Chuyen de Ham So P2 GTLN GTNN Tiem Can Ks Ham So

TRUNG TM BDKT V LTH 36/73 NGUYN HONGTRUNG TM GIA S NH CAO CHT LNG

ST: 01234332133-0978421673

CHUYN HM S 12LUYN THI

TT NGHIP TRUNG HC PH THNG, I HC, CAONG

Hue, thang 7/2012

* GTLN V GTNN ca hm s* Tim cn ca th hm s

* KSHS hm bc ba, trng phng, hu t

TRUNG TM GIA S NH CAO CHT LNG. ST:0978421673 -TP HU

Chuyn LTH Bin son:Trnnh C1

MC LCBi 3. Gi t ln nht v gi tr nh nht ca hm s

- Dng 1: Tm GTLN, GTNN ca hm s bng nh ngha- Dng 2: t n ph tm GTLL v GTNN- Dng 3:ng dng gii phng trnh, bt phng trnh, h phng trnh- Dng 4: Chng minh bt ng thc, tm GTLN v GTNN trn mt min

Bi 4. Tim cn ca th hm s- Dng 1: Tm tim cn ngang v tim cn ng bng nh ngha- Dng 2: Mt s bi ton lin quan n tim cn. Tm m tha iu kin K

cho trcCh : Tim cn xin (Tho lun)

- Dng 3: Cc bi ton lin quan n tim cn hm phn thcBi 5. Kho st hm sVn 1: Hm trng phng

- Dng 1: Kho st v v th hm s- Dng 2: Mt s bi ton lin quan n hm trng phng

Vn 2: Hm bc ba- Dng 1: Kho st v v th hm s- Dng 2: Mt s bi ton lin quan n hm bc ba

Vn 3: Hm phn thc hu t- Dng 1: Kho st v v th hm s- Dng 2: Mt s bi ton lin quan n hm phn thc hu t

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BI 3. GI TR LN NHT, GI TR NH NHTA. KIN THC CN NM1. nh ngha:

Gi s hm s f xc nh trn min D (D R).

a) 0 0( ) ,max ( ) : ( )D

f x M x DM f x x D f x M

b) 0 0( ) ,min ( ) : ( )D

f x m x Dm f x x D f x m

2. Tnh cht:a) Nu hm s f ng bin trn [a; b] th [ ; ][ ; ]max ( ) ( ), min ( ) ( )a ba b f x f b f x f a .

b) Nu hm s f nghch bin trn [a; b] th [ ; ][ ; ]max ( ) ( ), min ( ) ( )a ba b f x f a f x f b .



B. PHNG PHP GII BI TP

Cch 1: Thng dng khi tm GTLN, GTNN ca hm s trn mt khong. Tnh f (x). Xt du f (x) v lp bng bin thin. Da vo bng bin thin kt lun.

Cch 2: Thng dng khi tm GTLN, GTNN ca hm s lin tc trn mt on[a; b]. Tnh f (x). Gii phng trnh f (x) = 0 tm c cc nghim x 1, x2, , xn trn [a; b] (nuc). Tnh f(a), f(b), f(x1), f(x2), , f(xn). So snh cc gi tr va tnh v kt lun.

1 2[ ; ]max ( ) max ( ), ( ), ( ), ( ),..., ( )na bM f x f a f b f x f x f x 1 2[ ; ]min ( ) min ( ), ( ), ( ), ( ),..., ( )na bm f x f a f b f x f x f x

BI TP MU:Bi 1. Tm GTLL v GTNN (nu c) ca cc hm s sau:

3 1) 3xa y x trn on [0;2]

b) 2

23 1

1x xy x x

DNG 1: TM GI TR LN NHT V GI TR NH NHT CA HM S

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Hng dn:b) Bng bin thin

x 0 2 'y - 0 + 0 +

y 3 113

Da vo bng bin thin, hc sinh c th d dng xc inh GTLL,GTNN

Bi 2. Tm GTLL v GTNN (nu c) ca cc hm s sau:a) 2 4 3y x x b) 4 22y x x c) 4 22 2y x xHng dn:

b) Hm s xc nh trn Bng bin thin:

x -1 0 1'y - 0 + 0 - 0 +

y 0

Da vo bng bin thin:Hm t ga tr nh nht ti 1x ,

1Min y . Hm khng c gi tr ln nht

Bi 3. Tm gi tr ln nht, gi tr nh nht (nu c) ca cc hm s sau:

1 3

-1 -1



22xy x trn 0;Hng dn:Hm xc nh trn tp 0;

2 0;' 0 2

xy xBng bin thin

x 0 2 'y - +

y 8

Da vo bng bin thin ta thy hm t gi tr nh nht ti 0;2, 8x Min y

Hm khng c gi tr ln nhtBi 4. Tm gi tr ln nht, gi tr nh nht ca 2 5 6y x x trn on [-1;6]Hng dn:

Hm t gi tr nh nht ti x=-1; x=6 v t gi tr ln nht ti 52x

Bi 5. Tm gi tr ln nht gi tr nh nht ca cc hm s sau: 26 4y x x trn on [0;3]

Hng dn: Hm t gi tr ln nht ti x=3, nh nht ti x=0Bi 6. ( thi TSH 2003 khi B) . Tm gi tr ln nht, nh nht ca hm s 24y x x

Hng dn:Cch 1: Tp xc nh 2;2D ;

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2

21 ; 0 44xy y x xx

2 20 24

x xx x max 2 2min 2

yy

Cch 2: t 2sin , ;2 2x u u

2 sin cos 2 2 sin 2;2 24y u u u ; max 2 2 ; min 2y yBi 7. Tm gi tr ln nht, nh nht ca hm s 2

11

xy x trn on [-1;2]

Hng dn:Hm t gi tr nh nht ti x=-1 v t gi tr ln nht ti 1xBi 8. Tm gi tr ln nht, nh nht ca hm s 3 23 1y x x trn on [-2;1]Hng dn:Hm cho xc nh trn 2;1

t 3 2( ) 3 1, 2;1g x x x x ,

0'( ) 0 2 2;1xg x x

Do :

2;1 2;1

( ) 1; ( ) 19Max g x Ming x

Ta c: 2;1 ( ) 19;1 ( ) 0;19x g x g x

1 1(0). ( 1) 0 0;1 : ( ) 0g g x g x . Vy 2;1 2;1( ) 19; ( ) 0Max f x Min f xBI TP T LUYN:Bi 1. Tm GTLN, GTNN ca cc hm s sau:

a) 22

11

x xy x x b) 3 44 3y x x c)

4 23

1 ( 0)x xy xx x



d) 2 2y x x e) 21

2 2xy x x

f) 2

22 4 5

1x xy x g)

2 1 ( 0)y x xxBi 2. Tm GTLN, GTNN ca cc hm s sau:

a) 3 22 3 12 1y x x x trn [1; 5] b) 33y x x trn [2; 3]c) 4 22 3y x x trn [3; 2] d) 4 22 5y x x trn [2; 2]

e) 3 1

3xy x trn [0; 2] f)

11

xy x trn [0; 4]

g) 24 7 7

2x xy x trn [0; 2] h)

22

11

x xy x x trn [0; 1]

Bi 3. Tm GTLN, GTNN ca cc hm s sau:a) 2100y x trn [6; 8] b) 2 4y x xc) 22y x x

Bi 4. Tm gi tr ln nht v g tr nh nht ca hm s 3 2 72 90y x x x trnon [-5;5]Hng dn:Hm s cho xc nh trn 5;5t 3 2( ) 72 90, 5;5g x x x x x

Ta c :

6 5;5'( ) 0 4 5;5xg x x

Vi (4) 86; ( 5) 400; (5) 70g g gDo : 86 ( ) 400 0 ( ) 400 0 ( ) 400g x g x f xVy

5;5ax ( ) 400 5M f x khi x

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Bi 6. Tm gi tr ln nht, nh nht ca hm s sin2y x x trn on ;2Hng dn:

5'( ) 0 ; ;6 6 6f x x

Vy:

; ;2 2

5 3 5( ) ; ( )6 2 6 2 2Max f x khi x Min f x khi x



Khi t n ph, cn ch mt s iu sau: Nu t 2t x th 0t v gi s 1;1 0;1x t Nu sin 1;1cos

t x tt x

Nu 22sin 0;1ost x tt c x

BI TP P DNG:Bi 1. ( d b TSH 2003 khi B) Tm gi tr ln nht, nh nht ca

36 24 1y x x trn on 1;1 .Hng dn:t 2 0;1u x . Ta c 33 3 24 1 3 12 12 4y u u u u u

2

239 24 12 02 0;1

uy u uu

T ta c 4max 4;min 9y y

Bi 2. Tm gi tr ln nht v g tr nh nht ca hm s 6 4 29 13 4 4y x x x trnon [-1;1]Hng dn:t 2 0;1 , 1;1t x t x ta c:

3 2 9 1( ) 3 4 4f t t t t lin tc trn on [0;1]

DNG 2: S DNG PHNG PHP T N PH TM GTLL V GTNN

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12'( ) 0 3 0;12

tf t

t

0;1 1;1

[ 1;1]0;1

3 1 3 2( ) ( )4 2 4 21 1( ) 0 ( ) 04 4

Max f t khi t hay Max f x khi x

Min f t khi t hay Min f x khi x

Bi 3. Tm gi tr ln nht, nh nht ca hm s 4 2sin os 2y x c xHng dn:Hm cho xc nh trn 4 2 4 2sin os 2 sin sin 3y x c x x x

t 2sin , 0;1t x t . Xt hm 2( ) 3, 0,1f t t t t

Vy

0;1 0;1

11( ) 3; ( ) 4Max f x Min f x

Bi 4. Tm gi tr ln nht, nh nht ca hm s 2s inx 1

sin s inx 1y xHng dn:t sin , 1;1t x t

21( ) 1;11

tf t t t , ( )f t lin tc trn 1;1 , '( ) 0 0f t t

1;1

1;1

( ) ( ) 0 sin 1 2 ,2( ) ( ) 0 sin 0 ,

Max f x Max f t khi x x k kMin f x Min f t khi x x k k

Bi 5. Tm gi tr ln nht v gi tr nh nht ca hm s: 2 2sin os4 4x c xyHng dn:Cch 1:



2 2 2 2 2 2sin os sin 1 sin sin sin44 4 4 4 4 4

x c x x x xxy

t 2sin4 , 0;4xt t , xt hm s

2 4 , 1;4ty ttT suy ra c:

1;4 1;1( ) ( ) 5 ; ( ) ( ) 4Max f x Max f t Min f x Min f t

Cch 2:p dng bt ng thc trung bnh cng, trung bnh nhn ta c:

2 2sin os4 4 2 4 4.x c x ng thc xy ra khi 2 2sin os4 4 ,2 2

x c x kx k

2

2 2 2 22

sinsin os sin os

os4 1 4 1 4 1 0 4 4 54 1

xx c x x c x

c x

ng thc xy ra khi sin 0x hoc cos 0x

Vy 4 ; 54 2 2k kMiny khi x Maxy khi x

BI TP T LUYN:Bi 1. Tm GTLN, GTNN ca cc hm s sau:

a) 2sin 1sin 2

xy x b) 21

cos cos 1y x xc) 22sin cos 1y x x d) cos2 2sin 1y x x

Bi 2. Tm GTLN, GTNN ca cc hm s sau:

a) 2

4 21

1xy x x b)

2 24 4 3y x x x x

g) 2 24 2 5 2 3y x x x x e) 3 3sin cosy x x

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Phng php:Gi s f(x) l mt hm s lin tc trn min D v c min ( ) ; max ( )D Df x m f x M .

Khi :

1) H phng trnh ( )f x

x D c nghim m M.

2) H bt phng trnh ( )f x

x D c nghim M .

3) H bt phng trnh ( )f x

x D c nghim m .

4) Bt phng trnh f(x) ng vi mi x m .5) Bt phng trnh f(x) ng vi mi x M .

BI TP P DNG:Bi 1. Tm m phng trnh sau c nghim x 0; 1 3 :

2 2 2 1 (2 ) 0 (2)m x x x x Hng dn:

t 2t x 2x 2 . (2) 2t 2m (1 t 2),dox [0;1 3]t 1

Kho st 2t 2g(t) t 1 vi 1 t 2; '( ) 0g t . Vy g tng trn [1,2]

Do , ycbt bpt 2t 2m t 1 c nghim t [1,2] tm g t g1;2

2max ( ) (2) 3

Bi 2. Tm m phng trnh sau c 2 nghim phn b it:2 210 8 4 (2 1). 1x x m x x

Hng dn:

DNG 3: NG DNG VO GII PHNG TRNH, BT PHNGTRNH, H PHNG TRNH V H BT PHNG TRNH:



Nhn xt: 2 2 21 0 8 4 2(2 1) 2( 1) x x x x

(pt) 2

2 22 1 2 12 2 01 1

x xmx x . t 2

2 11

x tx

iu kin : 2< t 5 .

Rt m ta c: m= 22 2t t . Lp bng bin thin 124 5 m hoc 5 < 4 m

Bi 3.

