc i, cc tiu th ( ) ( )1 1 2 2; , ;+ +M x ax b N x ax b, trong x1 ; x2 l hai nghim ca phng trnh y = 0 v ta c th dng Vi-et c. V d 1: Cho hm s ( )= + + + 3 2 2 3 23 3 1y x mx m x m m Tm m hm s c cc i, cc tiu. Khi , hy vit phng trnh ng thng qua cc im .

Hng dn gii :

Ta c 2 2 2 23 6 3(1 ) 0 2 1 0 = + + = + =y x mx m y x mx m Hm s c cc i, cc tiu khi y = 0 c hai nghim phn bit 1 0, = > m Vy hm s lun c cc i, cc tiu vi mi gi tr ca m.

Chia y cho y ta c 21 23 3

= + +

my x y x m m

Gi ( ) ( )1 1 2 2; , ;A x y B x y l cc im cc tr, khi ( )

( )

1

2

21 1 1

22 2 2

12

3 31

23 3

= + +

= + +

x

x

my x y x m m

my x y x m m

Do ( ) ( ) ( )1 22

21 12

2 2

20 , : 2

2

= + = = = + = +x x

y x m my y A B d y x m m

y x m m

Vy, phng trnh ng thng qua hai im cc tr ca th hm s cho l 22= +y x m m.

V d 2: Cho hm s = +3 23 2y x x mx Tm m hm s c cc i, cc tiu v ng thng i qua cc im cc tr song song vi ng thng d: y = 4x + 3.

Hng dn gii : Ta c 23 6 = y x x m Hm s c cc i, cc tiu khi y = 0 c hai nghim phn bit ( )0 9 3 0 3, * > + > > m m

Chia y cho y ta c 1 1 2 2 23 3 3 3

= + +

mxy

mx y

Gi ( ) ( )1 1 2 2; , ;A x y B x y l cc im cc tr, khi phng trnh ng thng qua A, B l

( ) 2: 2 23 3

= + +

m my x

Theo bi ta c

22 4

3 3

2 33

/ / : 4 3

+ = = = +

m

mm

d y x

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i chiu vi (*) ta c m = 3 l gi tr cn tm.

V d 3: Cho hm s = +3 23 2y x x mx Tm m hm s c cc i, cc tiu v cc im ny cch u ng thng (d): y = x 1.

Hng dn gii : Ta c 23 6 = y x x m Hm s c cc i, cc tiu khi y = 0 c hai nghim phn bit ( )0 9 3 0 3, * > + > > m m

Chia y cho y ta c 1 1 2 2 23 3 3 3

= + +

mxy

mx y

Gi ( ) ( )1 1 2 2; , ;A x y B x y l cc im cc tr, khi phng trnh ng thng qua A, B l

( ) 2: 2 23 3

= + +

m mAB y x

Cc im cc tr cch u ng thng (d) : y = x 1 nn xy ra 1 trong 2 trng hp: TH1: ng thng i qua 2 im cc tr song song hoc trng vi ng thng (d)

2 32 1 ,

3 2 + = =

m m (tha mn)

TH2: Trung im I ca AB nm trn ng thng ( ) 1 2 1 21 12 2

+ + = = I Iy y x x

d y x

( ) ( )1 2 1 22 2 2

2 2 2 2 3 .2 6 03 3 3 3

+ + + = + + = =

m m m mx x x x m

Vy 3

0;2

= = m m l cc gi tr cn tm.

V d 4: Cho hm s = +3 23y x x mx Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua (d): x 2y 5 = 0.

Hng dn gii :

Ta c 23 6 = +y x x m Hm s c cc i, cc tiu khi y = 0 c hai nghim phn bit ( )0 9 3 0 3, * > >



Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng 1 5

:2 2

= d y x

/s : m = 0

Bi 4: Cho hm s 3 2 33 4= +y x mx m Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng d : y = x.

/s : 2

.2

= m

Bi 5: Cho hm s 3 23( 1) 9 2= + + + y x m x x m

Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng 1

:2

=d y x

/s : m = 1

Bi 6: Cho hm s 3 23= +y x x mx Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng : 2 5 0 =d x y /s : m = 0

Bi 7: Cho hm s 3 2 7 3= + + +y x mx x Tm m hm s c cc i, cc tiu v ng thng i qua cc i, cc tiu vung gc vi ng thng

: 3 7 0. =d x y

/s : 3 10

.2

= m

Bi 8: Cho hm s 3 2 2 23( 1) (2 3 2)= + + +y x m x m m x m m Tm m hm s c cc i, cc tiu v ng thng i qua cc i, cc tiu to vi ng thng

: 4 20 0+ =d x y gc 450.

/s : 3 15

.2

=m

Bi 9: Cho hm s 3 23 2= +y x x Tm m hm s c cc i, cc tiu v ng thng i qua cc i, cc tiu tip xc vi ng trn

2 2( ) : ( ) ( 1) 5 + =C x m y m .

/s : 4

2; .5

= = m m

MT S BI GII MU V CC TR HM BC BA CHN LC

Bi 1: Cho hm s 3 21 1

( 1) 3( 2)3 3

= + +y x m x m x

Tm m hm s c cc i, cc tiu ti x1 ; x2 sao cho 1 22 1.+ =x x

Gii :

TX : D = R

Ta c ( ) ( )2' 2 1 3 2y x m x m= + hm s c cc i cc tiu ti x1 ; x2 th phng trnh ' 0y = c 2 nghim phn bit x1 ; x2 hay

( ) ( )2 2' 1 3 2 5 7 0m m m m = = + > lun ng vi mi m

Theo nh l Vite ta c: ( )

( )1 2

1 2

2 1

3 2

x x m

x x m

+ =

=

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Nh vy ta c h:

( ) ( )( ) ( )

( )

1 2

1 2

1 2

2 1 1

3 2 2

2 1 3

x x m

x x m

x x

+ =

= + =

T ( ) ( )1 3v ta d dng gii c 21

3 2

4 5

x m

x m

= =

Thay vo ( )2 ta c:( ) ( ) ( ) ( )14 5 3 2 3 2 19 7316

m m m m = =

Vy ( )1 19 7316

m= l gi tr cn tm.

Bi 2: Cho hm s 3 2( 2) ( 1) 23

= + + +my x m x m x

Tm m hm s c cc i, cc tiu ti x1 ; x2 sao cho 1 2 1.<



Bi 4: Cho hm s 3 2 21 1

( 3)3 2

= + y x mx m x

Tm m hm s c cc i, cc tiu ti x1 ; x2 dng sao cho 2 21 2

5.

2+ =x x

Gii : TX : D = R

Ta c 2 2' 3y x mx m= + hm s c cc i, cc tiu ti x1 ; x2 dng th PT

2 2' 3 0y x mx m= + = c 2 nghim dng phn bit

x1 ; x2

( )2 21 2

21 2

4 3 0

0 3 2

3 0

m m

x x m m

x x m

= > + = > <

Ta c : ( ) ( )22 2 2 21 2 1 2 1 2 5 72 2 3 2 2x x x x x x m m m+ = + = = =

Kt hp K ta c 7

2m= l gi tr cn tm.

Bi 5: Cho hm s 3 3 2= +y x mx Tm m hm s c cc i, cc tiu ti A, B sao cho din tch tam gic ABC bng 3 2 , vi C(1 ; 1).

Ta c : 2 2' 3 3 0y x m x m= = = . hm s c cc i, cc tiu ti A, B 0(*)m >

Khi : 2 2 ( ; 2 2)

2 2 ( ;2 2)

x m y m m A m m m

x m y m m B m m m

= = + +

= = + +

T o ta c pt ng thng qua A,B l: 2 2 0mx y+ =

Theo bi ra ta c 32

2 11 1. ( ; ) 4 16 . 3 2

2 2 4 1ABC

mS AB d C AB m m

m

= = + =

+

21 4 . 2 1 3 2 (2 1) 18 2( *)2

m m m m m tm = = =

Vy m = 2 l gi tr cn tm.

