tai lieu on thi tot nghiep va dai hoc 2013
TRANSCRIPT
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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)
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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
PHN II. MT S DNG TON
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 = =
PHN II. MT S DNG TON
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.
Trang
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)
Trang
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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Dng 2.Vit phng trnh tip tuyn tha iu kin cho trc
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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
PHN II. MT S DNG TON
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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Dng 2. Mt s bi ton lin quan n kho st v v th hm s
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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)
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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 = .
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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)
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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 >
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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).