Tai Lieu on Thi Tot Nghiep Va Dai Hoc 2013

7/30/2019 Tai Lieu on Thi Tot Nghiep Va Dai Hoc 2013

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Ti liu n tp thi TN THPT nm hc 2012 2013

Ch 1.TNH N IU CA HM SPHN I. TM TT L THUYT

I.nh ngha: Cho hm s ( )y f x= xc nh trn D, vi D l mt khong, mt on hoc na khong.1.Hm s ( )y f x= c gi l ng bin trn D nu 1 2 1 2 1 2, , ( ) ( )x x D x x f x f x < II.iu kin cn hm s n iu: Gi s hm s ( )y f x= c o hm trn khong D

1.Nu hm s ( )y f x= ng bin trn D th '( ) 0,f x x D

2.Nu hm s ( )y f x= nghch bin trn D th '( ) 0,f x x D III.iu kin hm s n iu:

1.nh l 1. Nu hm s ( )y f x= lin tc trn on [ ],a b v c o hm trn khong (a,b) th tn tinht mt im ( , )c a b sao cho: ( ) ( ) '( )( )f b f a f c b a =

2.nh l 2. Gi s hm s ( )y f x= c o hm trn khong D1.Nu '( ) 0,f x x D v '( ) 0f x = ch ti mt s hu hn im thuc D th hm s ng bin trn D2.Nu '( ) 0,f x x D v '( ) 0f x = ch ti mt s hu hn im thuc D th hm s nghch bin trn 3.Nu '( ) 0,f x x D= th hm s khng i trn D

PHN II. MT S DNG TON

*Phng php : Xt chiu bin thin ca hm s ( )y f x=1.Tm tp xc nh ca hm s ( )y f x=2.Tnh ' '( )y f x= v xt du y ( Gii phng trnh y = 0 )3.Lp bng bin thin4.Kt lun

V d : Xt tnh bin thin ca cc hm s sau:

1.y = -x3+3x2-3x+1 4. y=3 2

2 1

x

x

+

2. y= 2x4

+5x2

-2 5.

2 2 2

1

x xy x

+ += +

3. y= (x+2)2(x-2)2 6.2

2

2 3

10

x xy

x

=

7. 2 6 10y x x= + 8.2 3

2 1

x xy

x

+=

+9.y= 2 1 3x x+ + 10.y=2x + 2 1x

11.y = x + cosx trn khong (0; ) 12. y= sin2x - 3 x trn khong (0;2

)

13.y= x.tanx trn khong ( ;2 2

) 14.y = -6sinx +4tanx -13x trn (0;)

V d:1.Tm m hm s y= 2x3-3mx2+2(m+5)x-1 ng bin trn R

2.Tm m hm s y=2

1

x x m

mx

+ ++

ng bin R

3.Tm m hm s y= 3mx+ 2 2x + ng bin trn R

4.Tm m hm s 3 2( ) 3 ( 2) 3y f x mx x m x= = + + nghch bin trn R

Trang

Dng 1.Xt chiu bin thin ca hm s ( )y f x=

Dng 2. Tm iu kin ca tham s hm s n iu trn mt khong cho trc .


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5. Tm m hm s 3 2 2( ) ( 1) ( 2)y f x x m x m x m= = + + + + nghch bin trn R

6. Tm m hm s ( ) ( )3 21

( ) 2 2 2 2 53

my f x x m x m x

= = + +

nghch bin trn R

7. Tm m hm s ( ) ( )3 21

( ) 1 3 23

y f x m x mx m x= = + + tng trn R

8.Tm m hm s y= 3x3-2x2+mx-4 tng trn (-1; + )9.Tm m hm s y= 4mx3-6x2+(2m-1)x+1 tng trn (0;2)

10.Tm m hm s y=2 6 2

2mx x

x+ +

gim trn [1; + )

11.Tm m hm s y=mx4 -4x2+2m-1 gim trn (0;3)12.Tm m hm s y= x3+3x2+(m+1)x+4m gim trn (-1;1)

13.Tm m hm s y=22 3

2 1

x x m

x

++

gim trn (1

;2

+ )

14.Cho hm s y=2 2 1

2

x mx m

x

+ +

a.Tm m hm s tng trn tng khong xc nhb.Tm m hm s gim trn khong (a;b) vi b-a =2

15.Tm gi tr ca tham s m hm s sau nghch bin trn mt on c di bng 13 2( ) 3y f x x x mx m= = + + +

16. Tm m hm s ( ) ( )3 21

( ) 1 3 43

y f x x m x m x= = + + + tng trn ( )0,3

17. Tm m hm s ( )3 2( ) 3 1 4y f x x x m x m= = + + + + gim trn ( )1,1

18. Tm m hm s4

( )mx

y f xx m

+= =

+gim trn khong ( ),1

19. Tm m hm s ( ) ( )3 21 1

( ) 1 3 23 3

y f x mx m x m x= = + + tng trn ( )2,+

20. Tm m hm s ( )( )

2 2

1 4 4 2( )1

x m x m my f xx m

+ + + = = ng bin trn ( )0,+

V d:

1.Gii phng trnh 3 23 4 7x x x x+ = + ( K x3+3x 0 0x )

2.Gii phng trnh x5+x3- 1 3x +4=0

3.Gii phng trnh21 22 2 ( 1)x x x x =

4. Gii phng trnh sinx =x

5.Tm m phng trnh c nghim 1x x m+ + =6.Tm phng trnh c nghim m 2 1x + - x = 0

7.Chng minh rng2

0 :1 cos2

xx x > < (HD xt hm s

2

( ) 1 cos2

xy f x x= = )

8.Chng minh rng2

0 : 12

x xx e x > > + + (HD xt hm s2

( ) 12

x xy f x e x= = )

9.Chng minh rng3

(0; ) : tan2 3

xx x x

> +

10.Chng minh rng : Nu 1x y+ = th 4 41

8

x y+ (HD xt hm s 4 4( ) (1 )y f x x x= = + )

Trang

Dng 3. S dng tnh n iu gii PT,BPT,BT


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11.Gii h phng trnh

3 2

3 2

3 2

2 1

2 1

2 1

x y y y

y z z z

z x x x

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

HD. Xt hm c trng 3 2( ) ,y f x t t t t= = + + . Chng minh hm s tng trn R .Sx y z

x y z

= = = = = =

12.Gii h phng trnh

3

3

3

sin6

sin6

sin6

y

x yz

y z

xz x

= + = +

= +

Ch 2. CC TR CA HM SPHN I. TM TT L THUYT

I.nh ngha: Cho hm s ( )y f x= xc nh trn D v 0x D

1. 0x c gi l mt im cc i ca hm s ( )y f x= nu tn ti mt (a,b) cha im 0x sao c

( , )a b D v { }0 0( ) ( ), ( , ) \f x f x x a b x< . Khi 0( )f x c gi l gi tr cc i ca hm s v0 0( ; ( ))M x f x c gi l im cc i ca hm s .

2. 0x c gi l mt im cc tiu ca hm s ( )y f x= nu tn ti mt (a,b) cha im 0x sao c( , )a b D v { }0 0( ) ( ), ( , ) \f x f x x a b x> . Khi 0( )f x c gi l gi tr cc tiu ca hm s v

0 0( ; ( ))M x f x c gi l im cc tiu ca hm s .

3.Gi tr cc i v gi tr cc tiu c gi chung l cc tr ca hm sII.iu kin cn hm s c cc tr : Gi s hm s ( )y f x= c cc tr ti 0x .Khi , nu ( )y f x=

o hm ti im 0x th 0'( ) 0f x = .III.iu kin hm s c cc tr :

1.nh l 1.(Du hiu 1 tm cc tr ca hm s )Gi s hm s ( )y f x= lin tc trn khong (a,b) cha im 0x v c o hm trn cc khon

0 0( , ) v ( , )a x x b . Khi :

+ Nu f(x) i du t m sang dng khi x qua im 0x th hm s t cc tiu ti 0x

+ Nu f(x) i du t dng sang m khi x qua im 0x th hm s t cc i ti 0x

2.nh l 2. (Du hiu 2 tm cc tr ca hm s )Gi s hm s ( )y f x= c o hm trn khong (a,b) cha im 0x , 0'( ) 0f x = v f(x) c o hm c

hai khc 0 ti im 0x . Khi :

+ Nu 0''( ) 0f x < th hm s t cc i ti im 0x

+ Nu 0''( ) 0f x > th hm s t cc tiu ti im 0x PHN II. MT S DNG TON

*Phng php1.(Quy tc 1)Tm cc tr ca hm s ( )y f x=1.Tm tp xc nh ca hm s2.Tnh '( )f x v gii phng trnh '( ) 0f x = tm nghim thuc tp xc nh3.Lp bng bin thin4.Kt lun

V d1: Dng quy tc 1 tm cc tr ca hm s

Trang

Dng 1. Tm cc tr ca hm s


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1. y =1

3x3+x2-3x+2 2.y = x4+2x2-3

2. y =3 1

2 4

x

x

+

4.y =2 3 3

1

x x

x

+

3. y= 22 4 5x x + 6. y=(2x+1) 29 x

7. y = 3 1x x+ + 8. y=2

2 3

1

x

x x

+

+ +9. y =

22 2

2 1

x x

x

+ ++

10. 4 26 8 25y x x x= + +

11. 2 2( 2) ( 2)y x x= + 12. 5 315 15 2y x x= +*Phng php 2.(Quy tc 2)Tm cc tr ca hm s ( )y f x=