Bi 6: Cho hm s 3 23( 1) 12 3 4= + + +y x m x mx m

Tm m hm s c cc i, cc tiu ti A, B sao cho tam gic ABC nhn O lm trng tm, vi 9

1; .2

C

Ta c : 2 2 2' 3 6( 1) 12 0 2( 1) 4 0 2 2 4 0y x m x m x m x m x mx x m= + + = + + = + =

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

2

xx x m x m x x m

x m

= = = =

hm s c cc i, cc tiu ti A, B 1(*)m khi :

Ta c: 3 2 3 2

2 9 (2;9 )

2 4 12 3 4 (2 ; 4 12 3 4)

x y m A m

x m y m m m B m m m m

= = = = + + + +

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V tam gic ABC nhn O lm trng tm nn 3 2

2 2 1 01

( *)929 4 12 3 4 0

2

+ = = + + =

mm tm

m m m m

Vy 1

2= m l gi tr cn tm

Bi 7: Cho hm s 3 2 32 3( 1) 6= + + +y x m x mx m Tm m hm s c cc i, cc tiu ti A, B sao cho 2.=AB

Ta c : 21

' 6 6( 1) 6 0 ( 1)( ) 0x

y x m m x x mx m

== + + = = =

hm s c cc i, cc tiu ti A, B 1(*)m khi :

Ta c: 3 3

3

2 2

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

3 ( ;3 )

x y m m A m mAB m m

x m y m B m m

= = + + =

= =

2 6 2 02 ( 1) ( 1) 2 ( 1) 1 ( *)2

mAB m m m tm

m

== + = = =

Vy m = 0 ; m = 2. l gi tr cn tm

Bi 8: Cho hm s 3 2 2 33 3( 1) 4 1= + + y x mx m x m m Tm m hm s c cc i, cc tiu ti A, B sao cho tam gic OAB vung ti O.

Ta c : 2 2 2 21 1

' 3 6 3( 1) 0 2 11 1

x m x my x mx m x mx m

x m x m

= = + = + = + = = = +

Hm s lun c cc i, cc tiu ti A, B :

Ta c: 3

3

1 ( ) 3 4 1 3 (1 ; 3)

1 ( ) 3 4 1 1 ( 1 ;1 )

x m y x m x m m A m m

x m y x m x m m B m m

= + = + = +

= + = + = + + +

V tam gic OAB vung ti O nn ta c :

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

mOAOB m m m m m m

m

= = + + + + = = =

Vy 1; 2.= =m m

Bi 9: Cho hm s 3 2 3 23( 1) 3 ( 2) 2= + + + + + +y x m x m m x m m Chng minh rng hm s lun c cc tr vi mi m, v khong cch gia cc im cc tr khng i.

Ta c :

3 2 3 2 32

3 2 3 2 3

( ) 3( )' 3 6( 1) 3 ( 2) 0

2 ( ) 3( ) 4

x m y x m x m m m m my x m x m m

x m y x m x m m m m m

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

= = + + + + = +

( y ' 0y = lun c 2 nghim phn bit)

Khong cch gia cc im cc tr l 4 16 2 5AB = + =

/s : 2 5.=AB

Bi 10: Cho hm s 3 2 21

( 1) 13

= + +y x mx m x

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Tm m hm s c cc i, cc tiu v yC + yCT > 2.

2 2 2 1' 2 1 0 ( ) 11

x my x mx m x m

x m

= += + = = = +

Hm s lun c cc i, cc tiu ti A, B

Ta c:

3 33

3 33

1 11 ( ) 1

3 3 3

1 51 ( ) 1

3 3 3

m mx m y x m x m

m mx m y x m x m

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

Ta c: yC + yCT > 2 3

3 32 2 2 2 2 6 03 3 0

mmm m m

m

> + > >

< < + + >

< (*).

Khi gi 1 1 2 2( ; ), ( ; )A x y B x y l ta 2 im cc tr.

Thc hin php chia y cho y ta c : 21 1 10 2 4 11 2

'3 9 9 9

m m m my x y x

+ + = + +

Ta c: 2 2

1 1 1 1 1

1 1 10 2 4 11 2 10 2 4 11 2'( )

3 9 9 9 9 9

m m m m m m my x y x x x

+ + + = + + =

2 2

2 2 2 2 2

1 1 10 2 4 11 2 10 2 4 11 2'( )

3 9 9 9 9 9

m m m m m m my x y x x x

+ + + = + + =

Vy vi k (*) ng thng i qua cc i, cc tiu ca hm s l : 210 2 4 11 2

9 9

m m my x

+=

b, Tng t cu a ta c: 22y x m m= + vi m R

Bi 12: Cho hm s 3 2 2(2 1) ( 3 2) 4= + + + y x m x m m x a) Tm m hm s c cc i, cc tiu.

b) Tm m hm s c cc i, cc tiu v cc im ny nm khc pha vi Oy.

Ta c : 2 2' 3 2(2 1) 3 2y x m x m m= + + + . hm s c 2 cc tr ' 0y = c 2 nghim phn bit.

2 2 2'

5 30' 0 (2 1) 3( 3 2) 0 10 5 0 (*)

5 30y

mm m m m m

m

> + > + + + > + >

<

Vi k(*) th hm s c 2 im cc tr ti 1 1 2 2( ; ), ( ; )A x y B x y . Theo nh ly Viet: 1 2

2

1 2

2(2 1)

3

3 2

3

mx x

m mx x

+ + =

+ =

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Theo k A, B nm khc pha vi Oy 2

1 2

3 20 1 2

3

m mx x m

+ = < < <

Kt hp (*) kt lun: 1 2m< <

Bi 13: Cho hm s 3 2 23= + +y x x m x m

Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng 1 5

:2 2

= d y x

Ta c 2 2' 3 6y x x m= + . hm s c 2 cc tr ' 0y = c 2 nghim phn bit

2'' 0 9 3 0 3 3y m m > > < < (*) .

Vi k(*) th hm s c 2 im cc tr ti 1 1 2 2( ; ), ( ; )A x y B x y .

+ Thc hin php chia y cho y ta c : 2 21 1 2 3

' 23 3 3 3

m m my x y x

+ = + +

+ T ta c pt ng thng qua cc im cc tr l: 2 22 3

23 3

m m my x

+= +

.

V A, B i xng nhau qua 1 5

:2 2

= d y x nn : . 1d AB

I d

k k

= ( vi I l trung im ca AB)

Gii 21 2

. 1 2 1 02 3

d ABm

k k m

= = =

( tha mn (*))

Vi 0m= ta c (0;0), (2; 4) (1; 2)A B I d

Vy 0m= l gi tr cn tm

(ch : y cc e c th gii c 2 k . 1

=

d AB

I d

k k ngn gn ta nn gii 1 k l . 1=

d ABk k v th li k

kia)

Bi 14: Cho hm s 3 2 33 4= +y x mx m Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng d : y = x.

Ta c : 20

' 3 6 02

== = =

xy x mx

x m. hm s c cc i cc tiu ti A,B 0m

Khi gi 2

3

3

(1; 2 )(0;4 ), (2 ;0)

( ;2 )

=

ABu m

A m B mI m m

( vi I l trung im AB)

V A,B i xng nhau qua :d y x= nn : 2

2

2 1

. 1 22 1 0

= = = =

d AB

I d m mm

u u m

Bi 15: Cho hm s 3 23( 1) 9 2= + + + y x m x x m

Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng 1

:2

=d y x

/s : m = 1

Bi 16: Cho hm s 3 23= +y x x mx

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Tm m hm s c cc i, cc tiu v cc im ny i xng nhau qua ng thng : 2 5 0 =d x y

/s : m = 0

Bi 17: Cho hm s 3 3= +y x mx m Tm m hm s c cc i, cc tiu. Khi chng minh rng cc im ny nm v hai pha ca trc Oy.

Ta c : 2 2' 3 3 0y x m x m= = = . hm s c 2 cc tr ' 0y = c 2 nghim phn bit 0(*)m >

Vi k(*) th hm s c 2 im cc tr ti 1 1 2 2( ; ), ( ; )A x y B x y . Theo nh ly Viet: 1 2

1 2

0x x

x x m

+ = =

Theo k A, B nm khc pha vi Oy 1 2 0( 0)x x m m = < >

Kt lun: Vi m>0 hm s c C, CT v cc im nm v 2 pha vi trc Oy

Bi 18: Cho hm s 3 23 2= +y x x mx Tm m hm s c cc i, cc tiu v ng thng i qua cc i, cc tiu song song vi ng thng

: 4 3 0.+ =d x y

Gii:

TX: R

o hm: 2' 3 6y x x m=

( )2' 0 3 6 0xy x x m= = ;

2' 3 3 3 9m m = + = +

' 0 3m > > . Nn 3m> th hm s c cc i, cc tiu.