2 2Tm tham so m e bat phng trnh 2 24 2 (1) co nghiem

tren 4;6x x x x m

Hng dn:

2

2

2

2 24, 4,6 th t 0;5ycbt tm m e bat phng trnh 24 co nghiem thc t 0;5Xet ham so f(t)= 24, lien tuc tren 0;5

at t x x xt t m

t t

0;5

Ta co: '( ) 0, 0;5 ( ) lien tuc va ong bien tren 0;5Vay bpt co nghiem thc tren oan 0;5 khi ax ( ) (5) 6

f t t f tm f t m f m m

Bi 4. Tm m h BPT: 2

3 23 02 2 4 0

x xx x x m m (1) c nghim.

Gii. (1) 3 20 3

2 2 4x

f x x x x m m (2).

Ta c:

2

23 4 4 0;23 4 4 2;3x x xf x x x x ;

(x) 0 23x . Hm khng c o hm ti 2x

Nhn BBTsuy ra: 0;3Max 3 21x f x f (2) c nghim th 20;3Max 4x f x m m 2 4 21m m 3 m 7

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Bi 5. Tm m PT: 22 2sin2 1 cosx m x (1) c nghim ,2 2x

Gii. Do ,2 2x ,2 4 4x nn t tg 1,12xt

22

1cos 1tx t ; 2

2sin 1tx t . Khi (1)

2 22 sin cos 1 cosx x m x

2 22 2 22

2 22 1 12 1 2 1 21 1t t tm f t t t mt t (2)

Ta c: 22 2 1 2 2 0 1; 1 2f t t t t t t (2) c nghim 1,1tth 1,1 1,1Min 2 Maxt tf t m f t

0 2 4 0 2m m . Vy (1) c nghim ,2 2x th 0;2m .@ Ch : bi trn ta s dng cng thc t tg 2

xt th 22

1cos 1tx t ;

22sin 1tx t . Cng thc ny trong SGK khng c. Tuy nhin, ta nn bit khi

no thy b em ra dng. Vic chng minh cng thc trn tng i d dng.

Bi 4. Gii phng trnh: 4 42 4 2x xGi : yu cu hc sinh phi nm cng thc tnh o hm ca hm ly tha(chng II-Giait tch 12Hng dn:t 4 42 4f x x x vi 2 4x

3 34 4

1 1 1 0 34 2 4f x x

x xNhn BBT suy ra: 3 2 2,4f x f x Phng trnh 4 42 4 2f x x x c nghim duy nht x 3Bi 5. Gii phng trnh: 3 5 6 2x x x



Hng dn:PT 3 5 6 2 0x xf x x . Ta c: 3 ln3 5 ln5 6x xf x 2 23 ln3 5 ln5 0x xf x x (x) ng binMt khc (x) lin tc v

0 ln3 ln5 6 0f , 1 3ln3 5ln5 6 0f Phng trnh (x) 0 c ng 1 nghim x0Nhn bng bin thin suy ra:Phng trnh 3 5 6 2 0x xf x x c khng qu 2 nghim.BI TP T LUYN:

Bi 1. Gii cc phng trnh sau: 5 5 1(1 ) 16x x

Bi 2. Tm m cc phng trnh sau c nghim:a) 3 6 (3 )(6 )x x x x mb) mmxxxx 2223 22 Hng dn:b)

(*)2

2 23 2 0

3 2 2 2x x

x x x mx m

mx

xxfx

xxmx

2123)(21

23)1(221

f(x) lin tc trn 1;2 v c 25( ) 0, 1;21f x xx

)(xf ng bin trn 2;1Bi ton yu cu 1 2(1) 2 (2) 4 3f m f m

Bi 3. Tm m cc bt phng trnh sau nghim ng vi mi x R:a) 22 1x x m c) 4 4 0mx x m

Bi 4. Cho bt phng trnh: 3 22 1 0x x x m .

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a) Tm m bt phng trnh c nghim thuc [0; 2].b) Tm m bt phng trnh tho mi x thuc [0; 2].

Bi 5. Tm m cc bt phng trnh sau:a) 3 1mx x m c nghim.b) ( 2) 1m x m x c nghim x [0; 2].

c) 2 2( 1) 1m x x x x nghim ng vi mi x [0; 1].Bi 6. Tm m BPT: 22 9m x x m c nghim ng xHng dn:

22 9m x x m 22 9 1m x x 22 9 1xm f x xTa c:

222 2

9 2 92 9 2 9 1

xf xx x

0 22 9 9 6x x

2

1 1lim lim 9 212x xf x

xx ;

2

1 1lim lim 9 212x xf x

xxNhn BBT ta c f x m , x 3 3Min 6 4 4x f x f m m

Bi 7. Tm m phng trnh: mxxxxx 1)1(4)1( c nghm

Hng dn:

t ( 1) 1xt x x khi pt cho ta m = t(t 1) suy ra 4m



(Phn nng cao-bi dng hc sinh gii -Trch ti liu ca Trn Phng v thamkho phn ti liu S Tng)

Phng Php:1. Cch ny da trc tip vo nh ngha GTLN, GTNN ca hm s.

Chng minh mt bt ng thc. Tm mt im thuc D sao cho ng vi gi tr y, bt ng thc va tm ctr thnh ng thc.

2. Xt bi ton tm GTLN, GTNN ca hm s f(x) trn mt min D cho trc.Gi y0 l mt gi tr tu ca f(x) trn D, th h phng trnh (n x) sau cnghim:

0( ) (1)

(2)f x yx D

Tu theo dng ca h trn m ta c cc iu kin tng ng. Thng thngiu kin y (sau khi bin i) c dng: m y0 M (3)V y0 l mt gi tr bt k ca f(x) nn t (3) ta suy ra c:

min ( ) ; max ( )D Df x m f x MBI TP MU:Bi 1. Tm gi tr nh nht ca hm s 24 2 1f x x x xGii. Gi y0 l 1 gi tr ca hm f(x) tn ti x0 sao cho y0 = 20 0 04 2 1x x x

2 2 2 20 0 0 0 0 0 0 0 0 04 2 1 2 4 2 1y x x x y y x x x x g(x0) = 2 20 0 0 03 2(1 ) 1 0x y x y . Ta c g(x) = 0 c nghim x0 = 2 2 20 0 0 0(1 ) 3(1 ) 2(2 1)y y y y = 0 02( 1)(2 1) 0y y

DNG 4: Chng minh bt ng thc, tm GTLL v GTNN ca hm strn mt min

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Do y0 = 2 2 20 0 0 0 0 0 03 ( 1) 3 3 0x x x x x x x nn

0 2y0 1 0 0 12y . Vi x = 12 th Minf(x) =

12

Bi 2. Cho 2 5 4 .y f x x x mx Tm cc gi tr ca m sao cho Min 1y

Gii. Ta c

21

22

5 4 ; x 1 4 :5 4 ; 1 4 :

x m x x Pf x x m x x PGi (P) l th ca y = f(x) (P) = (P1) (P2) khi (P) c 1 trong cc hnhdng th sau y

Honh ca cc im c bit trong th (P):

Honh giao im (P1), (P2) xA = 1; xB = 4 ; Honh nh (P1): 5 2Cmx .

Nhn vo th ta xt cc kh nng sau: Nu xC [xA, xB] m[ 3, 3] th Minf(x) = Minf(1), f(4).

Khi Minf(x) > 1

3 3(1) 1(4) 4 1

mf mf m

1 < m 3 (1)

Nu xC [xA, xB] m[ 3, 3] th Minf(x) = 1 152Cmf x f =

2 10 94

m m

A

BCP2

P1A

B C

P2P1A

BC

P1P2



Khi Minf(x) > 1 2[ 3,3] 3 5 2 310 13 0

m mm m (2)

Kt lun: T (1) v (2) suy ra Minf(x) > 1 1 5 2 3m

Bi 3. Cho , 0

1x yx y Tm gi tr nh nht ca S = 1 1

x yx y

Gii: 2

x yS y x x y x y x y x yy x

Mt khc, S = 1 1x y

x y = 1 1y xy x = 1 1 x yx y

Suy ra 2S 1 1x y 42 2 2 2

2xy x y 2S MinS = 2 .

Bi 4. ( 33 III.2, B thi TSH 1987 1995) Cho 2 2 1x y . Tm Max, Min ca A 1 1x y y x .Gii. 1. TmMaxA: S dng bt ng thc BunhiaCpski ta c

A 2 2 2 21 1 2 2 2 2 2x y y x x y x y .Vi 12x y th Max A 2 2

2. TmMinA: Xt 2 trng hp sau y Trng hp 1: Nu 0xy , xt 2 kh nng sau:+) Nu 0, 0x y th A>0 Min 0A+) Nu x 0, y 0 thA 2 2( ) (1 ) (1 ) 2x y x y x y = 2 22 2 1x y x yT 2 kh nng xt suy ra vi 0xy th Min A = 1

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Trng hp 2: Xt 0xy : t x y t 2 1 02txy 1,1t

2 2 21 2 1 1 1 1 2 1A x y xy x y y x xy x y xy x y xy 2 2 21 1 11 2 12 2 2

t t tt t 2 1 1 2 2 12t t 2 3 21 1 2 2 1 2 2 22A f t t t tTa c: 2 1 23 1 2 1 2 1 22 0 ; 2 12 2 3f t t t t t t tTh 1 2,t t vo phn d ca f t chia cho f t 1 22 19 3 2 ; 027f t f t .Nhn bng bin thin suy ra:

2 1 1A f t A f t suy ra 1 2 19 3 2Min 127A f t

xy ra 1x y t ; 21 12

txy

x, y l nghim ca 2 1 2 2 3 03 9u u 1 2 15 2 2, 6x y

Kt lun: Max A 2 2 ; 2 19 3 2Min 27ABi 5. Cho a,b,c 0 tha mn iu kin 3a b c 2

Tm gi tr nh nht ca 2 2 22 2 21 1 1S a b cb c aGii. Sai lm thng gp:

2 2 2 2 2 23 62 2 2 2 2 2

1 1 1 1 1 13. 3.S a b c a b cb c a b c a

t 1 t1 t2 1 0 0

1 1f t

1



62 2 26 2 2 2

1 1 13. 2 2 2 3. 8 3 2 Min 3 2a b c Sb c a Nguyn nhn:

1 1 1 3Min 3 2 1 3 2S a b c a b ca b c mu thun vi githit Phn tch v tm ti li gii :

Do S l mt biu thc i xng vi a, b, c nn d on Min S t ti 12a b c

S im ri :

12a b c

2 2 2

2 2 2

14

1 1 1 4a b c

a b c

1 44 16

Cch 1: Bin i v s dng bt ng thc Csi ta c

2 2 22 2 2 2 2 2

16 16 16

1 1 1 1 1 1... ... ...16 16 16 16 16 16S a b cb b c c a a

2 2 217 17 17

16 32 16 32 16 32

17 17 178 16 8 16 8 16

17 17 1716 16 1617 16 16 16

a b cb c aa b cb c a

3 17 17 17 17

8 16 8 16 8 16 8 5 5 5117 3 3 1716 16 16 16

a b cb c a a b c

5 1517 173 17 3 17 3 17

22 (2 2 2 ) 2 2 22 3a b c a b c

. Vi 12a b c th

3 17Min 2S

Cch 2: Bin i v s dng bt ng thc BunhiaCpski ta c

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2 2 2 22 2

2 2 2 22 2

2 2 2 22 2

1 1 1 1 41 417 171 1 1 1 41 417 171 1 1 1 41 417 17

a a a bb bb b b cc cc c c aa a

1 4 4 417S a b c a b c

1 1 1 1 15 1 1 14 4 4 417 a b c a b c a b c

6 3

31 1 1 1 15 1 1 1 1 45 16 3 34 4 4 4 417 17abc a b c a b c abc

1 45 1 1 45 3 173 3 24 4 217 173

a b c . Vi 12a b c th

3 17Min 2S

Cch 3: t 1 1 1 , ; , ; ,u a v b w cb c aDo u v w u v w nn suy ra :

2

22 2 22 2 21 1 1 1 1 1S a b c a b c a b cb c a

2 2

2 1 1 1 1 15 1 1 116 16a b c a b c a b c

23151 1 1 1 1 1 12 34 16a b c a b c a b c



3 3

23

1 135 11 1 13 32 16abc a b c abc 29 135 12 16 3a b c

9 135 18 135 153 3 1742 16 4 4 4 2 . Vi 12a b c th

3 17Min 2S

Bi 6. a) Lp bng bin thin v tm gi tr ln nht ca hm s 231

xy x b) Cho 1a b c . Chng minh rng: 2 2 21 1 1 10a b cGii. a) TX: D ; 2 21 3 1 10 103 31 1xy x yx x

2 223 / 3 /lim lim lim lim11 1x x x x

x x x x xy xxxx

.

Suy ra lim 1; lim 1x xy y . Nhn BBT

ta c 23 10 max 101

xy yxb) Theo phn a) th 10 ,y x 23 10. 1,x x x .c bit ha bt ng thc ny ti cc gi tr , ,x a x b x c ta c:

2

2

2

: 3 10. 1: 3 10. 1: 3 10. 1

x a a ax b b bx c c c

2 2 29 10. 1 1 1a b c a b c 2 2 210 1 1 1a b c

Cch 2. Trn mt phng ta Oxy t ;1 ; ;1 ; ;1OA a AB b BC c .

Khi ; 3OC OA AB BC a b c .

x 1/3 y + 0 0

y1

10

1

a a+b

a+b+c

C

AB

123

O x1

y

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Do OA AB BC OA AB BC OCT suy ra 2 2 21 1 1 10a b c



BI TP P DUNG:Bi 1. Cho , , 0,1x y z tho mn iu kin: 32x y z .