1.Tm tp xc nh ca hm s2.Tnh '( )f x v gii phng trnh '( ) 0f x = tm nghim ( 1, 2, 3...)ix i = thuc tp xc nh3.Tnh ''( ) v ''( )if x f x

4.Kt lun

+Nu ''( ) 0if x < th hm s t cc i ti im ix+Nu ''( ) 0if x > th hm s t cc tiu ti im ix

V d 2: Dng quy tc II tm cc tr ca hm s

1.y= 3x5-20x3+1 2. y = 25 6 4x x +

3.y = cos23x 4. y = sin cos2 2

x x

5.y = -2sin3x+3sin2x-12sinx 6. y= sin3x + cos3x ( 0 2x )

7. 29y x x= 8.3

2 9

xy

x=

9. 3 3y x x= 10. [ ]s inx cos , ,y x x = +

VD1: Tm iu kin ca m sao cho :1. y= x3-mx2+2(m+1)x-1 t cc i ti x= -1

2. y=2 1x mx

x m

+ ++

t cc tiu ti x=2

3. y= 4 2 22 2x mx m t cc i ti x= 2

VD2:Cho hm s y=

1

3 x3

-(7m+1)x2

+16x-m .Tm m a. Hm s c cc i v cc tiub. Hm s c cc im cc i v cc tiu ti x1,x2 (1; ) +

VD3:Cho hm s y= x3-mx2+(m+36)x-5 .Tm m a. Hm s khng c cc tr

b. Hm s t cc i ,cc tiu ti cc im x1,x2 v 1 2 4 2x x =

VD3:Cho hm s y=22 2 1

1

x mx m

x

+ + +

.Tm m hm s c cc i v cc tiu

VD4:Cho hm s y= 2x3-3(2m+1)x2+6m(m+1)x+1Tm m cc im cc i ,cc tiu i xng nhau qua ng thng y=x+2

Trang

Dng 2.Tm iu kin ca tham s hm s c cc tr tha mn iu kin cho trc

Dng 3. Mt s bi ton lin quan n im cc tr ca th hm s


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VD5: Cho hm s y= x3-3x2-mx+2 .Tm m a. Hm s c cc i ,cc tiu trong khong (0;2)b. Hm s c cc i ,c tiu v cc im cc i ,cc tiu cch u ng thng y=x-1

VD6:Cho hm s2 (3 1) 4

2 1

x m x my

x

+ +=

.Tm m hm s c cc i, cc tiu i xng nhau qua ng

thng : 1 0x y + + = .VD1: Cho hm s y= x3+mx2-x

a. CMR hm s c cc i cc tiu vi mi mb. Xc nh m ng thng i qua hai im cc tr ca th hm s song song vi ng thng(d) y=-2x

VD2:Cho hm s y=2 (3 2) 4

1

x m x m

x

+ + +

a. Tm m hm s c C,CT v C,CT v im M(-2;1) thng hngb. Tm m hm s c C,CT v trung im ca on ni 2 im C,CT cch gc O mt khong

bng 3VD3.Cho hm s 3 23 2y x x= + c th (C). Tm gi tr ca tham s m im cc i v im cc tiu c(C) v hai pha khc nhau ca ng trn : 2 2 22 4 5 1 0x y mx my m+ + = .

VD4.Cho hm s

4 2 4

2 2y x mx m m= + + .Tm gi tr ca tham s m hm s c cc i v cc tiu, ngthi cc im cc i, cc tiu lp thnh mt tam gic u .

VD5.Cho hm s2 2

1

x mxy

x

+ +=

.Tm im cc tiu ca th hm s nm trn Parabol (P) 2 4y x x= +

VD6.Cho hm s2 ( 2) 3 2

1

x m x my

x

+ + + +=

+a. Tm m hm s c cc i v cc tiu

b. Gi s hm s c gi tr cc i, cc tiu l yC , yCT . Chng minh rng :2 2

CD

1

2CTy y+ > .

VD7.Cho hm s 3 2 2(2 1) ( 3 2) 4y x m x m m x= + + + +

a. Tm m hm s c hai im cc i v cc tiu nm v hai pha khc nhau ca trc tungb. Tm m hm s c cc i cc tiu ng thi hai gi tr cc tr cng du

VD8.Cho hm s 3 22 3(2 1) 6 ( 1) 1y x m x m m x= + + + +a.Chng minh rng vi mi gi tr ca tham s m hm s lun t cc i v cc tiu ti 1 2,x x v

2 1x x khng ph thuc vo tham s m.b.Tm m 1CDy >

VD9.Cho hm s 3 21

( ) 13

y f x x mx x m= = + + .Chng minh rng vi mi m hm s cho lun c cc

cc tiu .Hy xc nh m khong cch gia hai im cc tr l nh nht .

VD10.Cho hm s

2 2

2( 1) 4( ) 2x m x m my f x x+ + + += = + .Tm m hm s c cc i cc tiu, ng thi cc

im cc tr ca th hm s cng vi gc ta O to thnh tam gic vung ti O. ( A 2007)

VD11.Cho hm s1

( )y f x mxx

= = + .Tm m hm s c cc i cc tiu v khong cch t im cc tiu

ca th hm s n tim cn xin bng1

2.(A 2005)

VD12.Cho hm s 3 2 2 2( ) 3 3( 1) 3 1y f x x x m x m= = + + .Tm m hm s c cc i cc tiu v cc icc tr cch u gc ta O. ( B 2007)

Trang


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VD13.Cho hm s2 ( 1) 1

( )1

x m x my f x

x

+ + + += =

+(Cm) . CMR vi mi m (Cm) lun c cc i cc tiu v

khong cch gia hai im cc tr bng 20 . ( B 2005)

VD14.Cho hm s 3 2( ) (2 1) (2 ) 2y f x x m x m x= = + + .Tm m hm s c cc i cc tiu v cc imcc tr c honh dng . ( C D 2009)VD15. Cho hm s 4 22( 1)y x m x m= + + (1) m l tham s

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

l im cc tr thuc trc tung, B,C l hai im cc tr cn li . ( B 2011)Ch 3. GI TR LN NHT V GI TR NH NHT CA HM S

PHN I. TM TT L THUYTI.nh ngha: Cho hm s ( )y f x= xc nh trn D

1.Nu tn ti mt im 0x D sao cho 0( ) ( ),f x f x x D th s 0( )M f x= c gi l gi tr l

nht ca hm s f(x) trn D, k hiu ax ( )x D

M M f x

=

Nh vyx D

0 0

, ( )ax ( )

, ( )

x D f x MM M f x

x D f x M

= =

2. Nu tn ti mt im 0x D sao cho 0( ) ( ),f x f x x D th s 0( )m f x= c gi l gi tr nhnht ca hm s f(x) trn D, k hiu ( )

x Dm Min f x

=

Nh vyx D

0 0

, ( )( )

, ( )

x D f x mm Min f x

x D f x m

= =II.Phng php tm GTLN,GTNN ca hm s : Cho hm s ( )y f x= xc nh trn D

Bi ton 1.Nu ( , )D a b= th ta tm GTLN,GTNN ca hm s nh sau:1.Tm tp xc nh ca hm s2.Tnh '( )f x v gii phng trnh '( ) 0f x = tm nghim thuc tp xc nh3.Lp bng bin thin

4.Kt lunBi ton 2. Nu [ ],D a b= th ta tm GTLN,GTNN ca hm s nh sau:

1.Tm tp xc nh ca hm s2.Tnh '( )f x v gii phng trnh '( ) 0f x = tm nghim 1 2, ...x x thuc tp xc nh3.Tnh 1 2( ), ( ), ( ).... ( )f a f x f x f b

4.Kt lun: S ln nht l [ ],ax ( )

x a bM M f x

= v s nh nht l [ ],

( )x a b

m Min f x

=

Bi ton 3.S dng cc bt ng thc thng dng nh : Cauchy, Bunhiacpxki, ..Bi ton 4.S dng iu kin c nghim ca phng trnh, tp gi tr ca hm s


V d: Tm GTLN,GTNN ( nu c ) ca cc hm s sau:

1. 4 2( ) 2y f x x x= = 2.3 1

( )3

xy f x

x

= =

trn [ ]0;2

3. 2( ) 4y f x x x= = + (B-2003) 4.2ln

( )x

y f xx

= = trn 31,e (B-2004)

5.2

1( )

1

xy f x

x

+= =

+trn [ ]1, 2 (D-2003) 6.

2

2

3 10 20( )

2 3

x xy f x

x x

+ += =

+ +(SPTPHCM2000)

Trang

Dng 1. Tm GTLN, GTNN ca hm s


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7. ( ) 5cos os5xy f x x c= = trn ,4 4

8.

3sin( ) 1

2 cos

xy f x

x= = +

+9. ( ) 1 s inx 1 osxy f x c= = + + + 10. ( ) 2cos 2 osx-3y f x x c= = +

11. 22 1 2y x x x x= + + + + 12. 2sin .cos sin cosy x x x x= +

13.22 1

1

x xy

x

+ +=

+trn ( 1, ) + 14. 2 4 3 3 1y x x x= + + trn on

130,

4

15. 3 21 34

y x x= trn [ ]2,4 16. 3 3sin os 3sin 2y x c x x= + +

VD1 .Cho hm s 2 2 4y x x a= + + .Tm a gi tr ln nht ca hm s trn [ ]2,1 t GTLN.