S dng php chia a thc, ta c ( ) ( ) ( )22 12 2 3 6 . '3 3 3 3 = + + + = +

x x

m m xy x x x m r q y

T suy ra, phng trnh ng thng qua cc i, cc tiu l ( )2

2 23 3

= = + +

x

m my r x ( )

ng thng : 4 3d y x= +

Yu cu tr thnh song song d hay

22 4

33 39

2 33

+ = = =

mm

mmm

(tha mn)

Vy 3m= l gi tr cn tm.

Bi 19: Cho hm s 3 2 7 3= + + +y x mx x Tm m hm s c cc i, cc tiu v ng thng i qua cc i, cc tiu vung gc vi ng thng

: 3 7 0. =d x y

Gii: TX: R

o hm: 2' 3 2 7y x mx= + +

( )2' 0 3 2 7 0xy x mx= + + = ;

2' 21m =

21' 0

21

> >

<

m

m. Nn ( ) ( ); 21 21; +m th hm s c cc i, cc tiu.

S dng php chia a thc, ta c

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( ) ( ) ( )2

214 2 73 3 2 7 . '3 9 9 3 9

= + + + + + = +

x x

m m x my x x mx r q y

T suy ra, phng trnh ng thng qua cc i, cc tiu l 214 2 7

33 9 9

= +

m my x ( )

ng thng : 3 7d y x=

Yu cu tr thnh vung gc d hay 2 214 2 2 3 10

3. 1 153 9 3 2

= = =

m mm (tha mn)

Vy 3 10

2= m l gi tr cn tm.

Bi 20: Cho hm s 3 2 2 23( 1) (2 3 2)= + + +y x m x m m x m m Tm m hm s c cc i, cc tiu v ng thng i qua cc i, cc tiu to vi ng thng

: 4 20 0+ =d x y gc 450.

Gii: TX: R

o hm: ( ) ( )2 2' 3 6 1 2 3 2= + +y x m x m m ; ( ) ( )2 2 2' 9 1 3 2 3 2 3 9 3 = + = +m m m m m 3 5

2' 0 .3 5

2

+> >



2' 3 6= y x x ( )' 0 0 = =xy x hoc 2x =

Suy ra 2 im cc tr ln lt l ( )0;2A v ( )2; 2B Phng trnh ng thng qua hai im cc tr: : 2 2 0d x y+ =

ng trn ( )( )tm ; 1

:bn knh 5

I m mC

R

+

=

Yu cu:d tip xc( )C hay ( )( ) 2 1 2; 55

+ + = =

m md d C R

23 1 5 4

3

= = =

mm

m (t/m)

Vy 2m= v 43

m= l cc gi tr cn tm

Bi 22: Cho hm s 3 2 2 22( 1) ( 4 1) 2( 1)= + + + +y x m x m m x m Tm m hm s c cc i, cc tiu v ng thng i qua cc i, cc tiu vung gc vi ng thng

9: 5.

2= +d y x

Gii:

o hm ( ) ( )2 2' 3 4 1 4 1= + + +y x m x m m Hm s c cc i, cc tiu ' 0y = c hai nghim phn bit.

Hay ( ) ( )2 2 2 2 3' 0 4 1 3 4 1 0 4 1 02 3

> + > + > + + >

>

mm m m m m

m

Nn ( ) ( ); 2 3 2 3;m + + th hm s c cc i, cc tiu. S dng php chia a thc, ta c

( ) ( ) ( )2 3 2

2 22 8 2 2 8 10 16 2 2 3 4 1 4 1 . '9 9 3 9

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

x x

m m m m m x my x x m x m m r q yT

suy ra, PTT qua cc i, cc tiu l ( )

2 3 22 8 2 2 8 10 16

9 9

+ + + + += =

x

m m m m my r x ( )

Yu cu tr thnh vung gc d hay 2

2 09 2 8 2. 1 442 9

= + + = + = =

mm mm m

m (tha mn)

Vy 0m= v 4m= l cc gi tr cn tm.

Bi 23: Cho hm s 3 23 3( 6) 1= + + +y x mx m x Tm m im A(3 ; 5) nm trn ng thng i qua cc im cc i, cc tiu.

Gii:

o hm: ( )2' 3 6 3 6y x mx m= + + . ( ) ( )2 2' 9 9 6 9 6m m m m = + =

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3' 0

2

m

m

> > <

. Nn ( ) ( ); 2 3;m + th hm s c cc i, cc tiu.

S dng php chia a thc, ta c

( ) ( ) ( ) ( ) ( )2 2 22 2 12 6 1 3 6 3 6 . '3 3 x xx m

y m m x m m x mx m r q y = + + + + + + + + = +

T suy ra, PTT qua cc i, cc tiu l ( ) ( )2 22 6 6 1y m m x m m= + + + + ( ) Yu cu tr thnh ( ) i qua A hay ( ) ( )2 2 25 3. 2 2 12 6 1 5 12 37 5m m m m m m= + + + + + + + =

24

5 12 32 0 8

5

= = =

mm m

m

Kt hp cc iu kin trn, ta c 4m= l gi tr cn tm.

Bi 24: Cho hm s 3 21

3 3= + + + my x mx x

Tm m hm s c cc i, cc tiu v cc im ny nm cng pha vi ng thng d : 2x + y = 0.

Gii:

o hm: 2' 2 1y x mx= + + .

Hm s c cc i, cc tiu 2' 0 1 0 1m m > > >

Nn ( ) ( ); 1 1;m + th hm s c cc i, cc tiu.

Khi , gi ta ca hai im cc tr ln lt l ( )1 1;x y , ( )2 2;x y .

Theo vi-t, ta c 1 2

1 2

2

. 1

x x m

x x

+ = =

Yu cu tr thnh ( )( )1 1 2 22 2 0x y x y+ + > ( )1 Mt khc, ta c:

( ) ( ) ( )2

22 2 2 1 . '3 3 3 x x

m x my x x mx r q y

= + + + + = +

Nn PTT qua cc i, cc tiu l 22 2

3

my x

=

( )22 2 2 2

1 1 2 2 1 2

2 2 2 2 1 11 2 2 0 4 1 0 1 0 2

3 3 3 3

m m m mx x x x x x m

+ + > + > +

Kt hp iu kin ta c 1

2

m

m

>

l cc gi tr cn tm


3= + + +y x x mx m

Tm m hm s c cc i, cc tiu v khong cch gia hai im ny bng 2 15.

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Gii:

o hm: 2 2y x x m= + + . Ta c ' 0 1 0 1m m > > <

Nn 1m< th hm s c cc i, cc tiu. Gi ta hai im cc tr l ( )1 1;x y v ( )2 2;x y .

Khi , theo vi-t 1 1

1 2

2x x

x x m

+ = =

.

Mt khc, ta c

( ) ( ) ( )22 2 2 1 2 . '3 3 3 3 x xm m x

y x x x m r q y = + + + + + = +

Nn PTT qua cc i, cc tiu l 2 2 2

3 3

m my x

= +

Yu cu tng ng ( ) ( ) ( ) ( )2

2 2 2 2

1 2 1 2 1 2 1 2

260 60

3

mx x y y x x x x

+ = + =

( ) ( )2 2

2 3 21 2

2 2 4 8 131 60 . 4 4 60 4 12 21 122 0 2

3 9

m m mx x m m m m m

+ + = = + + = =


Bi 26: Cho hm s 3 22 3( 1) 6 (1 2 )= + + y x m x m m x Tm m hm s c cc i, cc tiu v cc im ny nm trn ng thng d : 4x + y = 0.

Gii:

TX: R

o hm: ( ) ( )2' 6 6 1 6 1 2y x m x m m= + +

( ) ( ) ( ) ( )2 22' 9 1 36 1 2 9 9 6 1 9 3m m m m m m = = + = . 1' 03

> m

Nn 1

3m th hm s c cc i, cc tiu.

