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Bi ging trng tm Hm s Thy ng Vit Hng Tel: 0985.074.831 Facebook: LyHung95
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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Bi ging trng tm Hm s Thy ng Vit Hng Tel: 0985.074.831 Facebook: LyHung95
Tham gia kha LTH v Luyn gii dnh 9 im Ton tr ln trong k TSH 2014 www.moon.vn 1
NG VIT HNG
BI GING TRNG TM V
HM S Chng trnh Luyn thi i hc SAP
NM HC 2013 - 2014
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Bi ging trng tm Hm s Thy ng Vit Hng Tel: 0985.074.831 Facebook: LyHung95
Tham gia kha LTH v Luyn gii dnh 9 im Ton tr ln trong k TSH 2014 www.moon.vn 2
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 xx
b) 1 3( ) 2 .1
=
f xx x
c) ( 3)(3 2 )( ) .1
+ =
x xf xx
d) 4 2 2 1 5( ) .3 2 4 +
=
x xf x
e) 2 3 2( ) .
1 +
=
x xf x xx
f) 2 2( ) .3 1 2 1
+ =
+
x xf xx x
g) 21 1 2( ) .
1
= + + +
xf xx x x x
h) 1 2 3( ) .2 2
= + +
f xx 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
24 3 0.8 15
+ +
x x
x x f) ( )
3 23 3 0.2
+>
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 kf 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 2 101
+ +=
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 xx
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 x x x x x x x x x x x xx 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 xx
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 xx 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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Tham gia kha LTH v Luyn gii dnh 9 im Ton tr ln trong k TSH 2014 www.moon.vn 4
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 ( )* = cx
b
Nu a 0 th (1) l phng trnh bc hai c bit thc ( )2
2
4; 2
=
= =
b acb ac b b
+ nu > 0 th (1) c hai nghim phn bit 2
1;24
.
2 2
= =
b b b acx
a a
+ nu = 0 th (1) c nghim kp .2
=
bx
a
+ 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 xa
cP 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 acbS x x
x x a
cP x xa
Phng trnh c hai nghim m phn bit khi
2
1 21 2
1 2
4 00
0; 0
0
> >
= + =
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Phng trnh c hai nghim phn bit v u ln hn khi
( )( ) ( )
2 2
1 2 1 2 1 21 2
21 2 21 2 1 2
4 0 4 000
2 2 2
0 0 0
> > > >
+ > = + = > = + = > >
> + + > + + >
b ac b acb b
x x S x x S x xx ,x a a
x xc 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 2 21 2 1 2
4 0 4 000
2 2 2
0 0 0
> > > >
+ < = + = < = + = + + >
b ac b acb b
x x S x x S x xx ,x a a
x xc 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 2222
0000 0 0 0
= = = =
= = > = = >= = >= = > > > > > < < < + + < + + < < < + + < + + > > <
th (1) c 2 nghim phn bit
2
1;22 2 5 3
.
1m m m
xm
+=
+
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b) Phng trnh (1) c hai nghim phn bit khi ( )23
0 2 5 3 0 *12
m
m mm
> > > <
Gi hai nghim phn bit l x1 ; x2 vi x2 > x1.
Theo nh l Vi-t ta c 1 2
1 2
41
2 31
b mx x
a m
c mx x
a m
+ = = +
+
= =
+
Hai nghim u dng khi 1 21 2
1 04 00 11.
0 2 3 301 2
o
mm
x x mmvn
x x mm
m
< + > > + > + > < +
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 11 1
m m m mx x x x x x m m m
mmm mx x x x
mm 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 54 2 5
54
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 10 1 2 0 .
2P 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 2x 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
5 0.2
+ + =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.+ +
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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
3 520 3 1 0
3 52
+>
> + >
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Hm s c hai cc tr khi y = 0 c hai nghim phn bit
2
2 2 6 2 2 6005 50 5 4 4 0 0
+ < +
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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 2 10 5 4 0
4>
> + + > <
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 2
04 0
4=
= = =
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
= == =
+ >
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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
0 0
BS x x A
x xP x x C
A
>= + >
> > = > >
Hai im cc tr cng c honh m.
Khi ta c 1 21 21 2
000
0 0
BS x x A
x 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
+ > + + > > < <
+ < + + >
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Vy vi
7 3 52
7 3 52
+
>
> + >
> >
+ > > >
x x x x mx x m
x x mx x
m
8 808 5.3 55
+> >
< < < <
mm
mm
m
i chiu vi iu kin (*) ta c 7 3 582
m
< < l gi tr cn tm.
d) Ta c 1
21 2
2
160 3 2( 1) 3 0
16
= = = + = <
+ = =
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 5
6 5
= > > <
<
mmx m m
m
2 2
5 58 5.