VD2. Cho hm s 4 4( ) sin os sin .cosy f x x c x m x x= = + + .Tm m sao cho gi tr ln nht ca hm s bng 2.

VD3. Cho hm scos 1

cos 2

k xy

x

+=

+.Tm k gi tr nh nht ca hm s nh hn -1.

VD4. Tm cc gi tr ca tham s a,b sao cho hm s2

a +b( )

1

xy f x

x= =

+c gi tr ln nht bng 4 v gi tr nh

nht bng -1.

VD5.Cho hm s2( ) 2 4 2 1y f x x x a= = + + vi 3 4x .Xc nh a gi tr ln nht ca hm s t gi

tr nh nht .

VD1. Mt tm tn hnh vung cnh bng a. Ngi ta phi ct b bn hnh vung bng nhau bn gc gthnh mt b cha hnh hp ch nht khng np, cnh hnh vung ct i bng bao nhiu th b c th tch ln

nht . S. Cnh hnh vung ct i bng6

a

VD2. Tm cc kch thc ca hnh ch nht c din tch ln nht ni tip ng trn bn knh R cho trc.

S.Cc kch thc ca hnh ch nht l 2R (hnh vung)VD3. Trong cc khi tr ni tip hnh cu bn knh R, hy xc nh khi tr c th tch ln nht .

S.Hnh tr c chiu cao2

3

Rh = bn knh y

22

4

hr R=

VD4. Cho ng (C) c phng trnh 2 2 2x y R+ = .Hy tm cc im H trn (C) sao cho tip tuyn ti cthai trc ta ti A v B c di on AB nh nht .VD5. Tm hnh thang cn c din tch nh nht ngoi tip ng trn bn knh R cho trc .

VD6. Cho 2 2 1x y+ = . Tm Max, Min ca biu thc2

2

2( )

2 2 1

xy yP

xy x

+=

+ +.

S. 2 6 2 6,2 2

MaxP MinP+ = =

VD7.Cho , 0x y > v 1x y+ = .Tm Min ca biu thc1 1

x yP

x y= +

VD8.Cho hai s thc thay i x, y tha mn 2 2 2x y+ = .Tm GTLN, GTNN ca biu thc 3 32( ) 3P x y x= + ( C Khi A 2008)

VD9. Cho hai s thc thay i x,y tha mn 2 2 1x y+ = .Tm GTLN, GTNN ca biu thc2

2

2( 6 )

1 2 2

x xyP

xy y

+=

+ +( H Khi B 2008)

Trang

Dng 2.Tm GTLN,GTNN ca hm s c cha tham s

Dng 3.ng dng ca bi ton tm gi tr ln nht, gi tr nh nht ca hm s


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VD10.Cho hai s thc khng m x, y thay i v tha iu kin x + y = 1 .Tm gi tr nh nht v gi tr lnnht ca biu thc 2 2(4 3 )(4 3 ) 25P x y y x xy= + + +

( H Khi D 2009)Ch 4. NG TIM CN CA TH HM S

PHN I. TM TT L THUYT1.ng tim cn ng .ng thng (d): 0x x= c gi l ng tim cn ng ca th (C) ca hm s ( )y f x= nu

0lim ( )x x f x = + hoc 0lim ( )x x f x+ = +Hoc

0

lim ( )x x

f x

= hoc0

lim ( )x x

f x+

=

2.ng tim cn ngang .ng thng (d): 0y y= c gi l ng tim cn ngang ca th (C) ca hm s ( )y f x= nu

0lim ( )

xf x y

+= hoc 0lim ( )

xf x y

=

3.ng tim cn xin .ng thng (d) ( 0)y ax b a= + c gi l tim cn xin ca th (C) ca th hm s ( )y f x= n

[ ]lim ( ) ( ) 0x

f x ax b+

+ = hoc [ ]lim ( ) ( ) 0x

f x ax b

+ =

Ch : Cch tm tim cn xin ca th hm s ( )y f x=ng thng (d) ( 0)y ax b a= + l tim cn xin ca th hm s ( )y f x= khi v ch khi

[ ]( )

lim ; lim ( )x x

f xa b f x ax

x+ += = hoc [ ]

( )lim ; lim ( )

x x

f xa b f x ax

x = =


V d 1. Tm cc tim cn ngang v tim cn ng ca th hm s sau:

1.2 3

( )1

xy f x

x

+= =

+2.

2

2

2 3( )

4

x xy f x

x

+ += =

3. 33( ) 27xy f x

x= = + 4. 2( ) 5y f x x= = V d 2. Tm cc tim cn ca th hm s sau:

1.2

( ) 2 11

y f x xx

= = + ++

2.23 5 2

( )3 1

x xy f x

x

+ = =

+

3.3 2

2

2 5 1( )

1

x xy f x

x x

+ = =

+4.

22 5 1( )

2 3

x xy f x

x

+ = =

V d 3.Tm cc tim cn ca cc th hm s sau:

1.22 1

( )2 1

xy f x

x

+= =

2.

2

2 1( )

2

xy f x

x x

= =

+ +3. 2( ) 2 4 2y f x x x x= = + 4. 2( ) 3 2 4y f x x x= = +

V d 1.Tm gi tr ca tham s m sao cho:

1. th hm s2 2 1

( )x m

y f xx m

+ = =

+c tim cn ng qua im M(-3,1)

2. th hm s22 3 2

( )1

x mx my f x

x

+ += =

c tim cn xin to vi hai trc ta mt tam gic c

din tch bng 4.

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Dng 1. Tm cc tim cn ca th hm s

Dng 2. Tm cc tim cn ca th hm s c cha tham s


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V d 2. Cho ng cong (Cm):1 2

( ) 32 1

y f x xmx

= = + +

v ng thng (dm) 2y mx m= + . Xc nh

bit rng (Cm) c cc i cc tiu v tim cn xin ca n to vi ng thng (dm)mt gc c1

os5

c =

V d 3. Cho hm s2

( )1

x my f x

mx

+= =

.Tm m sao cho th hm s c tim cn ng, tim cn ngang v c

tim cn cng vi hai trc ta to thnh mt hnh ch nht c din tch bng 8.

V d 4. Cho hm s 3 5( )2

xy f xx

= =

c th (C). Tm ( )M C tng khong cch t M n hai tim c

ca (C) l nh nht ?

V d 5. Cho hm s1

( )1

xy f x

x

= =

+c th (C). Tm ( )M C khong cch t M n giao im hai tim

cn l nh nht ?Ch 5. PHNG TRNH TIP TUYN CA TH HM S

PHN I. TM TT L THUYT1.Bi ton 1. Tip tuyn ca th hm s ( )y f x= c th (C) ti mt im .Phng trnh tip tuyn ca th hm s ti 0 0( , ) ( )M x y C c dang : 0 0 0'( )( )y y f x x x = .

Trong 0'( )f x c gi l h s gc ca tip tuyn ti tip im 0 0( , )M x y .2.Bi ton 2. Tip tuyn ca th hm s ( )y f x= c th (C) c h s gc k cho trc.

1.Gi 0 0( , )M x y l tip im ca tip tuyn, ta c ( )M C 0 0( )y f x =

Phng trnh tip tuyn c dng 0 0 0( ) '( )( )y f x f x x x =

2.V h s gc ca tip tuyn bng k nn 0'( )f x k= , gii PT 0'( )f x k= tm c 0 0x y3.Kt lun .

Ch :Nu hai ng thng song song th hai h s gc bng nhau. Nu hai ng thng vung gc thtch hai h s gc bng -1

3.Bi ton 3. Tip tuyn ca th hm s ( )y f x= c th (C) i qua mt im ( , )A AA x y1.Lp phng trnh ng thng d i qua im A vi h s gc k.

d: ( )A Ay k x x y= + (1)2.d l tip tuyn ca th hm s khi v ch khi h phng tnh sao c nghim

( ) ( )

'( )A Af x k x x y

f x k

= + =

(I)

3.Gii h (I) tm k. Thay k vo (1) vit phng tnh tip tuyn .PHN II. MT S DNG TON

V d 1. Cho hm s 3 2( ) 4 6 4 1y f x x x x= = + c th (C).

a.Vit phng trnh tip tuyn ca (C) ti A c honh l 2.b.Vit phng trnh tip tuyn ca (C) bit tip tuyn song song vi ng thng (d) 4 1 0x y = .c.Chng minh rng trn (C) khng tn ti hai tip tuyn vung gc vi nhau.

V d 2.Cho hm s2

( )1

xy f x

x

= =

c th (C).

a.Vit phng trnh tip tuyn ca (C) ti M c tung bng 3.b.Vit phng trnh tip tuyn ca (C) bit tip tuyn vung gc vi gc phn t th hai.c.Vit phng trnh tip tuyn ca (C) bit tip tuyn i qua im A(0, -2)

V d 3.Cho hm s 4 2( ) 6y f x x x= = + .Vit phng trnh tip tuyn ca th hm s bit tip tuyn

vung gc vi ng thng1

16

y x= ( Khi D 2010)

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Dng 1. Vit phng trnh tip tuyn ca th hm s

Dng2 . Vit phng trnh tip tuyn ca th hm s tha mn iu kin cho trc


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V d 4. Cho hm s 3 2( ) 4 6 1y f x x x= = + c th (C). Vit phng trnh tip tuyn ca th hm s iqua im M(-1, -9). ( Khi B 2008)

V d 5.Vit phng trnh tip tuyn ca th hm s3 2

( )1

xy f x

x

= =

bit :

b. Tung tip im bng5

2c. Tip tuyn song song vi ng thng : 3 0x y + =

d. Tip tuyn vung gc vi ng thng : 4 10 0x y + =e. Tip tuyn i qua im M(2,0)

V d 1 Gi ( )mC l th hm s3 21 1( )

3 2 3

my f x x x= = + ( m l tham s ). Gi M l im thuc ( )mC c

honh bng -1.Tm m tip tuyn ca ( )mC ti M song song vi ng thng 5 0x y = .( Khi D 2005)

V d 2.Cho hm s 3 2( ) 3 1 ( )my f x x x mx C= = + + + .a.Tm m (Cm) ct ng thng y = 1 ti ba im phan bit A(0,1), B, C

b.Tm m cc tip tuyn ti B v C vung gc vi nhau .V d 3.Cho hm s 3 2( ) 3 9 5y f x x x x= = + + (C). Hy vit phng trnh tip tuyn ca th hm s bittip tuyn c h s gc nh nht .