S dng php chia a thc, ta c:

( ) ( ) ( ) ( ) ( )2 3 213 1 4 6 2 6 6 1 6 1 2 . '

3 6 x xx m

y m x m m m x m x m m r q y = + + + + + + = +

T suy ra, PTT qua cc i, cc tiu l ( ) ( )2 33 1 4 6 2xy r m x m m m= = + +

ng nht vi d c ( )2

3

11;

3 1 4 3 114 6 2 0 0; 1;2

= = = = + = = = =

m mm

mm m m m m m

(tha mn)



2 33

= +y x x x

Gi A, B l hai im cc tr ca hm s. Tm im M trn Ox sao cho tam gic ABM c din tch bng 2.

Gii:

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TX: R

o hm: 2' 4 3y x x= + . ' 0 1y x= = hoc 3x = .

T suy ra hai im cc tr ca hm s: 4

1;3

A

v ( )3;0B

( )2

2 4 2 133 1 0

3 3AB

= + =

. ng thng : 2 3 6 0AB x y+ = .

Gi ( );0M m l im cn tm.

Yu cu: ( ) ( ) 02 61 132 . ; 2 . 2 3 362 3 13

mmdt ABM AB d M AB m

m

= = = = = =

Vy ( )0;0M v ( )6;0M l cc im cn tm. Ch : Cng thc chia nhanh a thc bc 3:

Vi a thc ( ) ( ) ( )3 2

. '= + + + = +x x xP ax bx cx d r q P vi 0a

Th lun c ( )

( )

22 2

3 9 9

3 9

= + =

x

x

c b bcr x d

a a

x bq

a

DNG 2. CC TR CA HM A THC BC BN TRNG PHNG

Xt hm s ( )4 2 3 2 20

4 2 2 2 0

2

xy ax bx c y ax bx x ax b b

xa

== + + = + = + = =

DNG 1. TM IU KIN V S CC TR CA HM S

Hm s c mt cc tr khi y ch i du mt ln, tc l 02

ba

Hm s c mt cc tr khi y ch i du ba ln, tc l y = 0 c ba nghim phn bit 02

>ba

V d 1: Cho hm s = + 4 22 3 1y x mx m Tm m a) hm s c 1 cc tr. b) hm s c 3 cc tr.

Hng dn gii :

Ta c ( )3 2 20

4 4 4 0=

= = = =

xy x mx x x m y

x m

a) Hm s c mt cc tr khi m 0. b) Hm s c ba cc tr khi m > 0.

V d 2: Cho hm s ( )= + + 4 21 3 3 5y m x mx m Bin lun theo m s cc tr ca hm s cho.

Hng dn gii :

Ta c ( ) ( )3 2

2

04 1 6 2 ( 1) 3 0

( 1) 3 , 1

= = + = + = +

xy m x mx x m x m y

m x m

TH1 : 1 6 ; 0 0= = = =m y x y x Trong trng hp ny hm s c mt cc tr, v l im cc tiu.

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TH2 : ( ) 2 31, 11

=+m

m xm

+ Hm s c mt cc tr khi 3

0 1 01

< +m

mm

+ Hm s c ba cc tr khi 03

011

>> < +

mm

mm

Kt lun : Hm s c mt cc tr khi 1 0 m

Hm s c ba cc tr khi 0

1

> <

m

m

BI TP LUYN TP: Bin lun theo m s cc tr ca cc hm s sau :

a) 4 22 (2 1) 3.= + + +y x m x m

b) 4 2(1 ) (3 1) 2 5.= + + +y m x m x m

c) 2 4 2 3(3 2) 1.= + y m x mx m

DNG 2. TNH CHT CC TR CA HM S

TH1: Hm s c ba im cc tr A, B, C.

+ Tm iu kin tn ti ba im cc tr : ( )0 *2

>ba

+ Vi iu kin (*) ta c 2

3

0

02

2

= = = = = = =

A A

B B

C C

x x y

by x x y

a

bx x y

a

, t ( )0; ; ; ; ;2 2

A B C

b bA y B y C y

a a

Do hm chn vi x nn cc im B, C c yB = yC. Nhn xt : A Oy, B ; C i xng nhau qua Oy nn tam gic ABC lun l tam gic cn ti A. Ta xt mt s tnh cht c bn thng gp ca hm s :

Tnh cht 1: 3 im cc tr to thnh mt tam gic vung cn. Do tam gic ABC cn ti A nn ch c th vung cn ti nh A.

Khi o ta c iu kin ( ). 0, 1= AB AC

vi ; ; ;2 2

= =

B A C A

b bAB y y AC y y

a a

T ( ) ( )21 . 0 02

= + =

B A

bAB AC y y

a

Gi tr m tm c kt hp vi iu kin tn ti (*) cho ta kt qu cui cng ca bi ton. Ngoi ra ta cng c th dng iu kin Pitago cho tam gic cn

ABC : 2 2 2 2 22+ = =AB AC BC AB BC

Tnh cht 2: 3 im cc tr to thnh mt tam gic u.

Tam gic ABC u khi ( )2 2, 2= =AB BC AB BC

vi ; ; 2 ;02 2

= =

B A

b bAB y y BC

a a

T ( ) ( )2 222

+ =B Ab b

y ya a

Gi tr m tm c kt hp vi iu kin tn ti (*) cho ta kt qu cui cng ca bi ton.

Tnh cht 3: 3 im cc tr to thnh mt tam gic c mt gc bng 1200

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Tam gic ABC cn ti A nn 0120=BAC . Gi H l trung im ca ( )0; BBC H y

Ta c ( )0 2 2cos cos60 2 4 , 3= = = =AH AHHAB AB AH AB AHAB AB

vi ( ); ; 0;2

= =

B A B A

bAB y y AH y y

a, t ( ) ( ) ( )2 23 4

2

+ = B A B Ab

y y y ya


Tnh cht 4: 3 im cc tr to thnh mt tam gic c din tch S = So cho trc

Gi H l trung im ca ( )0; BBC H y . Khi ( )2 2 21

. 2 . 4 . , 42

= = =ABC o oS AH BC S AH BC S AH BC

vi ( )2 ;0 ; 0;2

= =

B A

bBC AH y y

a, t ( ) ( )223 4 .4

2

=

o B A

bS y y

a


Tnh cht 5: 3 im cc tr to thnh mt tam gic c bn knh ng trn ngoi ti p R cho trc

S dng cng thc din tch tam gic 2. .

14 4 24. .2

= = = =abc abc AB AC BC ABS R R RR S AHAH BC

Gii phng trnh trn ta c gi tr ca m, i chiu vi (*) cho ta kt lun cui cng.

Tnh cht 6: 3 im cc tr to thnh mt tam gic c trng tm G(0 ; ) cho trc

Ta c iu kin trong trng hp ny l 2 33

+ += + =A B C A B

y y yy y

V d 1: (H khi B - 2011) Cho hm s = + +4 22( 1)y x m x m, vi m l tham s. Tm m th hm s cho c ba im cc tr A, B, C sao cho OA = BC, vi O l gc ta , A l im cc tr thuc trc tung, B v C l hai im cc tr cn li.

Hng dn gii :

Ta c 3 22

04 4( 1) 4 ( 1) 0

1

= = + = + = = +

xy x m x x x m y

x m

Hm s c ba im cc tr khi phng trnh y = 0 c ba nghim phn bit ( )1 0 1, * + > > m m

Vi m > 1 th 1 1

22 2

23 3

0

0 1 ( 1)

1 ( 1)

= =

= = + = + + = + = + +

x y m

y x m y m m

x m y m m

Theo bi ta c ta cc im cc tr l ( ) ( ) ( )2 20; , 1; 1 , 1; 1+ + A m B m m m C m m m T ( )2 2 2 2 2 2 24 1 4 4 0

2 2 2

= += = = + =

=

mOA BC OA BC m m m m

m

Kt hp vi iu kin (*) ta c 2 2 2= m l cc gi tr cn tm. V d 2: (D b khi B - 2003) Cho hm s = +4 2 22 1y x m x , vi m l tham s. Tm m th hm s cho c ba im cc tr l ba nh ca mt tam gic vung cn.

Hng dn gii :

Ta c 3 2 2 22 2

04 4 4 0

= = = = =

xy x m x x x m y

x m

Hm s c ba im cc tr khi phng trnh y = 0 c ba nghim phn bit ( )2 0 0, * > m m

Vi m 0 th ( ) ( ) ( )1 1

4 4 42 2

43 3

0 1

0 1 0;1 , ;1 , ;1

1

= = = = = = =

x y

y x m y m A B m m C m m

x m y m

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Ta nhn thy tam gic ABC lun cn ti A. ABC vung cn th phi vung cn ti A.