3 247 1 10 25
< > + + < + +
m mm
mm m m m
i chiu vi iu kin (*) ta c 7 3 582
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 21 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 i chiu vi iu kin (*) ta c 3 1 3
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 2
1.
3x 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 )3
2
2
.
3
+ =
=
mx x
mx x
Khi ( ) ( )2 2 21 21 2 2 1 21 14 4(1 2 ) 4(2 11 93 ) = + > >> x x x x mx mx x x
2
3 29816 12 5 0
3 298
+>
>
<
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 m
x x m x x m x m m
x x x x x x m m m m
2 4 348 16 9 0 .4
+ = =m m m Vy 4 34
4
=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 0 5 71 (1) 5 7 04 52 1 1
2 3
= >< < = + > <
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Phng trnh honh giao im ca (Cm) v Ox: ( )3 2
21
3 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 21213 5 0
13 3 212
+
> + >
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 10 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 , 12
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 2 428 1 3 1 28 3 1 3 .3x 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 2 1 101; 3 1 2 12 2 3 1 12 3 1 10 3x 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 2 22 2 223 1; 1 2 12 2 3 1 1 12 3 1 2 6x 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
0
1120 cos2
= = AOB AOB
( ) ( )2 2
22 2
4 0( 4) 1 4 ( 4) 2 ( 4) 3 24 44 024 ( 4)4 0 12 2 3 2412 2 3 3 33
<
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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 02
. 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 12my 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 12my 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( ) 1
3 0 0m m
I mm 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
22
2 2
3 22 1222
2 3 5 412 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 62 ;
2m 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 m
mAB u m m m m m vn
+
= = + + =
BI TP LUYN TP: Bi 1: Cho hm s ( )
23 2 11 2(2 1) 3.
3 2m x
y 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 ( )2
3 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
1 23 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 , : 22
= + = = = +
= +x x
y x m my y A B d y x m my 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 2
3 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 2
3 3 = + +
m my x
Theo bi ta c
2 2 43 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 2
3 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 2
3 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 xd y x
( ) ( )1 2 1 22 2 22 2 2 2 3 .2 6 03 3 3 3
+ + + = + + = =
m m m mx x x x m
Vy 30;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, * > >
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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 : 42; .
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 13 2
x x m
x x m
+ =
=
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Nh vy ta c h:
( ) ( )( ) ( )
( )
1 2
1 2
1 2
2 1 13 2 2
2 1 3
x x m
x x m
x x
+ =
=
+ =
T ( ) ( )1 3v ta d dng gii c 21
3 24 5
x m
x m
=
=
Thay vo ( )2 ta c: ( ) ( ) ( ) ( )14 5 3 2 3 2 19 7316m m m m = = Vy ( )1 19 7316m = l gi tr cn tm. Bi 2: Cho hm s 3 2( 2) ( 1) 2
3= + + +
my x m x m x
Tm m hm s c cc i, cc tiu ti x1 ; x2 sao cho 1 2 1.<
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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 25
.
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 23 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
72
m = 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 91; .
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 ) 02
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 22 9 (2;9 )2 4 12 3 4 (2 ; 4 12 3 4)
x y m A mx 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 ( *)9 29 4 12 3 4 0
2
+ =
= + + =
m
m tmm m m m
Vy 12
= 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)( ) 0 xy x m m x x mx m
== + + = =
=
hm s c cc i, cc tiu ti A, B 1(*)m khi :
Ta c: 3 3
32 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 mx 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 0
2m
OAOB m m m m m mm
= = + + + + = =
=
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 3
23 2 3 2 3
( ) 3( )' 3 6( 1) 3 ( 2) 0
2 ( ) 3( ) 4x m y x m x m m m m m
y x m x m mx 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 ( ) 1
1x m
y x mx m x mx m
= += + = =
= +
Hm s lun c cc i, cc tiu ti A, B
Ta c:
3 33
3 33
1 11 ( ) 13 3 3
1 51 ( ) 13 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 9m m m my x y x+ + = + +
Ta c: 2 2
1 1 1 1 11 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 21 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 9m 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 23
mx x
m mx x
++ =
+
=
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Theo k A, B nm khc pha vi Oy 2
1 23 2 0 1 23
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 32
3 3m m my x
+= +
.