V d 4.Cho hm s1

( )1

xy f x

x

+= =

(C). Xc nh m ng thng d: y = 2x + m ct (C) ti hai im phn

bit A, B sao cho tip tuyn ca (C) ti A v B song song vi nhau.

V d 5.Cho hm s2

( )1

xy f x

x= =

+c th (C). Tm ta im M thuc (C) bit tip tuyn ca (C) ti M

ct hai trc Ox, Oy ti A,B v tam, gic OAB c din tch bng1

4.

( Khi D 2007)V d 6.Cho hm s

2( )

2 3

xy f x

x

+= =

+(C). Vit phng trnh tip tuyn ca th (C) bit tip tuyn ct trc

honh, trc tung ln lt ti A v B v tam gic OAB cn ti O.( Khi A 2009)

V d 7. Cho hm s2 1

( )2

x xy f x

x

+ = =

+c th (C). Vit phng trnh tip tuyn ca (C) bit tip tuyn

vung gc vi tim cn xin ca th hm s. ( Khi B 2006)

V d 8.Cho hm s2 2

( )1

x xy f x

x

+ += =

c th (C). Tm trn (C) cc im A tip tuyn ca th hm

s ti A vung gc vi ng thng i qua A v tm i xng ca th hm s.( i hc An Ninh 2001)

V d 9.Cho hm s1

( )1

xy f x

x

+= =

c th (C). Xc nh m ng thng : 2d y x m= + ct th (C) t

hai im phn bit A,B sao cho tip tuyn ca (C) ti A v B song song vi nhau.(C-SPTPHCM 2005)

V d 10.Cho hm s 3 2( ) 3 4y f x x x= = + c th (C). Vit phng trnh Parabol i qua cc im cc trca th (C) v tip xc vi ng thng 2 2y x= + ( i hc An Ninh 1999)

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V d 11. Cho hm s 3 21

( ) 3 13

y f x x x x= = + + . Vit phng trnh tip tuyn ca th hm s bit tip

tuyn c h s gc ln nht.

V d 12. Cho hm s4 3

( )1

xy f x

x

= =

c th (C). Vit phng trnh tip tuyn ca th (C) bit tip

tuyn to vi trc Ox mt gc 045 .

V d 13.Cho hm s3 7

( )

2 5

xy f x

x

= =

+c th (C). Vit phng trnh tip tuyn ca (C) bit :

a. Tip tuyn song song vi ng thng 2 2 0x y + =b. Tip tuyn to vi : 2y x = mt gc 045c. Tip tuyn to vi :y x = mt gc 060

V d 14. Cho hm s2 1

( )1

xy f x

x

= =

c th (C) v im M bt k thuc (C). Gi I l giao im hai tim

cn ca th (C). Tip tuyn ti M ct hai tim cn ti A v B.a. Chng minh rng M l trung im ca on ABb. Chng minh rng din tch tam gic IAB khng ic. Tm ta im M chu vi tam gic IAB nh nht.

V d 15. Cho hm s 12 1xy x += a. Kho st v v th (C) ca hm sb. Chng minh rng vi mi m ng thng y x m= + lun ct th (C) ti hai im phn bit A v B

Gi 1 2,k k ln lt l h s gc ca tip tuyn vi ( C) ti A v B .Tm m tng 1 2k k+ t gi tr ln nht .( Khi A 2011)

Phng php: Gi s ta cn bin lun s tip tuyn ca th hm s y = f(x) i qua ( , )A AA x y

1.Lp phng trnh ng thng d i qua im A vi h s gc k.d: ( )A Ay k x x y= + (1)

2.d l tip tuyn ca th hm s khi v ch khi h phng tnh sao c nghim( ) ( )

'( )A A

f x k x x y

f x k

= + =

(I)

3.S nghim ca h phng trnh ny chnh l s tip tuyn i qua im A .V d 1.Cho hm s 3( ) 3 (C)y f x x x= = .Tm trn ng thng x = 2 nhng im m t c th k ng btip tuyn n th (C) ca hm s .V d 2. Cho hm s 3( ) 3 (C)y f x x x= = .Tm trn ng thng y= 2 nhng im m t c th k ng btip tuyn n th (C) ca hm s .

V d 3.Cho ng thng (d):x = 2 v hm s3 2

( ) 6 9 1y f x x x x= = + c th (C). T mt im bt ktrn (d) c th c bao nhiu tip tuyn vi th (C).V d 4.Cho hm s 3 2( ) 3 2y f x x x= = + c th (C). Tm trn ng thng y = -2 cc im m t kc n th (C) ca hm s hai tip tuyn vung gc vi nhau.V d 5.Cho hm s 4 2( ) 2y f x x x= = c th (C)

f. Vit phng trnh tip ca (C) i qua gc ta O.g. Tm im M thuc (C) tip tuyn vi (C) ti M cn ct (C) ti hai im A v B sao cho A

l trung im ca MB.h. Tm im M trn trc tung sao cho qua M c th k c 4 tip tuyn n th (C)

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Dng 3.Bin lun s tip tuyn ca th hm s i qua mt im


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V d 6.Cho hm s 3 2( ) 3 4y f x x x= = + c th (C). Tm nhng im trn trc Ox sao cho t c th kc ba tip tuyn n th (C).V d 7.Cho hm s 3 2( ) 3 2 1y f x x x x= = + + c th (C). Tm trn ng thng 2 1y x= cc im kc hai tip tuyn n th (C).V d 8.Cho hm s 3 2( ) 3 2y f x x x= = + c th (C). Tm trn ng thng 3 2y x= + cc im k chai tip tuyn vung gc n th (C).

V d 9. Cho hm s

1

( ) 1

x

y f x x

+

= = c th (C).Vit phng trnh tip tuyn ca (C) bit khong cch tim I(1,1) n tip tuyn ny l ln nht.V d 10.Cho hm s 3 2( ) 3y f x x x= = + c th (C).Tm cc im thuc trc honh m t c th k ba tip tuyn n th (C), trong c hai tip tuyn vung gc vi nhau.

V d 11. Cho hm s ( )2

x my f x

x

+= =

. Tm m t im A(1,2) k c hai tip tuyn AB,AC n th

hm s sao cho ABC u ( Vi B, C l hai tip im ).V d 12.Cho hm s 3( ) 1 ( 1)y f x x m x= = + + c th (C).

a.Vit phng trnh tip tuyn ti giao im ca (C) v trc Oy.b.Tm m chn trn hai trc Ox, Oy mt tam gic c din tch bng 8.

Ch 6. S TNG GIAO CA HAI THPHN I. TM TT L THUYT

1.Giao im ca hai th. Cho hm s ( )y f x= c th 1( )C v hm s ( )y g x= c th 2( )C

+ Hai th 1( )C v 2( )C ct nhau ti im 0 0 0 0( ; ) ( ; )M x y x y l nghim ca h phng trn( )

( )

y f x

y g x

= =

+Honh giao im ca hai th 1( )C v 2( )C l nghim ca phng trnh ( ) ( )f x g x= (1)+Phng trnh (1) c gi l phng trnh honh giao im ca 1( )C v 2( )C

+S nghim ca phng trnh (1) bng s giao im ca 1( )C v 2( )C2.S tip xc ca hai ng cong. Cho hai hm s ( )y f x= v ( )y g x= c th ln lt l 1( )C v

2( )C v c o hm ti im 0x .

+Hai th 1( )C v 2( )C tip xc vi nhau ti mt im chung 0 0( , )M x y nu ti im chnc chung cng mt tip tuyn . Khi im M c gi l tip im.

+Hai th 1( )C v 2( )C tip xc vi nhau khi v ch khi h phng trnh sau c nghim

( ) ( )

'( ) '( )

f x g x

f x g x

= =

Nghim ca h phng trnh trn l honh ca tip im.

PHN II. MT S DNG TONV d 1.Cho hm s

2 1( )

1

xy f x

x

+= =

+c th (C) v ng thng (d) : y x m= +

i. Chng minh rng vi mi m, (d) v (C) ct nhau ti hai im phn bit .j. Gi s (d) v (C) ct nhau ti hai im A v B. Tm m di on AB nh nht.

V d 2.Cho hm s 3 2( ) 6 9 6 (C)y f x x x x= = + .nh m ng thng (d): 2 4y mx m= ct th(C) ti ba im phn bit.V d 3.Cho hm s 4 2( ) 2( 2) 2 3y f x x m x m= = + + ( )mC . nh m th ( )mC ct trc Ox ti bn iphn bit c honh lp thnh cp s cng.V d 4.nh m th hm s 3 2( ) 1y f x x mx m= = + ct trc Ox ti ba im phn bit .