Tm Max, Min ca biu thc: 2 2 2cosS x y zGii. Do , , 0,1x y z nn 2 2 2 30 2 2x y z x y z .V hm s cosy nghch bin trn 0, 2 nn bi ton tr thnh.1. TmMaxS hay tmMin 2 2 2x y z

22 2 2 2 2 2 2 2 2 31 1 1 13 4x y z x y z x y z .Vi 12x y z th MaxS =

3cos 42. TmMinS hay tmMax 2 2 2x y zCch 1: Phng php tam thc bc hai:

Khng mt tnh tng qut gi s 1, , ;12z Max x y z z . Bin i v nh gia v tam thc bc hai bin z

222 2 2 2 2 23 92 2 32 4x y z z x y xy z z z z f zDo th hm y = f(z) l mt parabol quay b lm ln trn nn ta c:

51 1Max Max ; 1 12 2 4f z f f f f .Vi 11; ; 02z x y th MinS =

5cos 4Cch 2: Phng php hnh hcXt h ta cc vung gc Oxyz. Tp hp cc im , ,M x y z tho mn iukin , , 0,1x y z nm trong hnh lp phng ABCDA BCO cnh 1 vi A(0, 1, 1);B(1, 1, 1); C(1, 0, 1); D(0, 0, 1); A(0, 1, 0); B(1, 1, 0); C(1, 0, 0).

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Mt khc do 32x y z nn , ,M x y z nm trn mt phng (P): 32x y zVy tp hp cc im , ,M x y z tho mn iu kin gi thit nm trn thit dinEIJKLN vi cc im E, I, J, K, L, N l trung im cc cnh hnh lp phng. GiO l hnh chiu ca O ln EIJKLN th O l tm ca hnh lp phng v cng ltm ca lc gic u EIJKLN. Ta c O M l hnh chiu ca OM ln EIJKLN. DoOM2 = 2 2 2x y z nn OM ln nht OM ln nht M trng vi 1 trong 6 nh E, I, J, K, L, N.T suy ra:

2 2 2 2 511 4 4x y z OK 2 2 2 5cos cos 4x y z

Vi 11; ; 02z x y th MinS =5cos 4

Bi 2. ( thi TSH 2007 khi B)

Cho , , 0x y z . Tm Min ca S 1 1 1

2 2 2y zxx y zyz zx xy

Gii: S dng bt ng thc Csi cho 9 s ta c

S 4 4 42 2 2 9 4 4 4

9 9 91 . Min2 2 2 2y y x y zz zx xx y z Syz yz zx zx xy xy x y z

Bi 3. ( thi TSH 2005 khi A)Cho , , 0x y z ; 1 1 1 4x y z . Tm Min ca S

1 1 12 2 2x y z x y z x y z

Gii: S dng bt ng thc Csi cho cc s a, b, c, d > 0 ta c:

y3/ 2

OE

1

1K

3/ 2J

M

z

x

I

L

N

3/ 21

O



4 41 1 1 1 14. .4. 16161 1 1 1

a b c d abcda b c d abcda b c d a b c d

16 161 1 1 12

16 161 1 1 12

16 161 1 1 12

1 1 1 1 1 116 4 16 Min 12 2 2

x x y z x x y z x y zx y y z x y y z x y zx y z z x y z z x y z

Sx y z x y z x y z x y z

Bi 4. Cho x,yR tha mn iu kin 22 2 2 2 2 2 1 4 0x y x y x yTm gi tr ln nht, nh nht ca biu thc S= 2 2x y

Gii. Bin i 22 2 2 2 2 2 2 22 1 4 0x y x y x y x y 22 2 2 2 23 1 4 0x y x y x 22 2 2 2 23 1 4x y x y xDo 4x2 0 nn 22 2 2 23 1 0x y x y 2 23 5 3 52 2x y

Vi x = 0, y = 3 52 , th 2 2 3 5Min( ) 2x y .

Vi x = 0, y = 3 52 , th 2 2 3 5Max( ) 2x y

Bi 5. Cho x2 + xy + y2 = 3. Tm gi tr ln nht v nh nht ca biu thc:S = x2 xy + y2

Gii Xt y = 0 x2 = 3 S = 3 l 1 gi tr ca hm s.Xt y 0, khi bin i biu thc di dng sau y

22 2 2

2 2 2 2/ ( / ) 1 1

3 ( / ) ( / ) 1 1x y x yS x xy y t tu ux xy y x y x y t t vi

xt y u(t2 + t + 1) = t2 t + 1 (u 1)t2 + (u + 1)t + (u 1) = 0 (*)

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+ Nu u = 1, th t = 0 x = 0, y = 3 u = 1 l 1 gi tr ca hm s+ Nu u 1, th u thuc tp gi tr hm s phng trnh (*) c nghim t = (3u 1)(3 u) 0 1 1 33 u .

Vy tp gi tr ca u l 1 ,33

1Min 3u ; Max u = 3

Min S = 1 1Min 3u t = 1 2 2

13x y x yx xy y

Max S = 9 Maxu = 3 t = 1 2 23, 3

3 3, 3x y x yx xy y x y

Bi 5. Tm gi tr ln nht ca hm s 2 2 2sin sin sin ( )S x y x y

Gii . 2 2 2sin sin sin ( )S x y x y = 21 cos2 1 cos2 1 cos ( )2 2x y x y

S

2 29 12 cos( )cos( ) cos ( ) cos( )cos( ) cos ( )4 4x y x y x y x y x y x y

S 2

29 91 1cos( ) cos( ) sin ( )4 2 4 4x y x y x y .

Vi 3x y k , (k) th 9Max 4S

BI TP T LUYN:Bi 1. Gi s ( ; ; ) / 0, 0, 0, 1D x y z x y z x y z . Tm gi tr ln nht cabiu thc: 1 1 1

x y zP x y z .

HD: 1 1 13 1 1 1P x y z

S dng bt ng thc Csi:



1 1 1( 1) ( 1) ( 1) 91 1 1x y z x y z

P 34 . Du = xy ra x = y = z =13 . Vy

3min 4D P .

Bi 2. Cho D = 5( ; ) / 0, 0, 4x y x y x y . Tm gi tr nh nht ca biu thc:

4 14S x y .

HD: 1 1 1 1 14 254x x x x y x x x x y

4 14( ) 254x y x y

S 5. Du = xy ra x = 1, y = 14 . Vy minS = 5.

Bi 3. Cho D = ( ; ) / 0, 0, 1x y x y x y . Tm gi tr nh nht ca biu thc:

2 2 11 1x yP x yx y x y .

HD: 2 2 1(1 ) (1 ) 21 1

x yP x yx y x y = 1 1 1 21 1x y x y .

S dng bt ng thc Csi:

1 1 1(1 ) (1 ) ( ) 91 1x y x y x y x y

1 1 1 9

1 1 2x y x y

P 52 . Du = xy ra x = y =13 . Vy minP =

52 .

Bi 4. Cho D = ( ; ) / 0, 0, 4x y x y x y . Tm gi tr nh nht ca biu thc:

2 22

3 4 24x yP x y .

HD: 21 124 8 8 2

x y y x yP x y (1)

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Theo bt ng thc Csi: 1 12 . 14 4x x

x x (2)

32 21 1 33 . .8 8 8 8 4y y y y

y y (3)

P 92 . Du = xy ra x = y = 2. Vy minP =92 .



BI 4. NG TIM CN CA HM S:A. KIN THC CN NH1. nh ngha:

ng thng 0x x gl ng tim cn ng ca th hm s ( )y f xnu t nht mt trong cc iu kin sau c tho mn:

0lim ( )x x

f x ; 0lim ( )x x f x ;

0lim ( )x x

f x ; 0lim ( )x x f x

ng thng 0y y gl ng tim cn ngang ca th hm s ( )y f xnu t nht mt trong cc iu kin sau c tho mn:

0lim ( )x f x y ; 0lim ( )x f x y ng thng , 0y ax b a gl ng tim cn xin ca th hm s ( )y f x nu t nht mt trong cc iu kin sau c tho mn:

lim ( ) ( ) 0x f x ax b ; lim ( ) ( ) 0x f x ax b

2. Ch :

a) Nu ( )( ) ( )P xy f x Q x l hm s phn thc hu t.

Nu Q(x) = 0 c nghim x0 th th c tim cn ng 0x x . Nu bc P(x) bc Q(x) th th c tim cn ngang. Nu bc P(x) = bc Q(x) + 1 th th c tim cn xin.b) xc nh cc h s a, b trong phng trnh ca tim cn xin , ta c th pdng cc cng thc sau:

( )lim ; lim ( )x x

f xa b f x axx

hoc ( )lim ; lim ( )x x

f xa b f x axx

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Cc tnh gii hn v cc ca hm s ( )( )f xy g x

0lim ( )x x f x 0lim ( )x x g x Du ca g(x) 0

( )lim ( )x xf xg x

L Tu 0

+ +L>0 0 - -

- +L



B. PHN LOI V PHNG PHP GII BI TP:

BI TP P DNG:Bi 1. Tm cc ng tim cn ca cc hm s sau:

2 1 3 4 4) ; ) ; ) ; )2 1 1 2 6x x xa y b y c y d yx x x x

Hng dn:a) Hm s cho xc nh trm \ {0} .Ta c:

lim ( ) 1 y=-1 la tiem can ngang cua o th ham so khix f x x

lim ( ) 1 1 la tiem can ngang cua o th ham so khix f x y x

0 0lim ( ) , lim ( ) 0 la tiem can ng cua o th ham so khi

0 va 0x x

f x f x xx x

( )lim 0 Ham khong co tiem can xien khix

f x xxCc cu khc lm tng tBi 2. Tm cc ng tim cn ngang v ng ca cc hm s sau:

2 22 22

2 2

2 5 1 4) ; ) ;2 4 3 4 72 3 1) ; )1 3

x x x xa y b yx x x xx x xc y d yx x

Hng dn:

DNG 1: TM TIM CN NGANG V NG CA TH HM SBNG NH NGHA

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2 221 1

2

1 12 2

21 1

)Ham co tiem can ngang la: 1;Tiem can ng la: 1. V:

2 3 2 3 1lim lim .1 111 1 2 3v lim 0; lim1 2 12 3 2 3 1lim lim .1 11

x x

x x

x x

c yx

x x x xx xxx x

x xx x x x

x xx

2

1 1

1

1 2 3v lim ; lim 3 01 1Tng t cho lim ( )

x x

x

x xx x

f x

Cc cu khc lm tng tBi 3. Tm TCN v TC ca th hm s:

33 2) )27 5xa y b yx x

Hng dn:

5 5

) Tap xac nh : D= ;52Ta co: lim lim 5

Vay, o th co tiem can ng 5. Mat khac: lim 0 : 0x x

x

by x

x y TCN y

Bi 4. Tm TCN v TC ca th hm s:

2

22 1 2 1) ; )2 1 2x xa y b yx x x

;2 1) xc y x

Hng dn:

0 0

c)Ham so xac nh tren \{0}lim ( ) 1 1 la tiem can ngang cua o th ham so khilim ( ) 1 1 la tiem can ngang cua o th ham so khilim ( ) ; lim ( )

x

x

x x

f x y xf x y xf x f x

22

0 la tiem can ng cua o th ham so khi 0va 0

( ) 1lim lim 0 ham so khong co tiem can xien khix x

x xxf x x xx x



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BI TP T LUYN:Bi 1. Tm tim cn ng v tim cn ngang ca th hm s sau:

22

3

2 3 2 7. . 1 - 43 -4. . 127

x x xa y b yx xxc y d y xx

Bi 2. Tm tim cn cc hm s

2

2 3 1. . .1 1 4

x x x xa y b y c yx x x

2

-2

-4

-5 5

2

-2

-4

-5 5

2

-2

-4

-5 5



BI TP P DNG:

Bi 1. Tm gi tr ca tham s m sao cho th hm s 2 2 1x my x m c

tim cn ng qua im M( -3;1)Hng dn:

Ta c: 2 2 1 12x my x m x m c tp xc nh l \ { }D m

lim lim TCD:x m x m

y y x m

T ta tm c 3m

Bi 2. Tu theo m, tm cc ng tim cn ca 2 2: 6xC y x x mHng dn:

2

1 2

1 2

Xet phng trnh 6 0 vi 99 :ham khong co tiem can ng. TCN: 09 : : 3; : 0

8 9 :Phng trnh (*) co hai nghiem phan biet ; .: ;: 0

18 : , 2 ...4

x x m mm ym TC x TCN y

m x xTC x x x xTCN ym y xx

....

Bi 3. Cho hm s : 3x 4y x 2 . Tm im thuc (C) cch u 2 ng tim cn .