V A, B i xng nhau qua 1 5
:2 2
= d y x nn : . 1d AB
I dk k
= ( vi I l trung im ca AB)
Gii 21 2
. 1 2 1 02 3
d ABmk 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 dk 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
33
(1; 2 )(0;4 ), (2 ;0)( ;2 )
=
ABu mA m B m
I 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 21 2
0x xx 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 0x
y 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 ( )22 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
+ =
= =
m
mm
mm (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 0x
y 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 73
3 9 9
= +
m my x ( ) ng thng : 3 7d y x=
Yu cu tr thnh vung gc d hay 2 214 2 2 3 103. 1 15
3 9 3 2
= = =
m mm (tha mn)
Vy 3 102
= 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 52
+>
>
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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
= =
=
m
mm
(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 y T
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 4
42 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
2m
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 my 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 85
= = =
m
m mm
Kt hp cc iu kin trn, ta c 4m = l gi tr cn tm.
Bi 24: Cho hm s 3 213 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 21 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
3my x
=
( )22 2 2 2
1 1 2 2 1 22 2 2 2 1 11 2 2 0 4 1 0 1 0 2
3 3 3 3m m m m
x x x x x x m
+ + > + > +
Kt hp iu kin ta c 12
m
m
>
l cc gi tr cn tm
Bi 25: Cho hm s 3 213
= + + +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 11 2
2x xx x m
+ =
=.
Mt khc, ta c
( ) ( ) ( )22 2 2 1 2 . '3 3 3 3 x xm m xy x x x m r q y = + + + + + = +
Nn PTT qua cc i, cc tiu l 2 2 23 3
m my x = +
Yu cu tng ng ( ) ( ) ( ) ( )2
2 2 2 21 2 1 2 1 2 1 2
260 603
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 23 9
m m mx x m m m m m
+ + = = + + = =
Vy 2m = l gi tr cn tm.
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' 0 3 > m
Nn 13
m 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 my 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)
Vy 1m = l gi tr cn tm.
Bi 27: Cho hm s 3 21 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: 41;3
A
v ( )3;0B
( )2
2 4 2 133 1 03 3
AB = + =
. 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. Cng thc chia nhanh a thc bc 3:
Vi a thc ( ) ( ) ( )3 2
.
'= + + + = +x x x
P ax bx cx d r q P vi 0a
Th lun c ( )
( )
22 23 9 9
3 9
= +
=
x
x
c b bcr x d
a a
x bqa
DNG 2. CC TR CA HM A THC BC BN TRNG PHNG
Xt hm s ( )4 2 3 2 20
4 2 2 2 02
x
y ax bx c y ax bx x ax b bx
a
== + + = + = + = =
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 2 04 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
=+
mm x
m
+ Hm s c mt cc tr khi 3 0 1 01
< +
mm
m
+ Hm s c ba cc tr khi 03 0
11>
> < +
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 ya
bx x y
a
, t ( )0; ; ; ; ;2 2
A B Cb bA y B y C ya 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 ya a
T ( ) ( )21 . 0 02
= + =
B AbAB AC y ya
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 ;0
2 2
= =
B A
b bAB y y BCa a
T ( ) ( )2 222
+ =B Ab by 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 AH
AB AB
vi ( ); ; 0;2
= =
B A B A
bAB y y AH y ya
, t ( ) ( ) ( )2 23 42
+ = B A B Ab y y y ya
Gi tr m tm c kt hp vi iu kin tn ti (*) cho ta kt qu cui cng ca bi ton. 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 ya
, t ( ) ( )223 4 .42
=
o B AbS y ya
Gi tr m tm c kt hp vi iu kin tn ti (*) cho ta kt qu cui cng ca bi ton. Tnh cht 5: 3 im cc tr to thnh mt tam gic c bn knh ng trn ngoi tip R cho trc
S dng cng thc din tch tam gic 2
. .
14 4 24. .2
= = = =abc abc AB AC BC ABS R R R
R 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 2 20
4 4( 1) 4 ( 1) 01
= = + = + =
= +
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 2 2 20
4 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 10 1 0;1 , ;1 , ;1
1
= =
= = =
= =
x y
y x m y m A B m m C m mx 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) 0 1= = = + = = =
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 2 04 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 10 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 4 302 4 3 3
= + = =
=
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 , 3
2= = = = = =
BH BHHAB AB BH BC AB BCAB AB
Ta c ( ) ( )2; , 2 ;0 ,= = AB m m BC m khi ( ) ( )43
03 3 4 1
3
= + = =
m
m m mm
i chiu vi iu kin (*) ta c 313
= 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 2 04 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 10 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. . .