T suy ra ( ) ( )4 4 2 8 2 6 0. 0 ; . ; 0 0 ( 1) 01

= = = + = = =

mAB AC AB AC m m m m m m m m

m

Kt hp vi iu kin (*) ta c 1= m l cc gi tr cn tm. V d 3: Cho hm s = + 4 22 1y x mx m , vi m l tham s. Tm m hm s c ba im cc tr ng thi cc im cc tr ca th to thnh mt tam gic a) c din tch bng 4 2 . b) u. c) c mt gc bng 1200

Hng dn gii :

Ta c ( )3 2 20

4 4 4 0=

= + = + = =

xy x mx x x m y

x m

Hm s c ba im cc tr khi phng trnh y = 0 c ba nghim phn bit, tc l m < 0, (*) Vi m < 0 th

( ) ( ) ( )2 2 22

0 1

0 1 0; 1 , ; 1 , ; 1

1

= = = = = = =

x y m

y x m y m m A m B m m m C m m m

x m y m m

Ta nhn thy A thuc Oy, B ; C i xng qua Oy nn tam gic ABC cn ti A.

a) Gi H l trung im ca ( )20; 1 BC H m m Khi , ( )2 21 . 4 2 . 8 2 . 128, 1

2= = = =ABCS AH BC AH BC AH BC

Ta c ( ) ( )22 ;0 ; 0; ,= = BC m AH m t ( ) 4 51 4 . 128 32 2 = = = m m m m i chiu vi iu kin (*) ta thy m = 2 l gi tr cn tm. b) Tam gic ABC u khi ( )2 2, 2= =AB BC AB BC

Ta c ( ) ( )2; , 2 ;0 ,= = AB m m BC m t ( ) 4 4 30

2 4 33

= + = =

=

mm m m m m

m

i chiu vi iu kin (*) ta c 3 3= m l gi tr cn tm. c) Tam gic ABC cn ti A nn c mt gc bng 1200 th 0120=BAC Gi H l trung im ca ( )20; 1 BC H m m Trong tam gic vung HAB c

( )0 2 23sin sin 60 3 2 3 , 32

= = = = = =BH BHHAB AB BH BC AB BCAB AB

Ta c ( ) ( )2; , 2 ;0 ,= = AB m m BC m khi ( ) ( )43

0

3 3 4 1

3

= + = =

m

m m mm

i chiu vi iu kin (*) ta c 3

1

3= m l gi tr cn tm.

V d 4: Cho hm s = + 4 22 1y x mx m , vi m l tham s. Tm m hm s c ba im cc tr ng thi cc im cc tr ca th to thnh mt tam gic c bn knh ng trn ngoi tip bng 2.

Hng dn gii :

Ta c ( )3 2 20

4 4 4 0=

= = = =

xy x mx x x m y

x m

Hm s c ba im cc tr khi phng trnh y = 0 c ba nghim phn bit, tc l m > 0, (*)

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Vi m > 0 th

( ) ( ) ( )2 2 22

0 1

0 1 0; 1 , ; 1 , ; 1

1

= = = = = + + + = = +

x y m

y x m y m m A m B m m m C m m m

x m y m m

Ta nhn thy A thuc Oy, B ; C i xng qua Oy nn tam gic ABC cn ti A.

Gi H l trung im ca ( )20; 1 + BC H m m Din tch tam gic ABC : ( )

2. . ., 1

2 4 2= = =ABC

AH BC AB BC AC ABS R

R AH

Ta c ( ) ( )2 4

2 2

2; ; 0;

= += = =

AB m mAB m m AH m

AH m

Khi , ( ) ( )( )4

3 22

1

1 2 2 1 0 1 1 0 1 5

2

=+ = + = + = =

mm m

m m m m mm m

i chiu vi iu kin (*) ta c 5 1

1;2

= =m m l cc gi tr tha mn yu cu bi ton

BI TP LUYN TP: Bi 1: Cho hm s 4 24 2 1= + +y x mx m , vi m l tham s. Tm m hm s c ba im cc tr ng thi cc im cc tr ca th to thnh mt tam gic

a) c din tch bng 3 2.

b) c trng tm l 2

0; .3

G

c) c bn knh ng trn ngoi tip bng 1. Bi 2: Tm m hm s 4 2 22 1= +y x m x c ba im cc tr A, B, C sao cho a) tam gic ABC u. b) 2 ,=OA BC trong O l gc ta , A l im cc tr thuc Oy, B ; C l hai im cc tr cn li. Bi 3: Tm m hm s ( )4 2 22 2 5 5= + + +y x m x m m c ba im cc tr v l ba nh ca mt tam gic vung cn. /s : m = 1. Bi 4: Tm m hm s 4 2 22= + + +y x mx m m c ba im cc tr ng thi cc im cc tr ca th to thnh mt tam gic c mt gc bng 1200.

/s : .= 3

1

3m

Bi 5: Cho hm s 4 2 42 2= + +y x mx m m c th (Cm) . Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c din tch bng 4.

/s : .= 5 16m Bi 6: Cho hm s 4 2 22 1= +y x m x (1) Tm m th hm s (1) c ba im cc tr A, B, C v din tch ca tam gic ABC bng 32. Bi 7: Cho hm s 4 22 1= + y x mx m c th (Cm) . Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c bn knh ng trn ngoi tip bng 1 .

/s : ; .= = 5 11

2m m

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DNG 1. TIP TUYN TI MT IM THUC TH HM S

Cng thc : Phng trnh tip tuyn ti im ( ) ( ) ( ); :o oM x y C y f x = l ( ) ( ) ( ) ( ) ( )o oo o o ox xy y x x y y y x x f x = + = + Cc lu :

+ Nu cho xo th tm yo = f(xo). + Nu cho yo th tm xo bng cch gii phng trnh f(x) = yo.

+ Tnh y = f(x). Suy ra y(xo) = f(xo). + Phng trnh tip tuyn l: y = f(xo).(x xo) + yo. Cc v d in hnh:

V d 1. Cho hm s y = x3 3x2 + 2. Vit phng trnh ti p tuyn trong cc trng hp sau: a) tip tuyn ti giao im ca th v trc Ox. b) tip tuyn ti giao im ca th v trc Oy. c) tip tuyn ti im c honh x = 2. d) tip tuyn ti im c tung y = 2.

Hng dn gii : a) Phng trnh honh giao im ca th v trc Ox :

( )( )3 2 2 13 2 0 1 2 2 01 3

xx x x x x

x

= + = =

=

Ta c 23 6 .y x x = Vi ( ) ( )1; 0 1 3 6 3 : 3 1 3 3.x y y y x y x= = = = = = + Vi ( ) ( ) ( ) ( )21 3; 0 1 3 3 1 3 6 1 3 3 4 2 3 6 6 3 6x y y= + = + = + + = + =

( ): 6 1 3 6 6 6 3.y x y x = = Vi ( ) ( ) ( ) ( )21 3; 0 1 3 3 1 3 6 1 3 3 4 2 3 6 6 3 6x y y= = = = + =

( ): 6 1 3 6 6 6 3.y x y x = + = + b) Phng trnh honh giao im ca th v trc Oy :

3 2 03 2

20

xy x x

yx

= = + ==

Ta c ( )23 6 2 12 12 0y x x y = = = T ta c phng trnh tip tuyn l ( ): 0. 2 0 0.y x y = + = c) Thay x = 2 vo hm s ta c y = 2. Ta c ( )23 6 2 12 12 24y x x y = = + = T ta c phng trnh tip tuyn l ( ): 24 2 2 24 46.y x y x = + = +

d) Vi y = 2 ta c ( )( )3 2 3 2 2 13 2 2 3 4 0 1 4 4 02

xx x x x x x x

x

= + = + = + + = =

Ta c ( )( )

22 12 12 24

3 61 3 6 9

yy x x

y

= + = =

= + =

T ta c phng trnh tip tuyn l ( )

( )1

2

: 24 2 2 24 46.