Hng dn:Gi M(x;y) (C) v cch u 2 tim cn x = 2 v y = 3

| x 2 | = | y 3 | 3x 4 xx 2 2 x 2x 2 x 2

x 1x x 2 x 4x 2

DNG 2: MT S BI TON LIN QUAN N TIM CN. TM THAM S MTHA IU KIN

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Vy c 2 im tho mn bi l : M 1( 1; 1) v M2(4; 6)

Bi 4. Cho hm s 2 11xy x . Tm nhng im trn (C) sao cho tng khong cch n

hai tim cn l nh nht.Hng dn:

00

0

00 0

0 0

2 1Goi M ; ( ).Goi A va B lan lt la hnh chieu cua M len tiem can ng12 1 1va tiem can ngang th MA= 1 ; 2 21 1

............Co hai iem M

xx Cxxx MB y x x

BI TP T LUYN:

Bi 1. Tm gi tr ca tham s m sao cho 2 72x my x m c tim cn ng qua im

( 7;1)M Bi 2. Tm m th ca cc hm s sau c ng hai tim cn ng:

a) 2 23

4 2(2 3) 1y x m x m b)

22

23 2( 1) 4

xy x m x

c) 23

2xy x x m d)

2 23

2( 2) 1xy x m x m

e) 2 21

2( 1) 2xy x m x m f) 2

32 2 1y x mx m

Bi 3. Cho hm s 2 1x my mx . Tm m sao cho th hm s c tim cn ng,

tim cn ngang v cc tim cn cng vi hao trc ta to thnh mt hnh chnht c din tch bng 8.Hng dn:

1 2 1ieukien : 0; 8 . 8 2m S mm m



Bi 4. Tm m th hm s 2 3 1 21mx m x my x

c tim cn xin ,

bit tip xc vi ng trn tm I(1;2), bn knh 2Hng dn:

ieukien : 0;Tiem can xien : 2 1 2 1 0

1tiep xuc vi ng tron I 1;2 , ban knh bang 2 ( ; ) 2 1

7

m y mx m mx y mm

d I m

Bi 5. Tm cc ng tim cn ca ng cong: 2 3( ) : 4xC y x x m

Bi 6. Ty theo gi tr ca tham s m. Tm tim cn ca th hm s sau: 311

xy mxHng dn:

3

1 1

* 0 - 1:ham khong co tiem can1* 1 ( ) lim ( ) 0 y=0 la tiem can ngang cua o th ham so1

1V lim ( ) lim ( ) o th ham so khong co tiem can ng30* 1

x

x x

m y xxm f x f xx

f x f xmm

3

3

1 ham so xac nh tren \ng thang y=0 la tiem can ngang cua o th ham so

1ng thang x= la tiem can ng cua o th ham so

m

m

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CH : TIM CN XIN (NHM 2: SEMINAR)MT S BI TP THO LUN:Bi 1. Tm tim cn xin ca th hm s:

2 3 22

2 -3 5 2 2 5 1. 2 1 . .1 3 1 1x x x xa y x b y c yx x x x

Bi 2. Tm tim cn ca cc th hm s sau:

2

2) 2 2) 1

a y x xb y x xHng dn:a) Hm s cho xc nh trm .Ta c:

( )lim 1; lim ( ) 11 la tiem can xien cua o th ham so khi

x xf xa b f x axx

y x x


x xf xa b f x axx

y x xb) Hm s cho xc nh trn ; 1 1; .Ta c:


x xf xa b f x axx

y x x


x xf xa b f x axx

y xNhn xt:1. Xt hm s 2( ) 0f x ax bx c a

Neu 0 th o th ham so khong co tiem can xiena



Neu 0 th o th ham so co tiem can xien khi2

va co tiem can xien khi2

ba y a x abx y a x xa

2. Xt hm s 2( ) 0f x mx n p ax bx c a th hm s co tiem can xien la ng thang y 2

bmx n p a x a

Bi 3. Tm tim cn ca th hm s sau: 22 4 2y x x x

Bi 4. Cho hm s 2 1 ( )1

x xy Cx .

a) Chng minh tch khong cch t mt im bt k trn (C) n hai ngtim cn l khng i

b) Khng c tip tuyn no ca (C) i qua giao im ca hai tim cn.Hng dn:

a) 0 00

3( ) ; 2 1M C M x x x

1

2

la : 1 0 la : 2 0

TCN xTC x y

. T : 1 2 3 2. 2d d pcm

b) 1 21;3I .

0 0 00

Gia s la tiep tuyen bat ky cua (C), luc o co dang:: '( )

6 0 : phng trnh nay vo nghiem. Vay khong co tiep tuyen nao1cua o th i qua I

y f x x x yI x

Bi 5. Cho hm s 2 23 2 2 ( )3

mx m xy Cx m .

1. Tm m gc gia hai tim cn bng 4502. Tm m ng tim cn xin ct hai trc ta ti cc im A,B to thnh

tam gic c din tch bng 4.

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Hng dn:6 2Ta co: -2 3

1 th ham so co tiem can 6m-2 0 m 3: 3 0; : 2 0

my mx x moTCN x m TC mx y

1. Gc gia hai tim cn bng 045 1 201 2

. 2os45 12.n nc mn n

2. Hm c tim cn xin 0 2.Khi o: A 0; 2 ; ;013

mB mm

1 . 4 22ABCS OAOB m

Bi 5. Tm gi tr ca tham s m sao cho th ham s 22 3 2

1x mx my x c

tim cn xin to vi cc trc to mt tam gic c din tch bng 4.

Bi 6. Tm m th ca cc hm s sau c tim cn xin:

a) 2 (3 2) 2 1

5x m x my x b)

2 (2 1) 3

2mx m x my x

Bi 7. Tnh din tch ca tam gic to bi tim cn xin ca th cc hm s sauchn trn hai trc to :

a) 23 1

1x xy x b)

23 4

2x xy x c)

2 7

3x xy x

Bi 8. Tm m tim cn xin ca th cc hm s sau to vi cc trc to mttam gic c din tch S ch ra:

a) 2 1

1x mxy x ; S = 8 b)

2 (2 1) 2 3

1x m x my x ; S = 8

c) 22 2(2 1) 4 5

1x m x my x ; S = 16 d)

22 2

1x mxy x ; S = 4



Bi 9. Chng minh rng tch cc khong cch t mt im bt k t rn th ca cchm s n hai tim cn bng mt hng s:

a) 2 1

1x xy x b)

22 5 4

3x xy x c)

2 7

3x xy x

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DNG 3: CC DNG TON LIN QUANNNG TIM CN HMPHN THCBi ton 1: Cc bi ton lin quan n khong cch

Bi 1. Cho hm s 1 ( )2xy Cx

a) Chng minh rng tch cc khong cch t im M thuc (C) n hai ngtim cn bng mt s khng i

b) Tm im M thuc (C) tng khong cch t M n hai ng tim cn tgi tr nh nht.

Hng dn:

a) 0 01;1 ( )2M x Cx

. Gi 1 2;d d ln lt l khong cch t M n tim

cn ng v tim cn ngang th 1 2. 1d d

b) 1 2 0012 22d d x x

Bi ton 2: Da vo tnh cht hai nhnh ca th (C) nm v hai pha (C) cang tim cn.

Bi 1. Cho hm s 2 ( )2 1xy Cx . Vi gi tr no ca m th ng thng

: 1md y mx m ct (C)a) Ti hai im thuc cng mt nhnh ca thb) Ti hai im thuc hai nhnh ca th

Hng dn:Phng trnh honh giao im ca ng thng ( )md v (C)

212 21 2 1 2 3 1 3 0 (1)

xxmx m x mx m x m



a) ng thng ( )md ct (C) ti hai im thuc cng mt nhanh ca th khi(1) c hai nghim 1 2,x x sao cho

1 2 1 201 1 (hoac - )2 2 3

mx x x x m

b) ng thng ( )md ct (C) ti hai im thuc cng hai nhanh ca th khi(1) c hai nghim 1 2,x x sao cho 1 20 0x x m

Bai 2. Cho ham so 2 ( )1xy Cx

a) Tm cac iem thuoc hai nhanh cua o th sao cho khoang cach cua chungla ngan nhat

b) Goi (d) la ng thang qua A(1;0) co he so goc k. Tm k e (d) cat (C) taihai iem M, N thuoc hai nhanh cua (C) sao cho 2AM AN

Hng dan:a) Goi P va Q lan lt la cac iem thuoc nhanh phai va nhanh trai cua o th

ham so th 3 31 ;1 0 ; 1 ;1 0 .P a a Q b ba b Ta co: 222 1 1 369 4 24PQ a b aba b ab

2 6 3364a b

MinPQ a bab ab

b) Phng trnh ng thang (d): 1y k x . Phng trnh honh giao

im ca ng thng ( )d v (C)

2121 1 2 1 2 0 (1)

xxk x x kx k x k

ng thng ( )md ct (C) ti hai im thuc hai nhanh ca th khi (1) c hainghim 1 2,x x sao cho 1 21 0x x k .

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1 1 2 2 1 1 2 1

1 2

Ta co: ; ; ; nen 1; ; 1;22 1 2 1 3

M x y N x y AM x y AN x yAM AN x x k

Bi ton 3: Tnh chat tiep tuyen tai mot iem tuy y thuoc (C)Bai 1. Cho ham so 2 1 ( )1

xy Cx

Goi I la giao iem cua hai ng tiem can, M la mot iem tuy y tren (C). Tieptuyen tai M cua o th (C) cat hai ng tiem can tai P va Q.

a) Chng minh rang M la trung iem cua PQ va dien tch tam giac IPQ khongoi

b) Tm tren o th (C) iem M sao cho 2 2IP IQ

Hng dan:

00 0

0 00

2 1; ( ). Phng trnh tiep tuyen tai M cat hai ng tiem can lan12lt tai hai iem 1; va Q 2 1; 21

) Ta co: 2 . Vay M la trung iem cua PQS

P Q M

IP

xM x CxxP xx

a x x x

00

1 1 2. . .2 1 22 2 1Q IP IQ xx

00

) Theo ket qua cau a) th M la trung iem PQ nen 202 2

b IP IQ IMxIM x



BAI TAP T LUYENBai 1. Cho ham so 2( ) : x mC y x m

. Goi I la giao iem cua hai ng tiem can.Tm m e ng tron tam I, ban knh 2R tiep xuc vi 2y x m .

Bai 2. Cho 7( ) : 2xC y x . Goi I la giao iem cua hai ng tiem can. Tm

( ) : 2 sao cho IM nho nhatM d y x .

Bai 3. Cho ham so 2 1( ) : 1xC y x . Goi I la giao iem hai ng tiem can. Tm

tren o th (C) iem M sao cho tiep tuyen tai M vi o th (C) cat hai ng tiemcan tai A va B thoa man 2 10IA IB

Bai 4 .Cho ham so 2 3( ) : 2xC y x . Goi I la giao iem hai ng tiem can. M la

iem bat k tren (C), tiep tuyen tai M vi o th (C) cat hai ng tiem can tai Ava B. Tm toa o iem M sao cho ng tron ngoai tiep tam giac IAB co dien tchnho nhat.

Bai 5. Cho ham so 2 4( ) : 1xC y x

. Tm tren o th (C) iem M sao cho tieptuyen tai M vi o th (C) tao vi hai ng tiem can mot tam giac co chu vi nhonhat

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BI 5: KHO ST HM SVN 1 : HM TRNG PHNG

Min xc nh : D=o hm:

3 2' 4 2 2 2y ax bx x ax b Phng trnh ' 0y hoc c mt nghim ( . 0a b ) hoc c 3 nghim phnbit. Do hm s hoc ch c mt cc tr hoc c ba cc tr.Gii hn:

42 4

khi 0lim lim 1 ax ax khi 0x xab cy ax a

Bng bin thin:Du ca 'y ph thuc vo du ca ( 0 0)a a haya v du ca a.b, do ta c bn trng hp bng bin thin khc nhau. th hm s: Do bn trng hp khc nhau v chiu bin thin nn th ca hm trng phng c bn dng sau y:

DNG 1: Kho st s bin thin v v th hm sy=ax4 +bx2+c ( 0)a



Hm s chn nn th nhn trc Oy lm trc i xng

a > 0 a < 0

y = 0 c 3 nghimphn bit ab < 0

y = 0 ch c1 nghim ab > 0

y

x0

y

x0

y

x0

y

x0

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MT S BI TP RN LUYN KHO ST HM S:Bi 1. Kho st s bin thin v v th hm s (trng hp c 3 cc tr):

44 2 23) ) 4 2

xa y x x b y x

Bi 2. Kho st s bin thin v v th hm s (trng hp c 1 cc tr):4

4 2 21 3 3) )2 2 4 2xa y x x b y x

BI TP T LUYN:Bi 1. Kho st s bin thin v v th hm s

4 2 4 2

4 2 4 2

1) 1 ) 4 202) 4 3 ) 2 1

a y x x b y x xc y x x d y x x

Bi 2. Kho st s bin thin v v th hm4 2 4 2

4 2 4 2

4 2 4 2

. - 2 b. 21 5. 6 1 . 32 2

. - 2 3 . 2 1

a y x x y x xc y x x d y x xe y x x f y x x



Mt s tnh cht ca hm trng phng1. Hm s lun c cc tr vi mi gi tr ca tham s sao cho 0a 2. Hm s t gi tr cc i, cc tiu (c ba cc tr)

2' 0 2 (2 ) 0y x ax b c ba nghim phn bit 02ba

3. th hm s lun nhn Oy l trc i xng.

4. Hm s c hai cc i v mt cc tiu 00ab

5. Hm s c mt cc i v hai cc tiu 00ab

6. Nu hm s c ba cc tr tr chng to thnh mt tam gic cn.7. th (C) ct Ox ti 4 im phn bit lp thnh cp s cng:

2

2

1 2 2 1 1 22 12 2

4 2( ) Ox , , , : 0 (*) co 4 nghiem tao thanh CSC

0, 0. Luc o: (*) 00 0 3( ) 0 ( ) 0

Giai he p

C A B C D AB BC CD hayax bx c

tat t x t at bt ct t t t t tycbt t tg t at bt c g t at bt c

2 1

1 2

1 2

9hng trnh :

t tS t tP t t

8. iu kin cn t mt im trn trc i xng k n th hmtrng phng (C) ba tip tuyn l ba tip tuyn phi c mt tiptuyn nm ngang.