, 12 4 2
= = =ABCAH BC AB BC AC ABS R
R AH
Ta c ( ) ( ) 2 42 2 2; ; 0; = += = =
AB m mAB m m AH m
AH m
Khi , ( ) ( )( )4 3 221
1 2 2 1 0 1 1 0 1 52
=+ = + = + =
=
mm m
m m m m mm m
i chiu vi iu kin (*) ta c 5 11;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 20; .
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
13
m
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 : ; .= = 5 112
m 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 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 = 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 03 220
xy x xyx
= = +
==
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 0 2
xx x x x x x x
x
= + = + = + + =
=
Ta c ( )( )
2 2 12 12 243 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 3xC yx
+=
ti C(1; 7). d) ( ) 3( 2):1
xC yx
=
ti im B c yB = 4
e) ( ) 1:2
xC yx
+=
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: 2
4 4C y x x= ti cc giao im ca (C) vi trc honh.
Bi 2: Cho hm s 1.1
=
+
xyx
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 ykx 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 12 2 2
:
:
d y k x md y k x m
= +
= +
+ d1 v d2 song song vi nhau th c cng h s gc : 1 21 2
d dk km m
=
+ d1 v d2 vung gc vi nhau th c tch h s gc bng 1 : 1 2 21
1. 1 .= = d d d
dk k k
k
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 22 3 1 0 3 2 1 .3 3 3
+ = = + = + = x y y x y x k
b) 1 3 15 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 50 1 4(1 ) 0 .4
< + + < < k k
Vy vi 45
< 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 khi50 1 4(1 ) 04(2) 0 (2) 1 0 1
> + + > >
=
k kg g k k
Vy vi 45
1
>
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 21 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 my x m y x
Tip tuyn ti im un c h s gc l 2 236. 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
3 4 3 2 8 0 423
=
= + = = =
u
mmk k m m m
m
Vy, vi 42;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 4526
= m th tip tuyn ti x = 2 vung gc vi .
V d 4: Cho hm s +=
1,
2xyx
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 21 2 3 3 312 2 2 ( 2)
+ += = = + =
x xy yx x x x
Gi ( ) ( ) 3 3; 1 ;1 .2 2
= + +
o o o o
o o
M x y C y M xx x
Ta c 21lim21lim 12
+=
+
=
x
x
x
x
x
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 12 3
2 ( 2)
+
= = =
oI MIM
I M o o
xy ykx x x x
Tip tuyn ti M vung gc vi ng IM khi 2 23 3
. 1 . 1( 2) ( 2)= = tt IM o ok k
x x
2 2 3 2 3( 2) 32 3 2 3
= = +
= = =
o o
o
o o
x xx
x x
+ Vi ( )3 32 3 1 1 1 3 2 3;1 32 3= + = + = + = + + +o o ox y Mx + Vi ( )3 32 3 1 1 1 3 2 3;1 32 3= = + = + = o o ox y Mx Vy c hai im M tha mn yu cu bi ton.
V d 5: Cho hm s ( )=+
2 1, .
1xy C
x
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 ( )23
.
1 =
+y
x Gi ( )2 1;
1
+
aM a Ca
Tip tuyn vi (C) ti M c h s gc: ( )23( ) .
1= =
+ttk y a
aGiao im hai ng tim cn I(1; 2).
ng thng IM c h s gc l ( )23
.
1
= =
+M I
IMM I
y ykx x a
Theo bi ta c ( ) ( ) ( )4
2 2
03 3. 9 . 9 1 1
21 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 kin50 1 4( 1) 04(1) 0 (1) 1 0 1
> >
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Khi xN ; xP l cc nghim ca phng trnh 212 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 23 2 2
3. 1 9 18 1 0
3 2 23
+
== + + =
=
m
k k m mm
i chiu vi iu kin ta c 3 2 23
m
= 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 23 2 2(3 6 )(3 6 ) 1 9 18 1 0
3M M N Nx x x x k k k = + + = =
i chiu vi iu kin (*) ta c 3 2 23
m
= 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=
xyx
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
=
xyx
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 tip tuyn vi (C) bit tip tuyn ny song song vi ng thng (d): y = 6x 1 b) Vit phng trnh tip tuyn vi (C) bit tip tuyn vung gc vi = +1( ) : 2.
9d 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
=
+
xyx
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 yk 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 tip tuyn trong cc trng hp sau:
a) Bit tip tuyn i qua
2 ; 13
A 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, 13
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 0 13=
+ = = =
xx x x x x x
x
Vi 20 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 kxk 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 xk x x x x x x x x
x 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 tip tuyn k t im ( )1; 2A n th hm s +=
2.