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V d 5.Cho hm s 4 2( ) (3 2) 3y f x x m x m= = + + c th ( )mC .Tm m ng thng y = - 1 ct th( )mC ti 4 im phn bit u c honh nh hn 2. ( Khi D 2009)

V d 6.Cho hm s 3 2( ) 3 4y f x x x= = + (C). Chng minh rng mi ng thng i qua im I(1,2) vi hs gc k (k>-3) u ct th hm s ti ba im phn bit I, A, B ng thi I l trung im ca AB.

( Khi D 2008)V d 7. Cho hm s 3( ) 3 2y f x x x= = + (C). Gi d l ng thng i qua im A(3,20) v c h s gc mTm m ng thng d ct th hm s ti ba im phn bit. ( Khi D 2006)

V d 8. Cho hm s2 1

( )1

xy f x

x+= =

+c th (C). Tm m ng thng 2y x m= + ct th (C) ti ha

im phn bit A, B sao cho tam gic OAB c din tch bng 3 ( O l gc ta )( Khi B 2010)

V d 9. Cho hm s 3 2( ) 2 (1 )y f x x x m x m= = + + . Tm m th hm s ct trc honh ti ba im

phn bit c honh 1 2 3; ;x x x tha mn iu kin2 2 2

1 2 3 4x x x+ + < . ( Khi A 2010)

V d 10.Cho hm s 3 21 2

( )3 3

y f x x mx x m= = + + . Tm m th hm s ct trc honh ti ba im

phn bit c honh 1 2 3; ;x x x tha mn iu kin2 2 2

1 2 315x x x+ + >

V d 11.Cho hm s 1( )1

y f x xx

= = +

c th (C). Tm gi tr ca tham s m ng thng d: y = m c

th (C) ti hai im phn bit A, B sao cho OA vung gc vi OB. (Vi O l gc ta )V d 12.Chng minh rng nu th hm s 3 2( ) axy f x x bx c= = + + + (C) ct trc honh ti ba im cchu nhau th im un nm trn trc honh.

V d 13. Cho hm s2 1

1

xy

x

+=

+a. Kho st s bin thin v v th (C ) hm s chob. Tm k ng thng 2 1y kx k= + + ct th (C) ti hai im phn bit A,B sao cho khong cch

t A v B n trc honh bng nhau. ( Khi D 2011)

Ch 7. KHO ST V V TH HM SPHN I. PHNG PHPCc bc chnh khi tin hnh kho st v v th hm s: ( )y f x=

1. Tm tp xc nh ca hm s2. Tnh gii hn v tm cc tim cn ca th hm s (Nu c)3. Tnh o hm y v gii phng trnh y = 04. Lp bng bin thin5. Nu kt lun v tnh bin thin v cc tr ca hm s6. Tm im un ca th hm s (i vi hm bc ba v hm trng phng )7. Tm cc im c bit thuc th hm s8. V th hm s v nhn xt


V d 1. Kho st v v th ca cc hm s sau:a. 3 2( ) 3 1y f x x x= = + b. 3 2( ) 2 3 12 13y f x x x x= = + c. 3( ) 3y f x x x= = + d. 3 2( ) 3 3 2y f x x x x= = + + +e. 3 2( ) 3 5 2y f x x x x= = + + f. 2( ) ( 3)y f x x x= = g. 3 2( ) 2 4 3y f x x x x= = + h. 3 2( ) 6 9 8y f x x x x= = + + +

V d 2. Kho st v v th ca cc hm s sau:a. 4 2( ) 3 6 2y f x x x= = + b. 2 4( ) 2y f x x x= =

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Dng 1. Kho st v v th hm s


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c. 4 2( ) 2 3y f x x x= = + d. 4 2( ) 2 3y f x x x= = + +

e. 4 21 1

( )2 2

y f x x x= = f. 4 2( ) 5 4y f x x x= = +

V d 3. Kho st v v th ca cc hm s sau:

a.2 1

( )2

xy f x

x

+= =

+b.

1( )

1

xy f x

x

+= =

c. ( )1

xy f x

x= =

+d.

1( )

2

xy f x

x

+= =

V d 4. Kho st v v th ca cc hm s sau:

a.2 1

( )2

x xy f x

x

+ = =

+b.

2 2 5( )

1

x xy f x

x

+ = =

c.2 2

( )1

x xy f x

x

= =

d.

2 3 3( )

2

x xy f x

x

+= =

e.2 1

( )1

x xy f x

x

+ += =

+f.

2 2 6( )

2 2

x xy f x

x

+= =

+

V d 1.Cho hm s 3( ) 3 1y f x x x= = + c th (C)a. Kho st v v th hm sb. Dng th (C) bin lun theo k s nghim ca phng trnh: 3 3 1 0x x k + =

V d 2. Cho hm s1

( )y f x mxx

= = + c th (Cm)

a. Kho st s bin thin v v th hm s khi1

4m =

b. Tm m hm s c cc tr v khong cch t im cc tiu ca (Cm) n tim cn xin ca (Cm)

bng1

2(Khi A Nm 2005)

V d 3.Cho hm s 3 2( ) 2 9 12 4y f x x x x= = + a. Kho st v v th hm s

b. Tm m phng trnh sau c 6 nghim phn bit :3 22 9 12x x x m + = (Khi A Nm 2006)

V d 4. Cho hm s2

( )1

xy f x

x

= =

c th (C).

a. Kho st v v th hm sb. Tm im trn th (C) tha :

1. C ta nguyn2. Cch u hai tim cn ca th hm s

3. Cch u hai im A(0;0) v B(2;2)4. Tng khong cch n hai tim cn l nh nhtV d 5.Cho hm s 3 2( ) 3 6y f x x x= =

a. Kho st s bin thin v v th hm s

b. Khi a thay i bin lun s nghim phng trnh:3 23 6x x a =

V d 6.Cho hm s 3 2 2 3 2( ) 3 3(1 )y f x x mx m x m m= = + + + a. Kho st v v th hm s khi m = 1 (C1)b. Tm k phng trnh 3 2 3 23 3 0x x k k + + = c ba nghim phn bitc. Vit phng trnh ng thng i qua hai cc tr ca th hm s (C1)

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CHUYN : PHNG TRNH MDng 1: Phng php a v cng c s

Dng cc php bin i a phng trnh cho v dng : ( ) ( )f x g xa a= (1) Nu c s a l mt s dng khc 1 th (1) ( ) ( )f x g x =

Nu c s a thay i (c cha bin hoc cha tham s) th[ ]

0(1)

( 1) ( ) ( ) 0

a

a f x g x

> =(t gp)

Bi 1 : Gii cc phng trnh sau

1. 2 8 1 32 4x x x + = S : { }2; 3

2.2 5 65 1x x =

3. 25 125x = S:3

2

4.3 1

4 7 160

7 4 49

x x =

5.2 56

22 16 2x x

= S : { }1;7

6.3

(3 2 2) 3 2 2x

= + S :

1

3

7. 1 15 6.5 3.5 52x x x+ + = S : { }1

8. 2 3 2 3 5 53 .5 3 .5x x x x+ + =

9.1

1 15 25x x

x x

+ =

10. 1 2 2 93 .2 12x x x =11. 1 2 3 1 23 3 3 9.5 5 5x x x x x x+ + + + ++ + = + + S : { }0

12. 13 .2 72x x+ = S : { }2

13. 1 22 .3 .5 12x x x = S : { }2

14. 2 53 9x x =15. 4 4 13 81x x = S : 1x

16.1

2 22 ( 4 2) 4 4 4 8x x x x x+ = + S :1

2

17. 6 4.3 2 4 0x x x + = S : { }0;2Bi 2 : Gii cc phng trnh sau

1.22 2 2 3( 1) ( 1)x xx x+ = S : { }2; 3

2. 3( 1) 1xx + = S : { }3

3.1 2 1 2

2 2 2 3 3 3x x x x x x

+ + = + S : 24.

3 1

1 3( 10 3) ( 10 3)x x

x x

+ ++ = S : 5

5. 8.3 3.2 24 6x x x+ = + (H Quc Gia HN-2000) S : { }1;3

6.2 2 22 4.2 2 4 0x x x x x+ + = (H D-2006) S : { }0;1

Dng 2 : Phng php t n pht ( ) , 0f xt a t= > vi a v ( )f x thch hp a phng trnh bin s x cho v phng trnh mi vi bint, gii phng trnh ny tm t (nh so iu kin t > 0) ri t tm x.Bi 1 : Gii cc phng trnh sau

1. 9 4.3 45 0x x = S : 2

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2. 22 2 6 0x x+ =3. 9 8.3 7 0x x + =4.

2 2

4 6.2 8 0x x + =5. 18 6.2 2 0x x + = S : 06. 1 15 5 26x x+ + = S : 1; -17. 17 7 6 0x x + = S : 1

8.2 2

sin cos9 9 10x x+ = S : 2k

9. 2 24 16 10.2x x + = S : 3; 1110. 2 25 5 24 2 4x x x x+ + + = (t t=

252 x x+ ) S : 2

11.2 3 3

8 2 12 0x

x x

+

+ = S : 3; 6log 8

12. (7 4 3) (2 3) 2 0x x+ + + = S : 0

13. (2 3) (2 3) 14x x+ + = S : 2

14.2 2 2

15.25 34.15 15.9 0x x x + =

15.1 1 1

6.9 13.6 6.4 0x x x

+ =S : 1; -1

16. 2 43.4 2.3 5.36x x x+ = S : 0; 1/2

17. 3(3 5) 16.(3 5) 2x x x++ + = S : 3 5( )2

log 4+

18.2 2 22 6 9 3 5 2 6 93 4.15 3.5x x x x x x+ + + + = S : 1; -4

Bi 2 : Gii cc phng trnh sau1. 3.8 4.12 18 2.27 0x x x x+ = (H A-2006) S : 12.