DNG 2: MT S BI TON LIN QUAN

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9. iu kin ca tham s th hm s 4 2 ( 0)y ax bx c a tip

xc vi Ox ti hai im phn bit:02

02

ba

by a

10. Phng trnh trng phng 4 2 0 ( 0) (*)ax bx c a t 2 , 0t x t lc phng trnh tr thnh 2 0 0at bt c a . Tathy rng: c 1 nghim dng ca (**) th s cho ra 2 nghim (1 m, 1dng) ca phng trnh (*).Vy: iu kin cn v phng trnh(*) c nghim l phng trnh(**) c t nht 1 nghim khng m.

Phng trnh (*) c 4 nghim (**) c 2 nghim dng phn

bit000

PS

Phng trnh (*) c 3 nghim (**) c 1 nghim dng v 1

nghim bng 0 00PS

Phng trnh (*) c 2 nghim (**) c 1 nghim dng

0P 002

S

Phng trnh (*) c 1 nghim (**) c nghim tha

1 2

1 2

0000002

PSt t

t t S



Phng trnh (*) v nghim (**) v nghim hoc c 2 nghim

m0000

PS

MT S BI TP IN HNHBi 1. Cho hm s 4 2( ) 2y f x x x

1. Kho st v v th (C) ca hm s.2. Trn (C) ly hai im phn bit A v B c honh ln lt l a v b . Tm

iu kin i vi a v b hai tip tuyn ca (C) ti A v B song song vi nhau.Hng dn:Ta c 3'( ) 4 4f x x x . Gi a, b ln lt l honh ca A v B.H s gc tip tuyn ca (C) ti A v B l

3 3'( ) 4 4 , '( ) 4 4A Bk f a a a k f b b b Tip tuyn ti A, B ln lt c phng trnh l:

' ' ( ) af' ay f a x a f a f a x f a ; ' ' ( ) f' by f b x b f b f b x f b b

Hai tip tuyn ca (C) ti A v B song song hoc trng nhau khi v ch khi: 3 3 2 24a 4a = 4b 4 1 0 (1)A Bk k b a b a ab b

V A v B phn bit nn a b , do (1) tng ng vi phng trnh:2 2 1 0 (2)a ab b

Mt khc hai tip tuyn ca (C) ti A v B trng nhau

2 2 2 2

4 2 4 21 0 1 0

' ' 3 2 3 2a ab b a ab ba bf a af a f b bf b a a b b

,

Gii h ny ta c nghim l (a;b) = ( -1;1), hoc (a;b) = (1;-1), hai nghim nytng ng vi cng mt cp im trn th l 1; 1 v 1; 1 .Vy iu kin cn v hai tip tuyn ca (C) ti A v B song song vi nhau l

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2 2 1 01

a ab baa b

Bi 2. Cho hm s 4 3 2x 2x 3 x 1 (1)y x m m .

1). Kho st s bin thin v v th (C) ca hm s (1) khi m = 0.2). nh m hm s (1) c hai cc tiu.

Hng dn:4 3 2x 2x 2 x 1y x m m (1)

o hm / 3 2 24 3 4 3 ( 1)[4 (4 3 ) 3 ]y x mx x m x x m x m

/ 210 4 (4 3 ) 3 0 (2)

xy x m x m

Hm s c 2 cc tiu y c 3 cc tr y/ = 0 c 3 nghim phn bit

(2) c 2 nghim phn bit khc 12(3 4) 0 4 .34 4 3 3 0

m mm m

Gi s: Vi 43m , th y/ = 0 c 3 nghim phn bit 1 2 3, ,x x x

Bng bin thin:x - x1 x2 x3 +y/ - 0 + 0 - 0 +y +

CTC

CT+

T bng bin thin ta thy hm s c 2 cc tiu.

Kt lun: Vy, hm s c 2 cc tiu khi 4 .3m

Bi 3.1.Kho st s bin thin v v th hm s y = x4 4x2 + 3



2.Tm a phng t rnh : 4 2 34 log 3 0x x a c 4 nghim thc phnbitHng dn:Theo th cu 1 bi ton yu cu tng ng 1 3log a < 3 3log 1a 31 log 1a

Bi 4. Cho hm s 4 2 22(1 ) 1y x m x m 1: Kho st s bin thin v v th hm s vi m=0 2: Tm m hm s c cc i cc,cc tiu v cc im cc tr ca th hms lp thnh tam gic c din tch ln nht.Hng dn:y'=4x3-4(1-m2)xLp lun hm s c cc i,cc tiu khi v ch khi 1m Ta cc im cc tr:A(0;m+1); B( 2 4 21 ; 2m m m m ) ; C(- 2 4 21 ; 2m m m m )

S ABC = 2 4 2 2 51 . ( ; ) 1 2 1 (1 ) 12 BC d A BC m m m m .Du bng xy rakhi m=0Vy m=0Bi 5.Cho hm s 4 25 4,y x x c th (C)1. Kho st v v th (C).2. Tm m phng trnh 4 2 2| 5 4 | logx x m c 6 nghim phn bit.Hng dn:

944

129log 12 144 124m m

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Bi 6. Cho hm s: 4 2 2( 10) 9y x m x .1.Kho st s bthin v v th ca hm s ng vi m = 02)Tm m th ca hs ct trc honh ti 4 im pbit 1 2 3 4, , ,x x x x tha :

1 2 3 4 8x x x x Hng dn:Phng trnh honh giao im ca (C m) v Ox.4 2 2( 10) 9 0x m x (1) t 2( 0)t x t Ptrnh tr

thnh: 2 2( 10) 9 0t m t (2) Ta c k:2 2

2

( 10) 36 09 0 ,

10 0,

mP mS m m

=> 0 < t1 < t2 , vi 2t x x t

V hs cho l hs chn v theo bi ta c : 1 2 1 2 1 24 2 . 16t t t t t t (3)

p dng Viet : 21 2 1 210 , 9b ct t m t ta a .

Ta c pt: m2 + 10 = 10 m = 0.Bi 7. Cho hm s 4 23 1 2 11y x m x m . Tm m (Cm) ct ng thng(d): 2 13y m ti 4 im c honh lp thnh cp s cng.Hng dn:Phng trnh honh giao im ca (C m) v Ox.

4 23 ( 1) 2 0x m x (1). t 2( 0)t x t . Ptrnh cho tr thnh:2 ( 1) 2 0t m t (2) Ta c k:

00 2 6 10

P mS

.



Khi phng trnh (2) c hai nghim dng 1 2;t t . Gi s 1 20 t t , khi phng (1) c 4 nghim phn bit c sp xp theo th t tng dn l

2 1 1 2, , ,t t t t . Bn nghim ny lp thnh cp s cng nn:

1 2 1 2 1 2 12 3 9 (1)t t t t t t t Theo nh l vi-t ta c: 1 2

1 2

1 (2)32. (3)3

mt t

t t

T (1) v (2) ta tm c : 1 2 9 11;30 30mmt t v t (3) cho ta:

10 61 ( )310 61 3

m loai

m

MT S BI T LUYN:Bi 1 (TNTHPT-2008). Cho hm s 4 22y x x

a. Kho st s bin thin v v th ca hm s.b. Vit phng trnh tip tuyn ca th hm s ti im c honh x = -2

Bi 2. Cho hm s 4 3 24 3( 1) 1y x mx m x a. Kho st s bin thin v v th hm s vi m =0b. Vi gi tr no ca m hm s c 3 cc tr

Bi 3 (H Lt - 2002)a. Gii phng trnh 4 22 1 0x x b. Kho st v v th hm s y = 4 22 1x x c. Bin lun theo m s nghim ca phng trnh 4 22 1 0x x m

Bi 4 (H Thi Nguyn - 2002) Cho hm s 4 2 m2 (C )y x mx a. Kho st v v th hm s vi m = 1b. Hy xc nh m hm s th hm s c 3 cc tr



Bi 5. (H Vinh - 2002)1. Kho st v v th hm s 4 25 4y x x 2. Xc nh m phng trnh 4 2 25 3 0x x m c 4 nghim phn bit.

Bi 6. Cho hm s4

2 924 4xy x

a. Kho st s bin thin v v th (C) ca hm sb. Bin lun theo k s giao im ca (C) vi th (P) ca hm s 22y k x

Bi 7. Cho hm s 4 2 3 22y x mx m m a. Kho st s bin thin v v th ca hm s khi m = 1b. Xc nh m th ( )mC ca hm s cho tip xc vi trc honh ti 2

imBi 8. (H Cn th - 2002). Cho hm s 4 22 2y x x m (Cm)

a. Kho st s bin thin v v th hm s vi m = 0b. Tm cc gi tr ca m th (Cm) ca hm s ch c hai im chung vi

Oxc. Chng minh vi mi m tam gic c 3 nh l ba cc tr l mt tam gic

vung cn.Bi 9. Cho hm s 4 2 22 1y x m x

a. Kho st v v th hm s vi m =1b. Tm m th hm s c ba cc tr l ba nh ca tam gic vung cn.

Hng dn:Hm s cho xc nh trn Ta c: 2 2' 4y x x m . Vi 0m hm c ba cc tr. Khi ta cc im cctr l 4 40;1 ; ;1 ; ;1A B m m C m m .D thy

. 0AB ACAC AB nn tam gic ABC vung cn

2 2 2 1AB AC BC m .



Vy, 1m l nhng gi tr cn tmBi 10. Cho hm s y = x4 2(2m2 1)x2 + m (1)

1/ Kho st s bin thin v v th ca hm s (1) khi m = 1.2/ Tm m th ca hm s (1) tip xc vi trc hanh.

Bi 11.1. Kho st s bin thin v v th ca hm s: y = x 4 6x2 + 52. Tm m phng trnh: x 4 6x2 log2m = 0 c 4 nghim phn bit trong 3nghim ln hn 1.Hng dn: Pt x4 6x2 + 5 = 5 + log2m Nhn vo th ta thy yu cu bi ton 0 < 5 + log2m < 5 1/32 < m < 1Bi 12. Cho hm s 4 28 9 1y x x

1. Kho st s bin thin v v th hm s2. Da vo th bin lun s nghim ca phng trnh:

4 28cos 9cos 0, 0;x x m x Hng dn:

4 2

4 2

4 21

at cos , phng tnh a cho tr thanh 8 9 0 (2)V x 0; nen t 1;1 .Ta co: (2) 8 9 1 1 (3).Goi (C ) : 8 9 1, 1;1 ; ( ) : 1So nghiem cua phng trnh (3) chnh

t x t t m

t t my t t t D y m

1

1

la so giao iem cua o th (C ) va (D).Chu y rang: o th (C ) giong vi o th (C) trong mien -1 t 1.Da vao o th (C) ta rut ra c ket luan....

Bi 12. Cho hm s 4 21 14y x mx m

a) Kho st hm s khi m=1

..

.

..xo

y

4

5

1-1 ..

.

..xo

y

4

5

1-1



b) Tm m hm s c 3 cc tr v ba cc tr ca th hm s to thnh mttam gic c din tch l 2

Bi 13. Cho (Cm): 4 22 3 1 2 1y x m x m . Tm m sao cho (Cm):a) Ct trc honh ti hai im A,B sao cho AB=4b) Ct : 2y ti 4 im c honh lp thnh cp s cng.



VN 2 : HM BC BA

PHNG PHP CHUNG:Tp xc nh: D=o hm:

2

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

y ax bx cy ax bx c

Nu (1) c hai nghim phn bit, hm s c cc i v cc tiu Nu (1) v nghim hay c nghim kp, th hm s n iu trn TX

Gii hn:3

2 30lim lim 1 0x x

khi ab c dy ax ax ax ax khi a

Bng bin thin:Du ca 'y ph thuc vo du ca 0 0a a haya v du ca 'y , do ta cbn trng hp bin thin khc nhau. th hm s: Do c bn trng hp khc nhau v chiu bin thin nn th cahm bc ba c bn dng sau ay:

a > 0 a < 0y = 0 c 2 nghimphn bit ' 0y ( C hai cc tr)

y = 0 v nghim hocc nghim kp

''

00

y

y

y

x0

I

y

x0

I

y

x0 I

y

x0I

DNG 1: KHO ST V V TH HM BC BA



( Khng c cc tr)

@ Mo nh: i vi trng hp th hm s khng c cc tr, v th cp v chnh xc ta nn tm im un (im m ti o hm cp hai bng 0) bit th un ln u?V ta d dng thy rng: th hm s nhn im un lm tm i xngBI TP P DNG:Bi 1. Kho st s bin thin v v th hm s sau: (Trng hp c cc tr)

3 2 3 2) 3 1 ) 2 3 2a y x x b y x x Bi 2. Kho st s bin thin v v th hm s: (Trng hp ' 0y c nghimkp)

3 2 3 21) 3 3 1 ) 13a y x x x b y x x x

Bi 3. Kho st s bin thin v v th hm s: (Trng hp ' 0y v nghim)3 2 3 2) 3 4 2 )a y x x x b y x x x

BI TP T LUYN: Kho st s bin thin v v th hm s sau:3 2 3 2

3 3 2

1 5) 2 1 ) 33 31 2 1) 3 )4 3 3

a y x x x b y x x x

c y x x d y x x



MT S LU KHI GII TON:Cho hm s 3 2ax ( )y bx cx d C

1. iu kin cn v th (C) c cc i v cc tiu ( c cc tr) l:' 2( ) 3ax 2 0y g x bx c c hai nghim phn bit

1. Phng trnh ng thng qua hai im cc tr. Ba im A, I, B thng hng(I l im un: im m ti y=0, A v B l hai im cc tr)

Gi s y = 0 c 2 nghim phn bit v y = k(Ax + B)y + r x + q vi k lhng s khc 0 th phng trnh ng thng qua 2 im cc tr l y = r x+ q. Vy phng trnh ng thng i qua hai im phn bit chnh lphn d trong php chia a thc ( ) : '( )f x f x

chng minh ba im A,I, B thng hng ta chng minh AB kAI

2. Qy tch cc tr, im un hm bc ba:T cc im A,B,I cha tham s m, ta tm c qu tch ca chnh cc im bng cch:

Kh tham s m Gii hn khong chy ca ta t iu kin tn ti m vi moih gi tr

tham s m mD Qy tch ca A,B, hay I l y = r x + q.