2 1xyx
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 s 22 1
+=
xyx
th h sau c nghim: ( )
( ) ( )2
2 ( 1) 2, 12 1
5, 2
2 1
+=
=
x k xx
kx
Thay (2) ln (1) ta c ( ) ( )( ) ( ) ( )2
22 5 ( 1) 2 2 2 1 5 1 2 2 1 0
2 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 :
( ) ( )( )
( )
( )( ) ( )2
2
1 52 12 2 2 1 2 1 5 1 52 1 2 21 , 2 . 2 1 2
2 2 2 1 22 2 152 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 kx x x x
Khi ( )2 2
21 52 5. 5. 30 25 0 15 10 22 1 10
= = + + = =
kk 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 30;2
A n th hm s 4 21 33 .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
+=
xyx
b) k t im O(0; 0) n th hm s 3( 1) .2
+=
xyx
Bi 4. CMR khng c tip tuyn no ca th hm s 1
=
+
xyx
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 xf 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) + + + + =
x
x m x m21
2 (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 2
3= =
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 x2 2( ) 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 + + = + =
x
f x x m x (3)22
( ) 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 mm
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) . Tm a t A k c 3 tip tuyn phn bit vi th (C).
Hng dn gii: Ta c y x x4 22 1= + . PT ng thng d i qua A a( ;0) v c h s gc k : y k x a( )=
d l tip tuyn ca (C) h phng trnh sau c nghim: x x k x a Ix x k
4 2
3
2 1 ( )( )
4 4
+ =
=
Ta c: kI Ax2
0( ) ( )1 0
=
= hoc x x k B
f x x ax
2
24 ( 1) ( )( ) 3 4 1 0 (1)
=
= + =
+ T h (A), ch cho ta mt tip tuyn duy nht l d y1 : 0= . + Vy t A k c 3 tip tuyn phn bit vi (C) th iu kin cn v l h (B) phi c 2 nghim phn bit x k( ; ) vi x 1 , tc l phng trnh (1) phi c 2 nghim phn bit khc 1 a
f
24 3 0( 1) 0
= >
a hoac a3 31 12 2
< >
V d 6: Cho hm s: xyx
21
+=
(C). Cho im A a(0; ) . Tm a t A k c 2 tip tuyn ti th (C) sao cho 2 tip im tng ng nm v 2 pha ca trc honh.
Hng dn gii: Phng trnh ng thng d i qua A a(0; ) v c h s gc k: y kx a= +
d l tip tuyn ca (C) H PT x
kx ax
kx 2
21
3( 1)
+= +
=
c nghim
PT: a x a x a2(1 ) 2( 2) ( 2) 0 + + + = (1) c nghim x 1 . qua A c 2 tip tuyn th (1) phi c 2 nghim phn bit x x1 2,
a a
aa
1 123 6 0
> = + > (*)
Khi ta c: a ax x x xa a1 2 1 2
2( 2) 2;1 1
+ ++ = =
v y yx x1 21 2
3 31 ; 11 1
= + = +
2 tip im nm v 2 pha i vi trc honh th y y1 2. 0<
x x1 2
3 31 . 1 01 1
+ + a 23
>
Kt hp vi iu kin (*) ta c: aa
23
1
>
.
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V d 7: Cho hm s xyx
11
+=
(C). Tm trn Oy tt c cc im t k c duy nht mt tip tuyn ti (C).
Hng dn gii: Gi oM y(0; ) l im cn tm. PT ng thng qua M c dng: oy kx y= + (d)
(d) l tip tuyn ca (C) o o o o
xkx y y x y x y
xx kk
xx
2
22
1( 1) 2( 1) 1 0 (1)
1 22 1;( 1)( 1)
+ = + + + + =
=
=
(*)
YCBT h (*) c 1 nghim (1) c 1 nghim khc 1
oo o
o o oo
y y x y k
x y y y x y k2
1 11 ; 1 81 2' ( 1) ( 1)( 1) 0 0; 1 22
= = = = = = + + = = = =
Vy c 2 im cn tm l: M(0; 1) v M(0; 1). V d 8: Cho hm s xy
x
31
+=
(C). Tm trn ng thng d y x: 2 1= + cc im t k c duy nht mt tip tuyn ti (C).
Hng dn gii: Gi M m m d( ;2 1)+ . PT ng thng qua M c dng: y k x m m( ) 2 1= + +
PT honh giao im ca v (C): xk x m mx
3( ) 2 11
+ + + =
[ ] [ ]kx m k m x mk m2 ( 1) 2 (2 4) 0 + + + = (*) tip xuc vi (C
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