2 222 2 3x x x x + = (H D-2003) S : -1; 23. ( 2 1) ( 2 1) 2 2 0x x + + = (H B-2007) S : 1; -1

4. 24.3 9.2 5.6

x

x x

=(H Hng Hi-1999) S : 4

5.2 22 1 2 22 9.2 2 0x x x x+ + + + = (H Thy Li-2000) S : -1; 2

6. 25 15 2.9x x x+ = (HSP Hi Phng-2000) S : 07. 3 1125 50 2x x x++ = (H Quc Gia HN-1998) S : 08. 2 2 23 2 6 5 2 3 74 4 4 1x x x x x x + + + + ++ = + (HV Quan H Quc T-1999) S : 1; 2; 5

9. cos cos( 7 4 3 ) ( 7 4 3) 4x x+ + = (H Lut HN-1998) S : k

10. 33( 1)

1 122 6.2 1

2 2x x

x x + = (H Y HN-2000) S : 1

Dng 3 : Phng php lgarit haBin i phng trnh cho v mt trong cc dng sau :

( ) ( ) logf x aa b f x b= =

( ) ( ) ( ) ( ) logf x g x aa b f x g x b= =

( ) ( ). ( ) ( ) log logf x g x a aa b c f x g x b c= + =Ch : Phng php ny thng p dng cho cc phng trnh cha php nhn, chia gia cc hm s m.VD. Gii cc phng trnh sau

1.2

3 .2 1x x = S : 30; log 22 4 22. 2 3x x = S : 32;log 2 2

3.2

5 6 35 2x x x + = S : 53;2 log 2+1

4. 3 .4 18x

x x

= S : 32; log 2

5. 228 36.3x

xx + = S : 34; 2 log 2 7 56. 5 7

x x

= S : 7 55

log (log 7)

Trang 1


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7. 53 log5 25x x = S : 5 log 54 38. .5 5 xx = S : 41

; 55

9. 9log 29. xx x= S : 91

10. 5 .8 500x

x x

= S : 53; log 2

Dng 4 : Phng php s dng tnh n iu ca hm s. Cch 1 : (D on nghim v chng minh nghim l nghim duy nht)

a phng trnh cho v dng ( ) ( )f x g x= (*)

Bc 1 : Ch ra

0

x l mt nghim ca phng trnh (*)

Bc 2 : Chng minh ( )f x l hm ng bin, ( )g x l hm nghch bin hoc ( )f x l hm ng bin,( )g x l hm hng hoc ( )f x l hm nghch bin, ( )g x l hm hng. T suy ra tnh duy nht nghim

Cch 2 :a phng trnh cho v dng ( ) ( )f u f v= , ri chng minh f l hm s lun ng bin (hoc lun nghchbin trn D). T suy ra ( ) ( )f u f v u v= = .V d 1: Gii phng trnh 3 4 0x x+ =

Cch 1 : 3 4 0 3 4 (*)x xx x+ = + = Ta thy 1x = l mt nghim ca phng trnh (*)

t : ( ) 3( ) 4

xf x xg x

= +=

Ta c : '( ) 3 . ln 3 1 >0 xxf x = + Suy ra ( ) 3xf x x= + l hm ng bin trn R.M ( ) 4g x = l hm hngVy phng trnh (*) c nghim duy nht l 1x =

Cch 2 : 3 4 0 3 4 (*)x xx x+ = + =Ta thy 1x = l mt nghim ca phng trnh (*)

Nu 1x > , ta c13 3 3

1

x

x > =

>3 3 1 4x x + > + = (v l)

Nu 1x < , ta c13 3 3

1

x

x

< =

Suy ra 1( ) 3xf x = l hm ng bin trn R '( ) 1 0g x x R= < Suy ra ( )g x l hm nghch bin trn RVy phng trnh (*) c nghim duy nht l 1x = .

Trang 1


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Vy pt (1) c 2 nghim l 0; 1x x= = .Bi 1 : Gii cc phng trnh sau

1. 12 3 17x x+ = S : 32. 3 4 5x x x+ = S : 2

3. 2( 3 2) ( 3 2) 10x

x x+ + = S : 2

4. 2 23.25 (3 10).5 3 0x xx x + + = S :{ }52;2 log 3

5. 2 (2 3) 2(1 2 ) 0x xx x+ + = S : { }0;26. 38 .2 2 0x xx x + = S : 27. (2.3 1) 3 2x xx = + S : 1

8.2 5 1 1 1

2 5 1x xe e

x x =

S : 2; 4

9. 3 2 2 32 3 .2 (1 3 ).2 2 0x x xx x x x+ + + + + = S : 010. 2 1 2 2 1 1 22 3 5 2 3 5x x x x x x + + ++ + = + + S : 1

Bi 2 : Gii cc phng trnh sau1. (2 3) (2 3) 4x x x + + = (Hc Vin Cng Ngh BCVT-1998) S : 1

2.21 22 2 ( 1)x x x x = (H Thy li-2001) S : 1

3. 12 4 1x x x+ = (H Bch khoa TPHCM-1995) S : 14. ( 3 2) ( 3 2) ( 5)x x x+ + = (Hc Vin Quan H Quc T-1997) S :

5. 3 5 6 2x x x+ = + (H S Phm HN-2001) S : { }0;1

CHUYN : PHNG TRNH LGARITDng 1: Phng php a v cng c sDng cc php bin i a phng trnh cho v dng

[ ] [ ]log ( ) log ( )a af x g x=0 1

( ) ( ) 0

a

f x g x

< = >

[ ]0 1

log ( )( )

a b

af x b

f x a

< =

=Bi 1 : Gii cc phng trnh sau

1. 2log (5 1) 4x + = S : 32. 3 9 27log log log 11x x x+ + = S : 7293. 3 3log log ( 2) 1x x+ + = S : 14. 22 2log ( 3) log (6 10) 1 0x x + = S : 2

5. 3 21

log( 1) log( 2 1) log2

x x x x+ + + = S : 1

6. 32 2log (1 1) 3log 40 0x x+ + = S : 48

7. 4 2log ( 3) log ( 7) 2 0x x+ + + = S : 1

8. 2 18

log ( 2) 6log 3 5 2x x = S : 3

9.3

1 82

2

log 1 log (3 ) log ( 1)x x x+ = S : 1 172

+

10.2

3 3log ( 1) log (2 1) 2x x + = S : 2

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11. 4 22 1

1 1log ( 1) log 2

log 4 2x

x x+

+ = + + S :5

2Bi 2 : Gii cc phng trnh sau

1. 2 21

log (4 15.2 27) 2log 04.2 3

x x

x+ + + =

(H D-2007) S : 2log 3

2. 4log ( 2).log 2 1xx + = (H Hu-1999) S : 23. 2 22 2 2log ( 3 2) log ( 7 12) 3 log 3x x x x+ + + + + = + (H Quc Gia HN-1998) S : 0;-5

4. 29 3 32log log .log ( 2 1 1)x x x= + (H Thy Li-1998) S : 1; 4

5. 2 3 2 3log log log .logx x x x+ = (H ng -1999) S : 1; 66. 5 3 5 9log log log 3.log 225x x+ = (H Y H Ni-1999) S : 3

7.2 3

4 82log ( 1) 2 log 4 log ( 4)x x x+ + = + + (H Bch Khoa HN-2000) S : 2;2 2 6

8. 2 2 22 3 2 3

log ( 1 ) log ( 1 ) 6x x x x+

+ + + + = (H Y Thi Bnh-1998) S : 4 3

9. 2 29 331 1

log ( 5 6) log log 32 2

xx x x

+ = + (HV BCVT-2000) S :

3

2Dng 2 : Phng php t n ph

Bin i phng trnh v dng ch cha mt loi hm s lgarit, t n ph t a phng trnh bin s x cho v phng trnh mi vi bin t, gii phng trnh ny tm t ri t tm x.Bi 1 : Gii cc phng trnh sau

1. 22 2log 2log 2 0x x+ = S :1

2;4

2. 2 23 log log (8 ) 1 0x x + = S : 2; 16

3. 2 11 log ( 1) log 4xx + = S :5

3;4

4. 2 2log 16 log 64 3xx + = S :1

34;2

5.2 2

3log (3 ).log 3 1xx = S : 1 23 6. 2 2log (2 ) log 2xx x x++ + = S : 2

7. 25 55

log log ( ) 1xx x+ = S :

11;5;

25

8. 2 2log 2 2log 4 log 8x x x+ = S : 2

9. 13 3log (3 1).log (3 3) 6x x+ = S : 3 3

28log 10;log

27

10. 2 21 2 1 3log (6 5 1) log (4 4 1) 2 0x xx x x x + + = S :1

4

11.2lg(10 ) lg lg(100 )4 6 2.3x x x = S : 1

10012. 2 2 2log 9 log log 32.3 xx x x= S : 213. log4(log2x) + log2(log4x) = 2 (t t= 4log x)