4. Xc nh tham s m th hm bc 3 ct trc honh trong tng trng hpc th:

a) (C) tip xc vi Ox th h sau c nghim 0' 0yy

b) (C) ct Ox ti 3 im phn bit

1 21 2

' 0 co 2 nghiem phan biet x ,( ). ( ) 0

y xy x y x

x"0C

x1

(C)yC

y

A ox2

x(H.3)yC

x0 x'0B

DNG 2: MT S BI TON LIN QUAN N HM BC BA



c) (C) ct Ox ti 2 im phn bit

1 21 2

' 0 co 2 nghiem phan biet x ,( ). ( ) 0

y xy x y x

d) (C) ct Ox t nht 1 im 3 2ax 0( 0)bx cx d a khng th v

nghime) (C) ct Ox ti 1 im duy nht

1 2

1 2

phng trnh y'=0 co nghiem kep hoac vo nghiem' 0 co hai nghiem phan biet x ,( ) ( ) 0

y xy x y x

f) Phng trnh 3 2ax 0( 0)bx cx d a c 3 nghim dng0 0. 0 . 0 hoac(0) 0 (0) 0

0 0

CD CT CD CT

CD CT

a ay y y yf fx x

(C)

Ax0 O x

y

(h.1a)

(C)

Ax0 x

y

(h.1b)x1 o x2yCT

yC

(C)yC

y

Ax0 o x1

Bx'0

(yCT= f(x0) = 0)x

(H.2)



g) Phng trnh 3 2ax 0( 0)bx cx d a c 3 nghim m0 0. 0 . 0 hoac(0) 0 (0) 0

0 0

CD CT CD CT

CT CD

a ay y y yf fx x

h) Phng trnh 3 2ax 0( 0)bx cx d a c 2 nghim dng:0 0

y'=0 co hai nghiem phan biet y'=0 co hai nghiem phan biet hoac. 0 . 00 0

CD CT CD CT

CT CT

a a

y y y yx x

i) Phng trnh 3 2ax 0( 0)bx cx d a c 2 nghim m:0 0

y'=0 co hai nghiem phan biet y'=0 co hai nghiem phan biet hoac. 0 . 00 0

CD CT CD CT

CD CT

a a

y y y yx x



5. Phng trnh bc 3 ct Ox lp thnh cp s cng tc (C) ct Ox ti 3 im phnbit cch u nhau.

1 3 2y'=0 co hai nghiem phan biet2 hay ( ) Ox , , : 0 : iem uon IDU

x x x C A B C AB BC f x Ox

6. Bin lun s nghim ca phng trnh : ax 3 + bx2 + cx + d = 0 (1) (a 0) khi x = l 1 nghim ca (1).

Nu x = l 1 nghim ca (1), ta cax3 + bx2 + cx + d = (x - )(ax2 + b1x + c1)

nghim ca (1) l x = vi nghim ca phng trnh ax 2 + b1x + c1 = 0 (2). Tac cc trng hp sau: nu (2) v nghim th (1) c duy nht nghim x = nu (2) c nghim kp x = th (1) c duy nht nghim x = nu (2) c 2 nghim phn bit th (1) c 3 nghim phn bit nu (2) c 1 nghim x = v 1 nghim khc th (1) c 2 nghim. nu (2) c nghim kp th (1) c 2 nghim

7. Tip tuyn : Gi I l im un. Cho M (C): 3 2ax ( 0)y bx cx d a . Nu M I th ta c ng 1 tip tuyn qua M. Nu M khc I v M ( )C th ta c ng 2 tip tuyn qua M. Bin lun s tip tuyn qua 1 im N khng nm trn (C) ta c nhiu

trng hp hn. Nu a>0: h s gc ca tip tuyn ti im un b nht; a



BI TP MU:Cho h ng cong bc ba (Cm) v h ng thng (Dk) ln lt c phng

trnh l : y = x3 + mx2 m v y = kx + k + 1.(I) PHN I. Trong phn ny cho m = 3. Kho st v v th (C) ca hm s.1) Gi A v B l 2 im cc i v cc tiu ca (C) v M l im bt k trn

cung AB vi M khc A , B . Chng minh rng trn (C) ta tm c hai imti c tip tuyn vung gc vi tip tuyn ti M vi (C).

2) Gi l ng thng c phng trnh y = 1. Bin lun s tip tuyn v i (C)v t E vi (C).

3) Tm E qua E c ba tip tuyn vi (C) v c hai tip tuyn vung gcvi nhau.

4) nh p trn (C) c 2 tip tuyn c h s gc bng p, trong trng hp nychng t trung im ca hai tip im l im c nh.

5) Tm M (C) qua M ch c mt tip tuyn vi (C).(II) PHN I I.Trong phn ny cho tham s m thay i.6) Tm im c nh ca (Cm). nh m hai tip tuyn ti hai im c nh ny

vung gc nhau.7) nh m (Cm) c 2 im cc tr. Vit phng trnh ng thng qua 2 im

cc tr.8) nh m (Cm) ct Ox ti 3 im phn bit.9) nh m :

a) hm s ng bin trong (1, 2).b) hm s nghch bin trong (0, +).

10) Tm m (Cm) ct Ox ti 3 im c honh to thnh cp s cng.11) Tm iu kin gia k v m (D k) ct (Cm) ti 3 im phn bit. Tm k (Dk)

ct (Cm) thnh hai on bng nhau.12) Vit phng trnh tip tuyn vi (C m) v i qua im (-1, 1).13) Chng minh rng trong cc tip tuyn vi (Cm) th tip tuyn ti im un c

h s gc ln nht.



BI GIIPHN I : m = 3

Kho st v v th (c gi t lm)

1) Gi n l honh ca M. V hm s t cc tiu ti x = 0 v t cc i ti x =2 nn 0 < n < 2; y' = 3x2 + 6x h s gc ca tip tuyn ti M l k1 = 3n2 +6n (0, 3] (v n (0, 2)). ng thng vung gc vi tip tuyn ti M c h sgc l k2 =

1

1k (vi 0 < k1 3). Honh ca tip tuyn vung gc vi tip

tuyn M l nghim ca 3x2 + 6x =1

1k (= k2) 3x

2 6x1

1k = 0. Phng

trnh ny c a.c < 0, k1 (0, 3] nn c 2 nghim phn bit, k1 (0, 3]. Vytrn (C) lun c 2 im phn bit m tip tuyn vung gc vi tip tuyn tiM.

2) E (e, 1) . Phng trnh tip tuyn qua E c dng y = h(x e) + 1 (D). (D)

tip xc (C) h3 23 3 ( ) 123 6

x n h x ex x h

c nghim.

Phng trnh honh tip im ca (D) v (C) l : x3 + 3x2 3 = ( 3x2 + 6x)(x e)+ 1 (1)

x3 + 3x2 4 = x( 3x + 6)(x e) (x 2)(x2 x 2) = 3x(x 2)(x e) x = 2 hay x2 x 2 = 3x2 3ex x = 2 hay 2x2 (3e 1)x + 2 = 0 (2)

(2) c = (3e 1)2 16 = (3e 5)(3e + 3)(2) c nghim x = 2 8 2(3e 1) + 2 = 0 e = 2

Ta c > 0 e < 1 hay e > 53 .

Bin lun :



i) Nu e < 1 hay 53 < e < 2 hay e > 2

(1) c 3 nghim phn bit c 3 tip tuyn.

ii) Nu e = 1 hay e = 53 hay e = 2

(1) c 2 nghim c 2 tip tuyn.

iii) Nu 1 < e < 53 (1) c 1 nghim c 1 tip tuyn.

Nhn xt : T th, ta c y = 1 l tip tuyn ti (2, 1) nn phng trnh (1) chcchn c nghim x = 2, e.

3) V y = 1 l tip tuyn qua E (e, 1), e v ng x = khng l tip tuyn nnyu cu bi ton. (2) c 2 nghim phn bit x1, x2 tha : y'(x1).y'(x2) = 1

51 3, (2)1 2

2 2( 3 6 )( 3 6 ) 11 1 2 2

e ex x la nghiem cua

x x x x

51 33 1

1 2 2. 11 2

9 . ( 2)( 2) 11 2 1 2

e hay eex x

x xx x x x

51 3

9[1 (3 1) 4] 1e haye

e

e = 5527 . Vy E55 ,127

4) Tip im ca tip tuyn (vi (C)) c h s gc bng p l nghim ca :



y' = p 3x2 6x + p = 0 (3)Ta c ' = 9 3p > 0 p < 3Vy khi p < 3 th c 2 tip tuyn song song v c h s gc bng p.Gi x3, x4 l nghim ca (3).Gi M3 (x3, y3); M4 (x4, y4) l 2 tip im. Ta c :

3 4 12 2x x b

a

3 3 2 23 4 3 4 3 4( ) 3( ) 6 12 2

y y x x x x

Vy im c nh (1, 1) (im un) l trung im ca M 3M4.5) Cch 1 : i vi hm bc 3 (a 0) ta d dng chng minh c rng :

M (C), ta c :i) Nu M khc im un, ta c ng 2 tip tuyn qua M.ii) Nu M l im un, ta c ng 1 tip tuyn qua M.

Cch 2 : Gi M(x0, y0) (C). Phng trnh tip tuyn qua M c dng :y = k(x x0) 3 20 03 3x x (D)

Phng trnh honh tip im ca (D) v (C) l :3 2 2 3 2

0 0 03 3 ( 3 6 )( ) 3 3x x x x x x x x ( 5 ) 3 3 2 2 20 0 03( ) ( )( 3 6 ) 0x x x x x x x x 2 2 20 0 0 00 3 3 3 6 0x x x xx x x x x x 2 20 0 0 02 (3 ) 3 0x x hay x x x x x 0 0 0( )(2 3) 0x x hay x x x x

003

2xx x hay x

Do , c ng 1 tip tuyn qua M (x0, y0) (C)



00 03 12

xx x

Suy ra, y0 = 1. Vy M(1, 1) (im un). Nhn xt : v x0 l 1 honh tip im nn pt (5) chc chn c nghim kp l

x0Phn II : Tham s m thay i. y' = 3x2 + 2mx6) (Cm) qua (x, y), m

y + x3 = m (x2 1) , m

2

31 0 1 1

1 10x x xhayy yy x

Vy (Cm) qua 2 im c nh l H(1, 1) v K(1, 1).V y' = 3x2 + 2mx nn tip tuyn vi (Cm) ti H v K c h s gc ln lt l:

a1 = y'(1) = 3 + 2m v a2 = y'(1) = 3 2m.2 tip tuyn ti H v K vung gc nhau.

a1.a2 = 1 9 4m2 = 1 m = 102 .

7) Hm c cc tr y' = 0 c 2 nghim phn bit. 3x2 = 2mx c 2 nghim phn bit.

x = 0 v x = 23m l 2 nghim phn bit.

m 0. Khi , ta c :22 1 1 '9 3 9y m x m x m y

v phng trnh ng thng qua 2 cc tr l :

229y m x m (vi m 0)

8) Khi m 0, gi x1, x2 l nghim ca y' = 0, ta c :



x1.x2 = 0 v x1 + x2 = 23m

y(x1).y(x2) = 2 21 22 29 9m x m m x m

= 2 21 22 ( )9m x x m =4 24

27m m

Vi m 0, ta c y(x1).y(x2) < 0

24 1 027m

2 27 3 34 2m m

Vy (Cm) ct Ox ti 3 im phn bi t.

1 21 2

' 0 2 ,( ). ( ) 0

y co nghiem phan biet x xy x y x

3 32m

Nhn xt :

i) Khi 3 32m th phng trnh y = 0 c 2 nghim m v 1 nghim dng.

ii) Khi 3 32m th phng trnh y = 0 c 2 nghim dng v 1 nghim m.

9) a) Hm ng bin trn (1,2) 3x2 + 2mx 0, x (1,2). Nu m 0 ta chonh 2 im cc tr l 0 v 23

m .

i) Nu m < 0 th hm ch ng bin trn 2 ,03m . Vy loi trng hp m < 0

ii) Nu m = 0 hm lun nghch bin (loi).



iii) Nu m > 0 th hm ch ng bin trn 20, 3m

Do , ycbt m > 0 v 2[1,2] 0, 3m

2 2 33m m

b) T cu a, ta loi trng hp m > 0.