: 9 1 2 9 7.

y x y x

y x y x

= + = +

= + = +

BI TP LUYN TP: Bi 1: Vit phng trnh tip tuyn ca (C) ti im c ch ra:

Bi 2:

TIP TUYN CA TH HM S

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a) ( ) 3 2: 3 7 1C y x x x= + ti A(0; 1). b) ( ) 4 2: 2 1C y x x= + ti B(1; 0).

c) ( ) 3 4:2 3

xC y

x

+=

ti C(1; 7). d) ( ) 3( 2):1

xC y

x

=

ti im B c yB = 4

e) ( ) 1:2

xC y

x

+=

ti cc giao im ca (C) vi trc honh, trc tung.

f) ( ) 3: 3 1C y x x= + ti im un ca (C).

g) ( ) 4 21 9: 24 4

C y x x= ti cc giao im ca (C) vi trc honh.

Bi 2: Cho hm s 1

.1

=+

xy

x

Vit phng trnh tip tuyn trong cc trng hp sau: a) tip tuyn ti giao im ca th v trc Ox. b) tip tuyn ti giao im ca th v trc Oy. c) tip tuyn ti im c honh x = 2. d) tip tuyn ti im c tung y = 3. e) chng minh rng mi tip tuyn ca th to vi hai tim cn mt tam gic c din tch khng i.

DNG 2. TIP TUYN BIT H S GC K

1) Mt s kin thc c bn v h s gc k ca mt ng thng d :

H s gc ca mt ng thng l tang (tan) ca gc hp bi ng thng v chiu dng trc Ox. K hiu k = tan. Nu ng thng d hp vi trc Ox (khng ni r chiu dng ca trc Ox) th k = tan. ng thng d i qua hai im M, N th h s gc

ca ng d c tnh bi

=

M Nd

M N

y yk

x x

ng thng d i qua im M(x1 ; y1) v c h s gc k th c phng trnh ( )1 1: .= +d y k x x y Trong trng hp tng qut, ng thng d c h s gc k th lun vit dng d: y = kx + m.

Cho hai ng thng 1 1 1

2 2 2

:

:

d y k x m

d y k x m

= + = +

+ d1 v d2 song song vi nhau th c cng h s gc : 1 2

1 2

d dk k

m m

=

+ d1 v d2 vung gc vi nhau th c tch h s gc bng 1 : 1 2 2

1

1. 1 .= = d d d

d

k k kk

o hm ti mt im xo thuc th hm s y = f(x) chnh l h s gc ca tip tuyn vi th ti im . Tc l ( ).=tt ok y x V d 1: Xc nh h s gc k ca cc ng cho di y ?

a) 2 1 2

2 3 1 0 3 2 1 .3 3 3

+ = = + = + = x y y x y x k

b) 1 3 1

5 3 0 5 3 .5 5 5

+ + = = = =x y y x y x k

c) 2 3 0 2 3 2.+ + = = =x y y x k

V d 2: Cho hm s = + +3 23 3.y x x x Mt ng thng d i qua A(2 ; 1) v c h s gc k. Tm k ng thng d v th hm s cho a) ct nhau ti duy nht mt im. b) ct nhau ti ba im phn bit. c) ct nhau ti ba im phn bit c honh dng.

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Hng dn gii : ng thng d qua A(2 ; 1) v c h s gc k nn c dng d : y = k(x 2) + 1. Phng trnh honh giao im ca hai th : 3 2 3 23 3 ( 2) 1 3 2 ( 2) + + = + + + = x x x k x x x x k x

22

2( 2)( 1) ( 2)

( ) 1 0, (1)

= = = =

xx x x k x

g x x x k

a) Hai th ct nhau ti duy nht mt im khi (1) v nghim 5

0 1 4(1 ) 0 .4

< + + < < k k

Vy vi 4

5< k th hai th cho ct nhau ti duy nht mt im.

b) Hai th ct nhau ti ba im phn bit khi (1) c hai nghim phn bit v khc 2.

iu xy ra khi5

0 1 4(1 ) 04

(2) 0 (2) 1 01

> + + > > =

k k

g g kk

Vy vi 4

51

>

k

k th hai th cho ct nhau ti ba im phn bit.

c) Do nghim x = 2 > 0 nn ba giao im c honh dng th (1) phi c hai nghim dng phn bit v khc 2.

Gi hai nghim l x1 ; x2. Khi ta c 1 2

1 2

0 1 01

0 1 0

+ > > < > >

x xk

x x k

Kt hp vi diu kin tn ti ba giao im cu b ta dc 4

15

< < k l gi tr cn tim.

V d 3: Cho hm s = + +3 22 3 1.y x mx mx a) Tm m tip tuyn vi th ti im un song song vi ng thng : 4x + y + 1= 0. b) Tm m tip tuyn vi th ti im x = 2 vung gc vi ng thng : 2x + 3y + 2= 0.

Hng dn gii :

a) Ta c

2

3 26 6

2 3 112 6 0

2

= += + +

= = =

y x mx my x mx mx m

y x m y x

Tip tuyn ti im un c h s gc l 2 23

6. 6 .2 4 2 2

= = + = +

u

m m m mk y m m m

ng thng c h s gc xc nh bi : 4 1 0 4 1 4. + + = = = x y y x k

Tip tuyn ti im un song song vi nn 2

22

34 3 2 8 0 4

23

== + = = =

u

mm

k k m m mm

Vy, vi 4

2;3

= = m m th tip tuyn ti im un ca th song song vi ng thng .

b) Tip tuyn ti x = 2 c h s gc l ( )2 24 12 13 24= = + + = +ttk y m m m

ng thng c h s gc xc nh bi 2 2 2: 2 3 2 0 3 2 2 .3 3 3

+ + = = = = x y y x y x k

Tip tuyn ti im x = 2 vung gc vi nn ( )2 45. 1 13 24 1 26 48 33 26

= + = + = = ttk k m m m

Vy, vi 45

26= m th tip tuyn ti x = 2 vung gc vi .

V d 4: Cho hm s +=

1,

2x

yx

c th l (C). Gi I l giao im ca hai tim cn ca th (C).

Tm im M trn th sao cho tip tuyn vi th ti M vung gc vi ng thng IM . Hng dn gii :

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Ta c 2

1 2 3 3 31

2 2 2 ( 2)

+ + = = = + =

x xy y

x x x x

Gi ( ) ( ) 3 3; 1 ;1 .2 2

= + +

o o o oo o

M x y C y M xx x

Ta c 21

lim21

lim 12

+ = + =

x

x

x

xx

x

, t ng x = 2 l tim cn ng v y = 1 l tim cn ngang.

im I l giao ca hai tim cn nn I(2 ; 1).

Tip tuyn ti M c h s gc l ( ) 23

( 2)= =

tt o ok y x

x

ng thng IM c h s gc 2

31 1

2 3

2 ( 2)

+ = = =

oI M

IMI M o o

xy yk

x x x x

Tip tuyn ti M vung gc vi ng IM khi 2 2

3 3. 1 . 1

( 2) ( 2)= =

tt IM o ok k

x x

2 2 3 2 3( 2) 32 3 2 3

= = + =

= =

o oo

o o

x xx

x x

+ Vi ( )3 32 3 1 1 1 3 2 3;1 32 3

= + = + = + = + + +o o o

x y Mx

+ Vi ( )3 32 3 1 1 1 3 2 3;1 32 3

= = + = + = o o o

x y Mx

Vy c hai im M tha mn yu cu bi ton.

V d 5: Cho hm s ( )=+

2 1, .

1x

y Cx

Tm im M thuc th (C) tip tuyn ca (C) ti M vi ng thng i qua M v giao im hai ng tim cn c tch h s gc bng 9.

Hng dn gii :

Ta c ( )2

3.

1 =

+y

x Gi ( )2 1;

1

+ a

M a Ca

Tip tuyn vi (C) ti M c h s gc: ( )2

3( ) .

1= =

+ttk y a

aGiao im hai ng tim cn I(1; 2).

ng thng IM c h s gc l ( )2

3.

1

= = +

M IIM

M I

y yk

x x a

Theo bi ta c ( ) ( )

( )42 203 3

. 9 . 9 1 121 1

== = + = = + + tt IM

ak k a

aa a

Vy c 2 im M tha mn bi l M(0; 3), M(2; 5).

V d 6: Cho hm s = + 3 22 3.y x x Mt ng thng d i qua M(1 ; 2) v c h s gc k. a) Tm k ng thng d v th hm s cho ct nhau ti ba im phn bit M(1 ; 2) ; A v B. b) Tim k tip tuyn ca th ti hai im A, B vung gc vi nhau.