Bi 2 : Gii cc phng trnh sau

1. 3 32 24

log log3

x x+ = (H Cng on-2000) S : 2

2. 22 2log ( 1) 6log 1 2 0x x+ + + = (Cao ng -2008) S : 1; 3

3. 94log log 3 3xx + = (H K Thut Cng Ngh TPHCM-1998) S : 3;

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4. 4 2 2 3log ( 1) log ( 1) 25x x + = (H Y HN-2000)

5. 2 2log 2 log 4 3x

x+ = (HV CNBCVT-1999) S : 1; 4

6. 15 25log (5 1).log (5 5) 1x x+ = (H S Phm HN-1998) S : 5 5

26log 6;log

25

7.2

2 2 2log 2 log 6 log 44 2.3x xx = (H S Phm TPHCM-2001) S :1

4

8. 2 22 1 1log (2 1) log (2 1) 4x xx x x ++ + = (H Khi A-2008) S : 52; 4

9. 2 23 7 2 3log (9 12 4 ) log (6 23 21) 4x xx x x x+ ++ + + + + = (H Kinh T Quc Dn-2001) S :1

4

10. 2 2log log 2(2 2) (2 2) 1x xx x+ + = + (H Quc Gia HN-2000) S : 0;1

11. 2 2 24 5 20

log ( 1).log ( 1) log ( 1)x x x x x x + = (HSP Vinh-2001) S : 1; 2020

log 4

log 4

1 1(5

2 5+

Dng 3 : Phng php m haa phng trnh cho v mt trong cc dng sau

( )0 1

log ( ) ( ) ( )a g xa

f x g x f x a

<

= =

log ( ) log ( )a bf x g x= t t= suy ra( )

( )

t

t

f x a

g x b

=

=. Kh x trong hpt thu c phng trnh theo n

gii pt ny tm t, t tm x.Bi 1 : Gii cc phng trnh sau

1. 3log (9 8) 2x x+ = + S : 30; log 8

2. 15log (5 20) 2xx ++ = S : 1

3. 33 23log (1 ) 2logx x x+ + = S : 4096

4. 3 22log tan log sinx x= S : 26 k +

5. 25 3log ( 6 2) logx x x = S : 9

6. 46 42log ( ) logx x x+ = (H Kin Trc TPHCM-1991) S : 16Bi 2 : Gii cc phng trnh sau

1. 2log (9 2 ) 3xx + = (H Hu-2000) S : 0; 3

2. 5 7log log ( 2)x x= + (H Quc Gia HN-2000) S : 5

3. 7 3log log ( 2)x x= + (H Thi Nguyn-2000) S : 49

4. 846 42log ( ) logx x x+ = (H Y HN-1998) S : 256

5. 3 22log cot log cosx x= (H Y Dc TPHCM-1986) S : 23

k +

Dng 4 : Phng php s dng tnh n iu ca hm s. Cch 1 : (D on nghim v chng minh nghim l nghim duy nht)

a phng trnh cho v dng ( ) ( )f x g x= (*) Bc 1 : Ch ra 0x l mt nghim ca phng trnh (*) Bc 2 : Chng minh ( )f x l hm ng bin, ( )g x l hm nghch bin hoc ( )f x l hm ng bin,

( )g x l hm hng hoc ( )f x l hm nghch bin, ( )g x l hm hng. T suy ra tnh duy nht nghim Cch 2 :

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a phng trnh cho v dng ( ) ( )f u f v= , ri chng minh f l hm s lun ng bin (hoc lun nghchbin trn D). T suy ra ( ) ( )f u f v u v= = .Bi 1 : Gii cc phng trnh sau

1. 5log ( 3) 4x x = S : 42. 2lg( 12) lg( 3) 5x x x x + = + + S : 53. 22 2log ( 3).log 2 0x x x x+ + = S : 2; 4

4.

2

3 3(log 3) 4 log 0x x x x

+ + = S : 35. 2 2 2ln( 1) ln(2 1)x x x x x+ + + = S : 0; 1Bi 2 : Gii cc phng trnh sau

1. 22 2log ( 1) log 6 2x x x x+ = (H ng -1997) S :1

;24

2.2

2

3 2

3log 3 2

2 4 5

x xx x

x x

+ += + +

+ +(H Ngoi Thng-2001) S : 1; 2

CHUYN :H PHNG TRNH M, LGARITBi 1 : Gii cc h phng trnh sau :

1.2 2

2 2

2 2log ( ) 1 log ( )

3 81x xy yx y xy

+

+ = + = (H A-2009) S : (2;2), (-2;-2)

2.

3 2

1

2 5 4

4 2

2 2

x

x x

x

y y

y+

= +

= +

(H D-2002) S : (0;1), (2;4)

3.1 4

4

2 2

1log ( ) log 1

25

y xy

x y

= + =

(H A-2004) S : (3;4)

4. 2 39 3

1 2 1

3log (9 ) log 3

x y

x y

+ =

= (H B-2005) S : (1;1), (2;2)

5.13 2

3 9 18

y

y

x

x

+ =

+ =S : 3

2( ; log 4)3

3 3

3 .2 9726.

log ( ) 3

x y

x y

= =

S : (5;2)

2

log log 27.

12

y xx y

x y

+ =

+ =S : (3;3)

3 3

4 328.log ( ) 1 log ( )

x y

y x

x y x y

+ = + =

S : (2;1)

41 log9.4096y

y x

x

= +

=S : (16;3), (1/64;-2)

4 2

4 3 010.

log log 0

x y

x y

+ =

=S : (1;1), (9;3)

Bi 2: Gii cc h phng trnh sau :

1.5

3 .2 1152

log ( ) 2

x y

x y

= + =

S : (-2;7)

2.2 2

1 1

1 1

log (1 2 ) log (1 2 ) 4

log (1 2 ) log (1 2 ) 2

x y

x y

y y x x

y x

+

+

+ + + + =

+ + + =S :

2 2( ; )

5 5

3.3 3log ( ) log 2

2 2

4 2 ( )

3 3 22

xy xy

x y x y

= +

+ + + =S : (1;3), (3;1)

Trang 2


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4.2 2

12 2x y xx y y x

x y+ + = +

= S : (-1;-1), (1;0)

5.2 2

ln(1 ) ln(1 )

12 20 0

x y x y

x xy y

+ + =

+ =S : (0;0)

6.2 1

2 1

2 2 3 1

2 2 3 1

y

x

x x x

y y y

+ + = +

+ + = +

S : (1;1)

CHUYN :BT PHNG TRNH MI. PHNG PHP

p dng cc phng php nh khi gii phng trnh m v kt hp vi tnh cht : Nu 1a > th ( ) ( ) ( ) ( )f x g xa a f x g x> > Nu 0 1a< < th ( ) ( ) ( ) ( )f x g xa a f x g x> > >II. CC DNG TON THNG GP

Dng 1: Phng php a v cng c sBi 1 : Gii cc bt phng trnh sau :1.

2 23 27x x+ < S : 3 1x <

4.

1

21 12

16

xx

++ >

S : 2x >

5. 1 2 1 22 2 2 3 3 3x x x x x x + + < + S : 2x >

6.2 2 23 2 3 3 3 42 .3 .5 12x x x x x x S : 1 4x x

7.3 1

1 3( 10 3) ( 10 3)x x

x x

+ ++ < S : 3 5 1 5x x < < < S :

72 3

2x x< < >

Bi 2 : Gii cc bt phng trnh sau :

1.

1 1

2 2 3 3

x x x x+

+ + (H Quc Gia HN-1996) S : 2x 2. 1 1( 2 1) ( 2 1)

x

x x+ + (Hc Vin Qun Y-1995) S :1 5 1 5

2 2x x

+< < >

3.2

1

2 133

x x

x x

(H Bch Khoa HN-1997) S : 2x

4. ( )2 1 1x

x x+ + < (H S Phm TPHCM-1976) S : 1x <

Dng 2 : Phng php t n phBi 1 : Gii cc bt phng trnh sau :

1. 9 2.3 3 0x x > S : 1x >

Trang 2


23/62


2. 2 6 72 2 17 0x x+ ++ > S : 3x > 3. 32 2 9x x+ S : 0 9x 4. 2.49 7.4 9.14x x x+ < S : 0 1x<

3. 1 2 1 23 2 12 0x

x x+ + < (HV CNBCVT-1998) S : 0x >

4.22.3 2

13 2

x x

x x

+

(HV Hnh Chnh QG-2001) S : 3

2

0 log 3x<

5. ( ) ( )2 2

2 15 1 2 3. 5 1x x x x

x x + +

+ ++ + < (H Phng ng-2000) S : 0 1x x< >

Dng 3 : Phng php s dng tnh n iu ca hm s.1. 13 2 3x x + > S : 1x >

2. 22 3 1x

x < + S : 2x S : 2x th log ( ) log ( ) ( ) ( ) 0a af x g x f x g x> > >

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

> > > >II. CC DNG TON THNG GP

Gii cc bt phng trnh sau :

1. 3log (2 1) 2x + < S :1

42

x <

3.3 2

log ( ) 12x

x

x

+>

+S : 1 2x<

8. 3log log (9 72) 1x

x (H B-2002) S : 9log 73 2x<

9. 2log (5 8 3) 2x x x + > (H Vn Lang-1997) S :1 3

2 5x x< < >

10. 22log 64 log 16 3x x+ (H Y H Ni-1997) S : 31 1

12 2

x x< <

11.2lg( 3 2)

2lg lg 2

x x

x

+>

+(H Kin Trc HN-1997) S :

3 33 1

6 2x

+< + + (H Dc HN-1997) S : 2 1 2x x < < < 0 l phng trnh mt c

tm I(-A ; -B; -C), bn knh 2 2 2R A B C D= + + 2) Giao ca mt cu v mt phng - Phng trnh ng trn:

Cho mt cu 2 2 2 2(S) : (x a) (y b) (z c) R + + = vi tm I(a ; b; c), bn knh R v mt phng(P): Ax + By + Cz + D = 0.