Khi m 0 ta c hm s nghch bin trn 2, 3m v hm s

cng nghch bin

trn [0, +).Vy hm nghch bin trn [0, +) th m 0.

Ghi ch : nn lp bng bin thin thy r rng hn.

10) y" = 6x + 2m , y" = 0 x = 3m

(Cm) ct Ox ti 3 im cch u nhau. y = 0 c 3 nghim phn bit v im un nm trn trc honh.

3 23 3 3 32 20 . 03 27 9

m mm m my m m

23 3

3 6222 1 027

mmm

11) Phng trnh honh giao im ca (C m) v (Dk) l

x3 + mx2 m = kx + k + 1 m(x2 1) = k(x + 1) + 1 + x3

x + 1 = 0 m(x 1) = k + 1 x + x2

x = 1 hay x2 (m + 1)x + k + m + 1 = 0 (11)



a) Do , (Dk) ct (Cm) ti 3 im phn bit (11) c 2 nghim phn bit khc 1

21 1 1 0( 1) 4( 1) 0

m k mm k m

(*) 22 3

2 34

k mm mk

b) V (Dk) qua im K(1,1) (Cm) nn ta c :

(Dk) ct (Cm) thnh 2 on bng nhau.

(Dk) qua im un32;3 27

m m m ca (Cm)

32 1 127 3

m mm k

32 27 279( 3)

m mk m (**)

Vy ycbt k tha (*) v (**).12) Phng trnh tip tuyn vi (Cm) i qua (1,1) c dng :

y = k(x + 1) + 1 (Dk)Vy, phng trnh honh tip im ca (D k) v (Cm) l :

x3 + mx2 m = ( 3x2 + 2mx)(x + 1) + 1 (12) m(x2 1) = ( 3x2 + 2mx)(x + 1) + 1 + x3

x + 1 = 0 m(x 1) = 3x2 + 2mx + 1 x + x2

x = 1 hay 2x2 + (1 m)x m 1 = 0 (13)

x = 1 12mx

y' (1) = 2m 321 1 1' 3 22 2 2

m m my m =14 (m

2 2m 3)



Vy phng trnh ca 2 tip tuyn qua (1, 1) l :y = (2m + 3)(x + 1) + 1

y = 14 (m2 2m 3)(x + 1) + 1

Nhn xt : C 1 tip tuyn ti tip im (1, 1) nn phng trnh (12) chc chn cnghim kp l x = 1 v phng trnh (13) chc chn c nghim l x = 1.

13) Cc tip tuyn vi (Cm) ti tip im ca honh x c h s gc l :h = 3x2 + 2mx

Ta c h t cc i v l max khi 2 3b mx a (honh im un)

Vy tip tuyn ti im un c h s gc ln nht.

Nhn xt :2 2 2

2 23 2 3 3 3 3m m mx mx x

MT S BI TP IN HNH:Bi 1. Cho hm s y = 4x3 + mx2 3x

1. Kho st v v th (C) hm s khi m = 0.2. Tm m hm s c hai cc tr ti x1 v x2 tha x1 = - 4x2

Hng dn:D = Ry = 12x2 + 2mx 3

Ta c: = m2 + 36 > 0 vi mi m, vy lun c cc tr1 2

1 2

1 2

4

614

x xmx x

x x

92m

Bi 2. Cho hm s 3 2( ) 3 1 1y f x mx mx m x , m l tham s1. Kho st s bin thin v v th ca hm s trn khi m = 1.



2. Xc nh cc gi tr ca m hm s ( )y f x khng c cc tr.Hng dn:+ Khi m = 0 1y x , nn hm s khng c cc tr.+ Khi 0m 2' 3 6 1y mx mx m Hm s khng c cc tr khi v ch khi ' 0y khng c nghim hoc c nghimkp

2 2' 9 3 1 12 3 0m m m m m 10 4m Bi 3. Cho hm s : 3 2 33 12 2y x mx m

1. Kho st hm s vi m=1.2. Xc nh m th hm s c cc i,cc tiu i xng vi nh au qua t:

y=xHng dn:

Tac 2 0' 3 3 3 ( ) 0 xy x mx x x m x m

ta thy vi 0m th y i du khi i qua cc nghim do vy hm s c C,CT

+ Nu m>0 hm s c C ti x=0 v 312MAXy m ;c CT ti x=m v 0MINy

+ Nu m



2) Cho (d ) c phng trnh y = x + 4 v im K(1; 3). Tm cc gi tr catham s m sao cho (d) ct (Cm) ti ba im phn bit A(0; 4), B, C sao cho tam gicKBC c din tch bng 8 2 .Hng dn:Phng trnh honh im chung ca (C m) v d l:

3 2 2

2

2 ( 3) 4 4 (1) ( 2 2) 00

( ) 2 2 0 (2)

x mx m x x x x mx mxg x x mx m

(d) ct (Cm) ti ba im phn bit A(0; 4), B, C phng trnh (2) c 2 nghimphn bit khc 0.

/ 2 1 22 0 ( )2(0) 2 0m mm m amg m

.

Mt khc: 1 3 4( , ) 22d K d Do :

218 2 . ( , ) 8 2 16 2562KBCS BC d K d BC BC 2 2( ) ( ) 256B C B Cx x y y vi ,B Cx x l hai nghim ca phng trnh (2).2 2 2

2( ) (( 4) ( 4)) 256 2( ) 256( ) 4 128

B C B C B C

B C B C

x x x x x xx x x x

2 2 1 1374 4( 2) 128 34 0 2m m m m m (tha K (a)).

Vy 1 1372m

Bi 5. Cho hm s: y = x3 + 3x2 + mx + 1 c (Cm); (m l tham s).1. Kho st s bin thin v v th hm s khi m = 3.2. Xc nh m (Cm) ct ng thng y = 1 ti 3 im phn bit C(0, 1), D,E sao cho cc tip tuyn ca (Cm) ti D v E vung gc vi nhau.

Hng dn:



Phng trnh honh giao im ca (C m) v ng thng y = 1 l:x3 + 3x2 + mx + 1 = 1 x(x2 + 3x + m) = 0

203 0 (2)

xx x m

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

209 4 0 40 3 0 0 9

mmmm

Lc tip tuyn ti D, E c h s gc ln lt l :kD = y(xD) = 23 6 ( 2 );D D Dx x m x m kE = y(xE) = 23 6 ( 2 ).E E Ex x m x m Cc tip tuyn ti D, E vung gc khi v ch khi: kDkE = 1.(3xD + 2m)(3xE + 2m) = 9xDxE+6m(xD + xE) + 4m2 = 19m + 6m (3) + 4m2 = 1;(v xD + xE = 3; xDxE = m theo nh l Vi-et).

4m2 9m + 1 = 0 m = 1 9 658 S: m = 1 19 65 9 658 8hay m

BI TP P DNG:Bi 1(TNTHPT 2008) .Cho hm s 3 22 3 1y x x

a) Kho st s bin thin v v th ca hm s.b) Bim lun theo m s nghim ca phng trnh 3 22 3 1x x m

Bi 2 (TN THPT- ln 2 2008). Cho hm s y = x3 - 3x2a. Kho st s bin thin v v th hm s cho.b. Tm cc gi tr ca m phng trnh 3 23 0x x m c 3 nghim phn

bit.



Bi 3 (TNTHPT - 2007).Cho hm s y= 3 3 2x x c th l (C) .a/ Kho st v v th hm s .b/ Vit phng trnh tip tuyn ti im A(2 ;4) .

Bi 4 (TNTHPT - 2006) .Cho hm s y= 3 23x x c th (C) .a/ Kho st v v th hm s .b/ Da vo th bin lun s nghim phng trnh : 3 23x x -m=0 .

Bi 5 (TNTHPT 2004- PB).Cho hm s y= 3 26 9x x x c th l (C) .a/ Kho st v v th hm s .b/ Vit phng trnh tip tuyn ti im c honh l nghim ca phng

trnh y=0c/ Vi gi tr no ca m th ng thng y=x+m 2-m i qua trung im ca

on thng ni cc i vo cc tiu .Bi 6 (TNTHPT 2004 - KPB).Cho hm s y= 3 2 33 4x mx m .

a/ Kho st v v th hm s khi m=1 .b/ Vit phng trnh tip tuyn ti im c honh x=1 .

Bi 7 (H- A- 2002). Cho hm s 3 2 2 3 23 3(1 )y x mx m x m m a. Kho st s bin thin v v th hm s vi m= 1b. Tm k phng trnh: 3 2 3 23 3 0x x k k c 3 nghim phn bit.c. Vit phng trnh ng thng qua 2 im cc tr ca th hm s (1).

Bi 8 (C SP MGTW- 2004). Cho hm s y = x3 - 3x2 + 4ma. Chng minh th hm s lun c 2 cc tr.b. Kho st v v th hm s khi m = 1

Bi 9 (H-B- 2007). Cho hm s 3 2 2 23 3( 1) 3 1y x x m x m a. Kho st s bin thin v v th hm s vi m =1b. Tm m hm s c cc i cc tiu v cc im cc tr cch u im O.

Bi 10 (H - D - 2004)Cho hm s y = x3 3mx2 + 9x + 1



a. Kho st v v th hm s vi m = 2b. Tm m nghim ca phng trnh y= 0 thuc ng thng y = x+ 1

BI TP T LUYN:Bi 1.Cho hm s y = (x -1)(x2 + mx + m)

a. Tm m th hm s ct trc honh ti 3 im phn bitb. Kho st s bin thin v v th ca hm s vi m= 4

Bi 2. Cho hm s 3 22 3( 1) 6( 2) 1y x m x m x a. Kho st v v th hm s vi m =2b. Vi gi tr no ca m hm s c cc i, cc tiu.

Bi 3. (H 2006- D. )Cho hm s 3 3 2y x x a. Kho st s bin thin v v th (C) ca hm s .b. Gi d l ng thng qua im A(3; 20) v c h s gc m. Tm m ng

thng d ct (C ) ti 3 im phn bit. (Gi ng thng d qua M(x 0;y0) ch s gc m c dng: y = m(x - x0) + y 0)

Bi 4. Cho hm s y = (x - m)3 - 3xa. Kho st v v th hm s vi m = 1b. Tm m hm s cho t cc tiu ti im c honh x = 0

Bi 5.Cho hm s y = (x -1)(x2 + mx + m)c. Tm m th hm s ct trc honh ti 3 im phn bita) Kho st s bin thin v v th ca hm s vi m= 4

Bi 6.Cho hm s y = 3 2 22 2x mx m x a. Kho st v v th hm s khi m =1b. Tm m hm s t cc tiu ti x = 1

Bi 7. Cho hm s y = x(x 3)2 (1)1. Kho st s bin thin v v th ca hm s (1)

2. Tm tt c cc gi tr ca a ng thng (d): y = ax + b khng th tip xc vi th ca hm s (1).



VN 3: HM HU T

PHNG PHP KHO ST HM NHT BIN, ; 0ax b dy x ad bccx d c

Tp xc nh: \ /D d c o hm: 2' .

ad bcy cx d

Nu 0ad bc hm s ng bin trn D Nu 0ad bc hm s nghch bin trn D

Gii hn, tim cn:

lim la tiem can ngang cua o th ham so

lim la tiem can ngang cua o th ham sox

dx c

a ay yc cdy x c

Bng bin thin:

th:

DNG 1: Kho st s bin thin v v th hm s hu t

a

c ac

x'y

y

ac

ac

x'y

y



MT S BI TP RN LUYN KHO ST HM S:Kho st s bin thin v v th hm s sau:

2 1 2) ) )1 1 22 2 1 3 2) ) )2 2 1

x x xa y b y c yx x xx x xd y e y f yx x x

0

ad bc > 0

x

y

0

ad bc < 0

x

y



Mt s ch 1. Hm s c tim cn ng v tim cn ngang.2. th hm s nhn giao im hai ng tim cn lm tm i xng3. Khng c bt k tip tuyn no ca th hm s i qua giao im hai

ng tim cn

4. Gi M l im ty trn ax( ) : 0bC y ad bccx d v (T) l tiptuyn ti M vi (C).

1 2

1 2

Ha ( ) : va MK (d ) : theo th t o.Xac nh giao iem A=(T) (d ); =(T) (d ) (neu co) th:

d aMH d x yc cB

AB lun nhn M lm trung im Din tch tam gic AIB khng i Tch s MH.MK khng i Din tch t gic IHMK khng i M,N nm v hai nhnh phn bit ca th hm s th cc honh

ca xM, xN nm v hai pha tim cn ngBI TP MU:

Bi 1. Cho hm s 11xy x .

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

DNG 2: Kho st v v hm s nht bin (htb1/1)



b) M(x0; y0) l mt im bt k thuc (C). Tip tuyn ca (C) ti M ct hai ngtim cn ti A v B. Gi I l giao im ca hai tim cn. Chng minh din tch tamgic IAB khng ph thuc vo v tr im M.Hng dn:1. Tp xc nh: D = \ 12. S bin thin:a) Chiu bin thin:

+ 22' ( 1)y x => ' 0 1y x => HS nghch bin trn mi KX

b) Cc tr: HS khng c cc trc) Gii hn v tim cn:+ lim 1; lim 1 1x xy y y l TCN+

1 1lim ; limx x

y y => 1x l TCd) Bng bin thin3. th: V ng, p

b) M(x0; y0) l mt im bt k thuc (C). Tip tuyn ca (C) ti M ct hai ngtim cn ti A v B. Gi I l giao im ca hai tim cn. Chng minh din tch tamgic IAB khng ph thuc vo v tr im M.