Hng dn gii : a) ng thng d qua M(1 ; 2) v c h s gc k nn c dng d : y = k(x 1) 2. Phng trnh honh giao im ca hai th : 3 2 3 22 3 ( 1) 2 2 1 ( 1) + = + = x x k x x x k x

22 2

1( 1)( 1) ( 1)

1 ( ) 1 0, (1)

= + + = + + = = + =

xx x x k x

x x k g x x x k

Hai th ct nhau ti ba im phn bit khi (1) c hai nghim phn bit v khc 1.

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Ta c iu kin5

0 1 4( 1) 04

(1) 0 (1) 1 01

> >

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Khi xN ; xP l cc nghim ca phng trnh 2 12 0

2+ = = =

N P

N P

x xx x m

x x m

H s gc ca tip tuyn ti N, P ln lt l k1 v k2 tha mn 2

12

2

3 33 3

= =

N

P

k xk x

Tip tuyn ca (C) ti N v P vung gc vi nhau khi 21 2

3 2 2

3. 1 9 18 1 03 2 2

3

+== + + =

=

mk k m m

m

i chiu vi iu kin ta c 3 2 2

3m

= l cc gi tr cn tm.

V d 9: Cho hm s = +3 2 3 4y x x c th l (C) v ng thng d i qua A(2; 0) c h s gc k. Xc nh k d ct (C) ti ba im phn bit A, M, N sao cho tip tuyn ca (C) ti M v N vung gc vi nhau.

Hng dn gii : Phng trnh ng thng (d): y = k(x 2). Phng trnh honh giao im ca (C) v d: 3 2 23 4 ( 2) ( 2)( 2 ) 0x x k x x x x k + = =

( )22

( ) 2 0, 1Ax x

g x x x k= = = =

d ct (C) ti 3 im phn bit A, M, N th (1) phi c 2 nghim phn bit, khc 2 0 9

0(2) 0 4

kf

> <

(*)

Theo nh l Viet ta c: 1

2M N

M N

x x

x x k

+ = =

Cc tip tuyn ti M v N vung gc vi nhau khi ( ) ( ). 1 . 1M NM N x xk k y y = =

2 2 2 3 2 2(3 6 )(3 6 ) 1 9 18 1 03M M N N

x x x x k k k = + + = =


3m

= l cc gi tr cn tm.

BI TP LUYN TP: Bi 1. Cho hm s 3 2( 2) 3.= + +y x m x mx a) Tm m tip tuyn vi th ti im c honh x = 1 song song vi ng (d): y = 2x 1. b) Tm m tip tuyn vi th ti im c honh x = 0 vung gc vi ng (d): 4x 3y = 0. Bi 2. th hm s y = x4 + 2mx2 2m + 1 Tm m cc tip tuyn vi th ti A(1; 0), B(1; 0) vung gc vi nhau.

Bi 3. Cho hm s 3 23 2,y x x x= + + + c th l (C) v mt ng thng d i qua A(1; 3) c h s gc k. a) Tm k ng thng d ct (C) ti ba im phn bit cng c honh m. b) Tm k d ct (C) ti ba im phn bit A, B, C sao cho tip tuyn vi (C) ti hai im B, C vung gc vi nhau.

Bi 4. Cho hm s y = x4 + mx2 m 1. Tm m tip tuyn vi th ti A song song vi ng thng (d): y = 2x, vi A l im c nh c honh dng ca th hm s.

Bi 5. Cho hm s ( )3 1

.+

=+

m x my

x m

Tm m tip tuyn ti giao im ca th hm s vi trc Ox song song vi ng thng (d): y = x 5.

Bi 6. Cho hm s 2

,1

=x

yx

c th l (C). Gi I l giao im ca hai tim cn ca th (C).

Tm im M trn th sao cho tip tuyn vi th ti M vung gc vi ng thng IM.

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Bi 7. Cho hm s 2 1

,1

=

xy

x c th l (C). Gi I l giao im hai tim cn ca (C). Tm im M thuc (C)

sao cho tip tuyn ca (C) ti M vung gc vi ng thng IM.

/s: M(0; 1) v M(2; 3).

2) Cch vit phng trnh tip tuyn ca th khi bit h bit h s gc k hoc tnh song song, vung gc ca tip tuyn:

Gi M(xo; yo) l ta tip im ca tip tuyn v th (C). Theo ngha hnh hc ca o hm th ta c ( ).= ok y x Gii phng trnh trn ta c nghim xo, thay vo hm tm tung yo Vi cc cp ta (xo; yo) tm c ta c phng trnh tip tuyn cn tm l ( ) .= +o oy k x x y V d: Cho hm s y = x3 3x + 7, c th l (C).

a) Vit phng trnh ti p tuyn vi (C) bit tip tuyn ny song song vi ng thng (d): y = 6x 1

b) Vit phng trnh ti p tuyn vi (C) bit tip tuyn vung gc vi = +1( ) : 2.9

d y x

Hng dn gii : Ta c y = 3x2 3. Gi M(xo; yo) l ta tip im ca tip tuyn v th (C), h s gc ca tip tuyn ti M l

( ) 23 3.= = tt o ok y x x a) Tip tuyn song song vi 2: 6 1 3 3 6 3.= = = = tt d o od y x k k x x

Vi ( )3 7 : 6 3 7 6 7 6 3.= = = + = + o ox y pttt y x y x Vi ( )3 7 : 6 3 7 6 7 6 3.= = = + + = + +o ox y pttt y x y x Vy c hai tip tuyn tha mn yu cu bi l 6 7 6 3.= + y x

b) Tip tuyn vung gc vi 21

: 2 . 1 9 3 3 9 2.9

= + = = = = tt d tt o od y x k k k x x

Vi ( )2 9 : 9 2 9 9 9.= = = + = o ox y pttt y x y x Vi ( )2 5 : 9 2 5 9 23.= = = + + = +o ox y pttt y x y x Vy c hai tip tuyn tha mn yu cu bi l 9 9; 9 23.= = +y x y x

BI TP LUYN TP: Bi 1. Cho hm s y = x3 3x2 + 2 Vit phng trnh tip tuyn vi th bit tip tuyn vung gc vi ng (d): 5y 3x + 4 = 0.

Bi 2. Cho hm s 4 21

2 .4

= +y x x

a) Vit phng trnh tip tuyn ti giao im ca th v trc Ox. b) Vit phng trnh tip tuyn ca th bit tip tuyn song song ng thng y = 15x+ 2012.

Bi 3. Cho hm s 2 1

.2

=+

xy

x

a) Vit phng trnh tip tuyn ti giao im ca th v trc Ox. b) Vit phng trnh tip tuyn ti giao im ca th v trc Oy. c) Vit phng trnh tip tuyn bit tip tuyn song song vi ng thng (d): y = 5x 1. d) Vit phng trnh tip tuyn bit tip tuyn vung gc vi ng thng (d): x + 20y 3 = 0.

DNG 3. TIP TUYN I QUA MT IM N TH HM S

Cho hm s y = f(x) c th l (C). im A(xA ; yA) khng thuc th. Vit vit cc phng trnh tip tuyn k t A n th ta thc hin nh sau : + Gi d l ng thng i qua A v c h s gc k ( ): = +A Ad y k x x y

+ ng thng d l tip tuyn ca th (C) khi h sau c nghim : ( ) ( )

( )( ) , 1

( ), 2

= + =

A Af x k x x y

k f x

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+ Ta gii h phng trnh trn bng cch th (2) ln (1). Gii (1) c x ri thay li vo (2) tm k, t ta c phng trnh dng d chnh l tip tuyn cn tm.