+ d(I, (P)) > R: (P) v (S) khng c im chung+ d(I, (P)) = R: (P) tip xc (S)+ d(I, (P)) < R: (P) ct (S) theo ng trn c tm H l hnh chiu ca I xung (P), bn kn2 2

r R d= B. BI TP.

Bi 1: Xc nh tm v bn knh ca ng trn (C):2 2 2

2x 2y z 9 0

x y z 6x 4y 2z 86 0

+ =

+ + + =Bi 2: Cho (S): x2 + y2 + z2 -2mx + 2my -4mz + 5m2 + 2m + 3 = 0

a) nh m (S) l mt cu. Tm tp hp tm I ca (S)b) nh m (S) nhn mt phng (P): x + 2y + 3 = 0 lm tip din

c) nh m (S) ct d:

x t 5

y 2t

z t 5

= + =

= +

ti hai im A, B sao cho AB 2 3=

Bi 3: Vit phng trnh mt cu (S) c tm thuc Ox v tip xc vi hai mt phng (Oyz)v (P): 2x + y - 2z + 2 = 0.

Bi 4. Trong khng gian vi h trc to cc vung gc Oxyz, cho bn im A(1;2;2), B(-1;2;1), C(1;6;-1), D(-1;6;2)

a. CMR: ABCD l t din c cc cp cnh i bng nhau.b. Tnh khong cch gia AB v CD.c. Vit phng trnh mt cu ngoi tip t din ABCD.

Bi 5. Cho im I(1;2;-2) v mt phng (P): 2x + 2y + z + 5 = 0.a. Lp phng trnh mt cu (S) tm I sao cho giao ca (S) v mt phng (P) l ng trn c chu vi

bng 8

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b. CMR. Mt cu (S) tip xc vi ng thng (): 2x 2y = 3 zc. Tnh din tch thit din ca hnh lp phng ct bi mt phng (CMN).

Bi 6. Trong khng gian vi h trc to cc vung gc Oxyz cho hai ng thng (d 1) (d2) c phng

trnh ( )

=

=

=

4

2

:1

z

ty

tx

d ( )

=++

=+

012344

03:

2

zyx

yxd

a. CMR: (d1) v (d2) cho nhau.b. Tnh khong cch gia (d1) v (d2).c. Vit phng trnh mt cu (S) c ng knh l on vung gc chung ca (d 1) v (d2).

Bi 7. Trong khng gian vi h trc to cc vung gc Oxyz cho hai mt phng song song c phngtrnh tng ng l: ( ) 0122:1 =+ zyxP ( ) 0522:2 =++ zyxPV im A(-1;1;1) nm trong khong gia hai mt phng . Gi (S) l mt cu qua A v tip xc vi c haimt phng (P1), (P2)

a.CMR: Bn knh ca hnh cu (S) l mt hng s v tnh bn knh .b.Gi I l tm hnh cu (S). CMR: I thuc mt ng trn c nh xc nh tm v tnh bn knh ng

trn .BI TP TNG HP.

Bi 1. (H-2007D). Cho A(1;4;2), B(-1;2;4) v ng thng (d):x 1 y 2 z

1 1 2

+= =

a) Vit phng trnh ng thng i qua trng tm G ca tam gic OAB v vung gc vi

mp(OAB)b) Tm ta im M thuc sao cho: MA2 + MB2 nh nht

Bi 2. (H-2007B). Cho mt cu (S): x2 + y2 + z2 - 2x + 4y + 2z 3 = 0 v mp (P): 2x y + 2z 14 = 0a) Vit phng trnh mp (Q) cha Ox v ct (S) theo mt ng trn c bn knh bng 3b) Tm ta im M thuc (S) sao cho khong cch t M n (P) ln nht

Bi 3. (H-2007A). Cho hai ng thng 1x y 1 z 2

d : 2 1 1 += = v 2

x 1 2t

d : y 1 tz 3

= +

= + =a) Chng minh hai ng thng cho nhaub) Vit phng trnh ng thng d vung gc vi mt phng (P): 7x + y 4z = 0 v ct c hai ng

thng d1, d2.

Bi 4. (H-2008A). Cho A(2;5;3) v ng thng d:x 1 y z 2

2 1 2

= =

a) Tm ta hnh chiu ca A trn db) Vit phng trnh mp (P) cha d sao cho khong cch t A n (P) ln nht

Bi 5. (H-2005B). Cho hnh lng tr ng ABC.A1B1C1 vi A(0;-3;0), B(4;0;0), C(0;3;0); C1(0;0;4)

a) Vit phng trnh mt cu c tm A v tip xc vi mp(BCC1B1)b) Gi M l trung im ca A1B1. Vit phng trnh mt phng (P) i qua A, M v song song vi BC1Mt phng (P) ct A1C1 ti N. Tnh di MN

Bi 6. (H-2010D)1. Chng trnh Chun: Cho hai mt phng (P): x + y + z 3 =0 v (Q): x y + z 1 = 0. Vit phng trnhmt phng (R ) vung gc vi (P) v (Q) sao cho khong cc htu72 O n (R) bng 2.

2. Chng trnh Nng cao: Cho hai ng thng 1

x 3 t

d : y t

z t

= + = =

v 2x 2 y 1 z

d :2 1 2

= = . Xc nh ta

im M thuc d1 sao cho khong cch t M n d2 bng 1

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Bi 7. (H-2010A)

1. Chng trnh Chun: Cho ng thngx 1 y 2 z 3

:2 3 2

+ = = v mt phng (P): x 2y + z = 0. Gi C

giao im ca v (P), M l im thuc . Tnh khong cch t M n (P) bit MC = 6

2.Chng trnh Nng cao: Cho A(0;0;-2) v ng thngx 2 y 2 z 3

:2 3 2

+ + = = . Tnh khong cch t A

n . Vit phng trnh mt cu tm A, ct ti hai im B v C sao cho BC = 8.

Bi 8. Vit phng trnh ng thng qua im A(3;2;1) ct v vung gc vi ng thng () c phngtrnh:

1

3

42

+== zyx

Bi 9. Cho (P): 012 =++ zyx v ( )3

2

2

1:

+== zyxd . Vit phng trnh ng thng qua giao im ca

(d) v (P) vung gc vi (d) v nm trong (P).

Bi 10. Cho A(2;-1;1) v ( )

=+

=+

022

04:

zyx

zy

a) Vit phng trnh mt phng (P) qua A v vung gc vi ().b) Xc nh to im B i xng vi A qua ().

Bai 11. Cho ( )

1

: 9 2

12

x t

d y t

z t

= + = + =

v ( )3

6

1

2

2

5:

== zyx

a) CMR: (d) v () thuc mt mt phng.b) Vit phng trnh mt phng .c. Vit phng trnh hnh chiu song song ca (d) theo () ln mt phng (P) 01223 = zyx

Bi 12. Cho ( )3

1

2

1

7

3:1

==

zyx; ( )

1

9

2

3

1

7:

== zyx

a) Hy vit phng trnh chnh tc ca ng thng (3) i xng vi (2) qua (1) b) Vitphng trnh chnh tc ca ng phn gic gc A.Bi 13. Cho A(0;0;-3), B(2;0;-1) v mt phng 01783 =+ zyx

a) Tm to giao im I ca ng thng AB v mt phng (P)b) Tm to ( )PC sao cho tam gic ABC u.

Bi 14. Cho cc im A(-2;1;0), B(-2;0;1), C(1;-2;-6), D(-1;2;2)a) Chng minh 4 im A, B, C, D lp thnh t din.b). Tnh th tch khi t din ABCD.c). Vit phng trnh mt phng (ABC), (BCD). Vit phng trnh tham s ca CD.d) Vit phng trnh ng thng qua d qua A vung gc vi (BCD).e) Tm ta im A i xng vi A qua (BCD)

f). Tnh khong cch gia AB v CD.g) Tm trn CD mt im I sao cho I cch u (ABC) v (ABD).

Bi 15. Cho A(0;1;2), B(2;3;1), C(2;2;-1)a) Vit phng trnh mt phng (P) qua A, B, C. Chng minh rng O cng nm trn mt phng (P).b) Chng minh rng t gic OABC l hnh ch nht, tnh din tch hnh ch nht.c) Tnh th tch hnh chp S.OABC bit S(9;0;0).

Bi 16. Cho A(1;3;-2), B(13;7;-4) v ( ) 0922: =+ zyxa) Gi H l hnh chiu vung gc ca A trn . Xc nh H.b) Cho K(5;-1;1). CMR: A, I, K, H to thnh t din. Tnh th tch t din.

Bi 17. Cho mt phng (P): 16x 15y 12z + 75 =0a) Lp phng trnh mt cu (S) tm l gc to O, tip xc vi mt phng (P).

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b) Tm to tip im H ca mt phng (P) vi mt cu (S).c) Tm im i xng ca gc to O qua mt phng (P).

Tai Lieu on Thi Tot Nghiep Va Dai Hoc 2013

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