+ 00 0 0 00

1; ( ) ( 1)1xM x y C y xx

+ PTTT ti M c dng: 00200

12 ( ) 1( 1)xy x x xx ()

+ Giao im ca 2 tim cn: (1;1)I

+ A = () TC => A= 00

31; 1xx

+ B = () TCN => B = 02 1;1x + IA =

0

41x

+ IB = 02 1x

+ SIAB = 12 .IA.IB = 4 (vdt) khng ph thuc v tr M

Bi 2. Cho hm s 2 12xy x c th l (C)

1. Kho st s bin thin v v th ca hm s2. Chng minh ng thng d: y = -x + m lun lun ct th (C) ti hai im phnbit A, B. Tm m on AB c di nh nht.Hng dn:2. Honh giao im ca th (C ) v ng thng d l nghim ca phng trnh

222 1

2 (4 ) 1 2 0 (1)xx x mx x m x m

Do (1) c 2 21 0 ( 2) (4 ).( 2) 1 2 3 0m va m m m nn ngthng d lun lun ct th (C ) ti hai im phn bit A, BTa c yA = m xA; yB = m xB nn AB2 = (xA xB)2 + (yA yB)2 = 2(m2 + 12) suyra AB ngn nht AB2 nh nht m = 0. Khi 24AB

Bi 3. Cho hm s 2 21xy x (C)



1. Kho st hm s.2. Tm m ng thng d: y = 2x + m ct th (C) ti 2 im phn bit A, B

sao cho AB = 5 .Hng dn:2. Phng trnh honh giao im: 2x 2 + mx + m + 2 = 0 , (x - 1) (1)d ct (C) ti 2 im phn bit PT(1) c 2 nghim phn bit khc -1 m2 - 8m -16 > 0 (2Gi A(x1; 2x1 + m) , B(x2; 2x2 + m. Ta c x1, x2 l 2 nghim ca PT(1).

Theo L Vit ta c 1 2

1 2

22

2

mx xmx x

.

AB2 = 5 2 21 2 1 2( ) 4( ) 5x x x x 21 2 1 2( ) 4x 1x x x m2 - 8m - 20 = 0 m = 10 , m = - 2 ( Tha mn (2))KL: m = 10, m = - 2.

Bi 4. Cho hm s: 21xy x (C)

1. Kho st v v th (C) hm s2. Cho im A( 0; a) 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:Gi k l h s gc ca t i qua A(0;a). PT t d c dng y= kx+a (d)

d l tip tuyn vi ( C ) khi v ch khi h PT 221

31

x kx axk

x

c nghim

Pt (1-a)x2 +2(a+2)x-(a+2)=0 (1) c nghim x 1

Theo bi ra qua A c 2 tip tuyn th pt (1) c 2 nghim x1 ; x2 phn bit



1 2 1 (*)' 3 6a aa

. Theo nh l Viet:

1 2

1 2

2 21

21

ax x aax x a

Suy ra: 1 21 2

3 31 ; 11 1y yx x . hai tip tuyn nm v hai trc ca Ox th

1 22. 0 3y y a . Kt hp vi iu kin (*) ta c

2 13 a

Bi 5. Cho hm s 2x 3y x 2 c th (C).

1. Kho st s bin thin v v th ca hm s (C)2. Tm trn (C) nhng im M sao cho tip tuyn ti M ca (C) ct hai

tim cn ca (C) ti A, B sao cho AB ngn nhtHng dn:

0 0 20 0

0200

0

1 1Lay iem M ;2 . Ta co: '( )2 21 1Tiep tuyen (d) tai M co phng trnh la: 2 22

2Giao iem (d) vi tiem can ngA 2;2 2Giao iem (d) vi tiem c

x C f xx xy x x xx

x

00

2 220 02 2

0 0

2an ngang 2 2;2 21 14 2 8. Dau"=" xay ra khi 2 ...2 2

B x x

AB x xx x

BI TP P DNG:Bi 1. Cho hm s 2 11

xy x

1. Kho st s bin thin v v th (C) ca hm s.2.Tm im M thuc (C) sao cho tip tuyn ti M ct trc Ox, Oy ln l t ti A, Btho mn:OA=3OB.



Hng dn:2. Gi s tip tuyn d ca (C) ti im M(x 0;y0) ct Ox, Oy ln lt ti A,B sao choOA=3OB.Do tam gic OAB vung ti O nn tanA=OB/OA=1/3. V vy h s gc ca ngthng d bng 1/3 hoc -1/3.H s gc ca d ti M(x0;y0) l:

0 02 20 00

0

3 3 1' 0 ' 31 12 (2) 14 ( 4) 3

y x y xx xx yx y

Khi c 2 tip tuyn tho :1 1 1( 2) 13 3 31 1 13( 4) 33 3 3

y x y x

va y x y x

Bi 2. Cho hm s 2 32xy x

1. Kho st s bin thin v v th (C) ca hm s.2. Cho M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ngtim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tmto im M sao cho ng trn ngoi tip tam gic IAB c din tch nhnht.

Hng dn:

Ta c: 00 00

2 3; , 22xM x xx

, 0 20

1'( )2

y xx

Phng trnh tip tuyn vi ( C) ti M c dng: 0

0200

2 31: ( ) 22xy x x xx

To giao im A, B ca v hai tim cn l: 0 00

2 22; ; 2 2;22xA B xx



Ta thy 0 02 2 2

2 2A B

Mxx x x x , 0

0

2 32 2

A BM

xy y yx suy ra M l trung

im ca AB.Mt khc I = (2; 2) v tam gic IAB vung ti I nn ng tr n ngoi tip tam gicIAB c din tch

S =2

2 2 200 0 2

0 0

2 3 1( 2) 2 ( 2) 22 ( 2)xIM x xx x

Du = xy ra khi 020 20 0

11( 2) ( 2) 3xx x x

Do c hai im M cn tm l M(1; 1) v M(3; 3)

Bi 3. Cho hm s 112

x

xy

1. Kho st s bin thin v v th (C) ca hm s .2. Tm ta im M thuc (C) sao cho khong cch t im )2;1(I ti tip

tuyn ca (C) ti M l ln nht .Hng dn:

Nu )(132;

00 CxxM

th tip tuyn ti M c phng trnh

)()1(3

132 02

00xxxxy hay 0)1(3)2()1()(3 0

200 xyxxx

Khong cch t )2;1(I ti tip tuyn l

202

0

40

04

0

00

)1()1(9

6)1(9

1619

)1(3)1(3

xxx

xx

xxd . Theo bt ng thc

Csi 692)1()1(9 2

020

xx , vy 6d . Khong cch d ln nht bng 6khi

3131)1()1(9

02

02

020

xxxx .



Vy c hai im M : 32;31 M hoc 32;31 MBi 4. Cho hm s 2 11

xy x (1).

1) Kho st v v th (C) ca hm s (1).2) Tm im M thuc th (C) tip tuyn ca (C) ti M vi ng th ng

i qua M v giao im hai ng tim cn c tch h s gc bng - 9.Hng dn:Ta c I(- 1; 2). Gi 0 2

0 0

3 3( ) ( ;2 )1 ( 1)M I

IMM I

y yM C M x kx x x x

H s gc ca tip tuyn ti M: 0 203'( ) 1Mk y x x

. 9M IMycbt k k

Gii c x0 = 0; x0 = -2. Suy ra c 2 im M tha mn: M(0; - 3), M(- 2; 5)Bi 5. Cho hm s 12

2 x

xy

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm nhng im trn th (C) cch u hai im A(2 , 0) v B(0 , 2)

Hng dn:Pt ng trung trc an AB : y = xNhng im thuc th cch u A v B c hong l nghim ca pt :

xxx

122

251

251012

x

xxx

Hai im trn th tha ycbt :

251,2

51;251,2

51



BI TP LUYN TP:Bi 1.

1. Kho st s bin thin v v th (C) ca hm s 2 11xy x

2. Vit phng trnh tip tuyn ca (C), bit khong cch t im I(1;2) ntip tuyn bng 2 .

Hng dn:*Tip tuyn ca (C) ti im 0 0( ; ( )) ( )M x f x C c phng trnh

0 0 0'( )( ) ( )y f x x x f x Hay 2 20 0 0( 1) 2 2 1 0x x y x x (*)*Khong cch t im I(1;2) n tip tuyn (*) bng 2

04

0

2 2 21 ( 1)x

x

gii c nghim 0 0x v 0 2x

*Cc tip tuyn cn tm : 1 0x y v 5 0x y Bi 2. Cho hm s 2 41

xy x .

1. Kho st s bin thin v v th (C) ca hm s.2. Tm trn th (C) hai im i xng nhau qua ng thng MN bit M( -3; 0)

vN(-1; -1).Hng dn:

Gi 2 im cn tm l A, B c 6 6;2 ; ;2 ; , 11 1A a B b a ba b

Trung im I ca AB: I 2 2;2 1 1a b a b

a b

Pt ng thng MN: x + 2y +3= 0



C : . 0AB MNI MN

=> 0 (0; 4)2 (2;0)

a Ab B

Bi 3. Cho hm s : 1x21xy

(C)1. Kho st v v th hm s.2. Vit phng trnh tip tuyn vi (C), bit tip tuyn i qua giao im cang tim cn v trc Ox.Hng dn:

Giao im ca tim cn ng vi trc Ox l

0,21A

Phng trnh tip tuyn () qua A c dng

21xky

Bi 4. Kho st s bin thin v v th (C) ca hm s : 3x 4y x 2 .

Tm im thuc (C) cch u 2 ng tim cn .

Hng dn:Gi M(x;y) (C) v cch u 2 tim cn x = 2 v y = 3

| x 2 | = | y 3 | 3x 4 xx 2 2 x 2x 2 x 2

x 1x x 2 x 4x 2

() tip xc vi (C) /x 1 1k x2x 1 2x 1 k co nghiem2x 1

)2(k1x2

3)1(2

1xk1x21x

2

Th (2) vo (1) ta c pt honh tip im l



213 xx 1 2

2x 1 2x 1

1(x 1)(2x 1) 3(x )2 v

1x 2 3x 1 2

5x 2 . Do 121k . Vy phng trnh tip tuyn cn tm l:

1 1y x12 2

Bi 5. Gi s l tip tuyn ti im 0;1M ca th hm s 2 11xy x (C)Hy tm trn (C) nhng im c honh ln hn 1 m khong cch t n lngn nht.Hng dn:Khong cch t mt im trn (C) ti ng thng l ngn nht khi v ch khiim l tip im ca th (C) vi tip tuyn l song song vi ng thng .Ta c: 2

3' ; ' 0 31

y yx

.

Phng trnh tip tuyn ca (C) l : 3 1y x .Gi 0 0 0; ( ), 1N x y C x c khong cch ti ngn nht, th th 0x l nghimca phng trnh 0 00

02 5'( ) 3 0 (loai)

x yy x x

Bi 6. Tm tt c cc gi tr m th hm s 22x my mx c t nht mtim cch u hai trc ta , ng thi honh v tung ca im tri dunhau.Hng dn: Nhng im cch u hai trc ta c honh v tung tri dunhau s nm trn ng thng y x . Gi s ;M x y l im tha mn bi th



ta c phng trnh: 2 0 *2 2x x mx mx x x

. Phng trnh (*) c t nht

mt nghim khc 2 khi v ch khi11 4 0 42 0 2

m mm x

Bi 7. Tnh khong cch gia hai nhnh ca th hm s 2 2xy Cx Hng dn: Gi s 1 2;C C l hai nhnh ca th hm s v

1 2

2

2 22

2 2; ; ;2 24 2 2' , ; .Tiep tuyen vi (C) tai M co phng trnh:2 22

4 2y= 4 2 2 022

a bM a C N b Ca bb ay MN b a b ax

ax a x a y aaa

Ta thy MN l khong cch ca hai nhnh 1 2;C C khi v ch khi tip tuyn tiM,N vi 1 2;C C song song vi nhau v chng vung gc vi ng thng chaMN. iu tng ng

2 2

2

4 4 (1)2 2

2 22 4 0 (2)2 2

a bb ab a a b a

4T (1) b=-4-a thay vao (2) ta c -2 2 32 0 0 (do 2)4 va 4;4 4 2

a a ab MN MN

BTTT: Tnh khong cch gia hai nhnh ca th hm s 2 12xy Cx



Bi 8. Gi A, B l giao im ca ng thng 16y x vi th hm s11

xy x .

Tm im M thuc ng phn gic gc phn t th nht sao cho MA MB nhnhtHng dn: Ta A, B l nghim ca h phng trnh

11 16 2; ; 3;3 21

1

y xA Bxy x

D thy A v B nm cng pha i vi ng phn gic : 0.x y Gi ' ;A a bl im i xng ca A qua : 0.x y th th

12 .1 .1 0 13 1' ;231 322 3 02 27 7M la giao iem cua A'B va M ;5 5

a ba A

b ba



TI LIU THAM KHO1. Trn nh C (2012). Bi ging cp tc luyn thi i hc.2. Trn S tng (2011). Chuyn hm s.3. Nguyn Ph Khnh, Nguyn Tt Thu (2011) . Chuyn hm s4. Cc thi ca b gio dc v o to.5. thi th cc trng 2012.

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