V d 1. Vit phng trnh ti p tuyn trong cc trng hp sau:

a) Bit ti p tuyn i qua

2; 1

3A n th hm s y = x3 3x + 1

b) K t A(0; 4) n th hm s ( )= 222 .y x Hng dn gii :

a) Gi d l ng thng qua 2

; 13

A v c h s gc k 2

: 1.3

=

d y k x

d l tip tuyn ca th hm s y = x3 3x + 1 th h sau c nghim: ( )

( )

3

2

23 1 1, 1

3

3 3, 2

+ =

=

x x k x

k x

Th (2) ln (1) ta c ( )3 2 3 2 023 1 3 3 1 2 2 013

= + = = =

xx x x x x x

x

Vi 2

0 3 : 3 1 3 13

= = = = +

x k d y x y x

Vi 1 0 : 1.= = = x k d y b) Gi d l ng thng qua A(0; 4) v c h s gc k : 4. = +d y kx

d l tip tuyn ca th hm s ( )22 4 22 4 4= = +y x x x th h sau c nghim: ( )

( )

4 2

3

4 4 4, 1

4 8 , 2

+ = +

=

x x kx

k x x

Ta c ( ) 4 2 30

1 44

= = =

xx x kx

k x x

Vi ( )0, 2 0 : 4= = =x k d y

Vi 3

3 3 3 323

04

4 4 4 8 3 4 0 4 24 8

3 3

= = = = = = = =

xk x x

k x x x x x x x xx xk x x

+ Nu x = 0 th ta c d : y = 4.

+ Nu 2 8 8 16 16

: 4.3 3 3 3 3 3 3 3

= = = = +x k d y x

+ Nu 2 8 8 16 16

: 4.3 3 3 3 3 3 3 3

= = + = = +x k d y x

V d 2. Vit phng trnh ti p tuyn k t im ( )1; 2A n th hm s +=2

.2 1x

yx

Hng dn gii : Gi d l ng thng qua A(1; 2) v c h s gc k : ( 1) 2. = d y k x

d l tip tuyn ca th hm s2

2 1

+=

xy

x th h sau c nghim:

( )

( )( )2

2( 1) 2, 1

2 15

, 22 1

+ = =

xk x

x

kx

Thay (2) ln (1) ta c ( )

( )( ) ( ) ( )222 5

( 1) 2 2 2 1 5 1 2 2 1 02 1 2 1

+ = + + + =

xx x x x x

x x

2 110 5 02

= = x x

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Vi ( )

( )21 5 5 5

: 1 22 2 2 3 2 2 32 1

= = = =

x k d y x

Vi ( )

( )21 5 5 5

: 1 22 3 2 2 3 2 22 1

= = = = + +

x k d y x

Nhn xt : Ngoi cch gii trn, ta cn c th thc hin bin i h (1), (2) mt cch linh hot hn nh sau :

( ) ( )( )

( )

( )( )

( )22

1 52 1

2 2 2 1 2 1 5 1 52 1 2 21 , 2 . 2 1 22 2 2 1 22 2 15

2 1

+= + = =

x k kx kx x

x xk

x

1 5 1 5 1 5 5 1 5. . 2

2 2 2 1 2 2 1 2 2 1 2 2 2 1 10

+ = = =

k k k

x x x x

Khi ( )2 2

21 52 5. 5. 30 25 0 15 10 22 1 10

= = + + = = k

k k k k kx

T ta c cc tip tuyn cn tm l ( )( )15 10 2 1 2.= y x BI TP LUYN TP: Bi 1. Vit phng trnh tip tuyn trong cc trng hp sau: a) bit tip tuyn i qua A(2; 0) n th hm s y = x3 x 6 b) bit tip tuyn i qua A(3; 0) n th hm s y = x3 + 9x c) bit tip tuyn i qua A(1; 2) n th hm s y = x3 3x d) bit tip tuyn i qua A(1; 3) n th hm s y = 3x 4x3

e) bit tip tuyn i qua 4 4

;9 3

A n th hm s 3 21

2 3 43

= + +y x x x

f) bit tip tuyn i qua A(1; 4) n th hm s y = 2x3 + 3x2 5 Bi 2. Vit phng trnh tip tuyn

a) k t O(0; 0) n th hm s 4 21 1

.2 2

= y x x

b) k t 3

0;2

A n th hm s 4 21 3

3 .2 2

= +y x x

Bi 3. Vit phng trnh tip tuyn:

a) k t im A(6; 5) n th hm s 2

.2

+=

xy

x

b) k t im O(0; 0) n th hm s 3( 1)

.2

+=

xy

x

Bi 4. CMR khng c tip tuyn no ca th hm s 1

=+x

yx

i qua giao im I ca 2 ng tim cn.

Bi 5. Tm im M thuc th hm s y = 2x3 + 3x2 12x 1 sao cho tip tuyn ca th ti im M i qua gc to O(0; 0).

DNG 4. BI TON TM S TIP TUYN K T MT IM N TH

Cch gii chung: Vi dng ton ny, thng thng th im A ch bit honh hoc tung . Khi , tm s tip tuyn c th k c t A n th ta phi bin lun h (1), (2).

V d 1: Cho hm s y x x33= (C). Tm trn ng thng (d): y x= cc im M m t k c ng 2 tip tuyn phn bit vi th (C).

Hng dn gii: Gi M m m d( ; ) . PT ng thng qua M c dng: y k x m m( )= .

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l tip tuyn ca (C) h PT sau c nghim: x x k x m mx k

3

23 ( ) (1)3 3 (2)

=

= (*)

Thay (2) vo (1) ta c: x mx m3 22 3 4 0 + = xmx

3

22

3 4=

(**)

T M k c ng 2 tip tuyn vi (C) (**) c 2 nghim phn bit

Xt hm s x

f xx

3

22( )

3 4=

. Tp xc nh D R

2 3 2 3\ ;3 3

=

x xf x

x

4 2

2 26 24( )(3 4)

=

; xf xx

0( ) 02

= = =

Da vo BBT, (**) c 2 nghim phn bit mm

22

= =

. Vy: M( 2;2) hoc M(2; 2) .

V d 2: Cho hm s y x x3 3 2= + . Tm trn ng thng d y: 4= cc im m t k c ng 2 tip tuyn vi (C).

Hng dn gii: Gi M m d( ;4) . PT ng thng qua M c dng: y k x m( ) 4= +

l tip tuyn ca (C) h PT sau c nghim: x x k x mx k

3

23 2 ( ) 4 (1)

3 3 (2)

+ = +

= (*)

Thay (2) vo (1) ta c: x x m x m2( 1) 2 (3 2) 3 2 0 (3) + + + + =

xx m x m2

12 (3 2) 3 2 0 (4) = + + + =

YCBT (3) c ng 2 nghim phn bit + TH1: (4) c 2 nghim phn bit, trong c 1 nghim bng 1 m 1=

+ TH2: (4) c nghim kp khc 1 m m2 23

= =

Vy cc im cn tm l: ( 1;4) ; 2 ;43

; (2;4) .

V d 3: Cho hm s y x x m x m3 22 ( 1) 2= + + (Cm). Tm m t im M(1;2) k c ng 2 tip tuyn vi (Cm).

Hng dn gii: PT ng thng qua M c dng: y k x( 1) 2= + . l tip tuyn ca (Cm) h PT sau c nghim:

x x m x m k x

x x m k

3 2

22 ( 1) 2 ( 1) 2

3 4 1

+ + = +

+ =

f x x x x m3 2( ) 2 5 4 3( 1) 0= + = (*) qua M k c ng hai tip tuyn n (Cm) th (*) c ng 2 nghim phn bit

Ta c f x x x f x x x22( ) 6 10 4 ( ) 0 1;3

= + = = =

Cc im cc tr ca (Cm) l: A m B m2 109(1;4 3 ), ; 33 27

.

Do (*) c ng 2 nghim phn bit m

A OxB Ox

m

4310981

= =

.

V d 4: Cho hm s y x x3 23 2= + (C). Tm trn ng thng (d): y = 2 cc im m t k c 3 tip tuyn phn bit vi th (C).

Hng dn gii: Gi M m d( ;2) ( ) . PT ng thng i qua im M c dng : y k x m( ) 2= +

l tip tuyn ca (C) h PT sau c nghim x x k x m x x k

3 2

23 2 ( ) 2 (1)

3 6 (2)

+ = +

+ = (*).

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Thay (2) v (1) ta c: x m x mx x x m x3 2 22 3( 1) 6 4 0 ( 2) 2 (3 1) 2 0 + + = + =

xf x x m x (3)2

2( ) 2 (3 1) 2 0

= = + =

T M k c 3 tip tuyn n th (C) h (*) c 3 nghim x phn bit

(3) c hai nghim phn bit khc 2 m mf

m

50 13(2) 0 2

> < > .

Vy t cc im M(m; 2) (d) vi m m

m

513

2

< >

c th k c 3 tip tuyn vi (C).

V d 5: Cho hm s ( ) ( )y x x2 21 . 1= + Cho im A a( ;0) . T