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Chn im riBn cht im ri l chn nhng h s ph hp cho qu trnh s dng bt
ng thc c thun
li v du bng c th xy ra. ri thnh cng th chng ta cn bit 1 siu:
Thng thng du bng xy ra vi nhng bin c vai tr nh nhau s bng nhau. Bin
c vai tr
nh nhau th khi i ch v tr ca bin bi ton khng thay i. VD: Cho 2 4a
b c+ + = .Tm
Min 3 3 43M a b c= + + th a,b l cc bin c vai tr nh nhau.
Do du bng ca BT C-si xy ra ti cc bin bng nhau nn ta thng s dng
pp tch: Tcha thnh m s: ....
a a aa
m m m= + + + , tch . ...ma a a a=
H s ph thuc vo cc yu t sau :+Du bng ca BT+Mc ch ca BT (h bc, kh
bin, to gi thit,)+T l h s trong biu thc cn tm cc tr v gi thit
Du bng thng xy ra trng vi du bng cc BT trong gi thit (nu c) Ta
ch cn chn h s
Bi ton chn lc:
Cho , , 0a b c > v 21 2 8 12ab bc ca+ + . Tm min1 2 3
Ma b c
= + +
Tduy: Rt n gin l tgi thit m CS ta c ?m n pa b c .Ri CS M sao cho
cng to rac mu m n pa b c .Nhng ni l vy cn lm th sao ?Bi gii:
M d nhn ta t1 1 1
, ,a b cx y z
= = = gi thit s thnh 2 8 21 12 zx y z xy+ + v
2 3M x y z = + +
Gith rix y
zm n
= = s c
12 8 2 8 21
21
21 8 21 2 2 8
12 z 2 8 21 2 8 21 (2 8 21)
?
m n m n
n m m n
x y x yxy x y z m n z m n z
m n m n
x y z
+ +
+ + +
+ + = + + + +
n lt M ri
12 2 3
32 3 2 3 ( 2 3)
m n m nx y x yM x y z m n z m n z
m n m n
+ + = + + = + + + +
Bi ton s thnh cng khi : 2 :3 (21 8 ) : (21 2 ) : (2 8 )m n n m m
n= + + + . Ta s vit thnh2 2
2
2 8 24 632 3 2 63
21 8 21 2 2 8 8(3 )16 4 6 63
m mn nm n mn
n m m n mn mn m
+ = + -= = =
+ + + -+ = +
.
Th vo thng th2 s c2
2 22 63 2 63
16 4 6 638(3 ) 8(3 )
m mm m
m m
- -+ = +
- - . Gith ng c m u
vo gii m ly FX s cho ngay9 15
2 8m n= = . Ty ta cng tm ra c
5 23, ,
4 3x y z = = =
Important Note:Haizz!c xong li gii trn cng thy au cu, nhngng lo,
mc chcui cng ca bi vit cng chl cc bn lm c cc bi dhn hoc bngtrn th
tc
gi mng lm ri !
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I.Tnh cht h bc H bc n bin CS m bin mx v m-n hng sa
( ) ( )... ... mm m m m n nx x x m xa a a a -+ + + + + + +
H bc a bin CS a bin mx v b bin ny s to ra n thc tch xy c t l bc
l ambn
)
... ... ( )m n m m m n n n am bna bax bx x x x y y y a b x y++ =
+ + + + + + +
Nhn xt:-H bc cho php ta m t ( , , ...)f a b c t ( , , ...)g a b
c trong bc ca g ln hn f
-im ri gip ta chn m,n,a Ch :
-Hai dng nu trn chmang tnh tiu biu cn trong qu trnh lm bi xin
nhcho nguyn tc:Mi su c thtch thnh m snhnhau v CS cng h sa (a thc bc
khng) ->khi bc sauCS sthay i.
VD1.Cho , , 0, 3a b c a b c> + + = Tm Min 3 3 364M a b c= +
+
Do a,c c vai tr nh nhau nn ta gi s du bng xy ra ti , 2 3 (*)a c
m b n m n= = = + = .Khi
3 3 3 23 ( )a m m m a+ + |=> dng c gii thit th h s ca a,b,c
bng nhau tc3 3 3 23 ( )a m m m c+ + | 2 23 64.3 (**)m n=
3 3 3 264 64 64 64.3 ( )b n n n b+ + | T (*) v (**) rt th ta c
m=24/17 v n=3/17
VD2.Cho 2, , 0, 2 4 3 13a b c a b c> + + = Tm Min 2 2 3M a b
c= + +
Gi s du bng xy ra ti 2, , 2 4 3 13 (*)a m b n c p m n p= = = + +
= .Ta CS nh sau2 2
2 2
3 3 3 2
2 ( )
2 ( )
3 ( )
a m m a
b n n b
c p p p b
+ +
+ +
+ +
=> dng c gii thit th22 2 3
(**)2 4 3
m n p= =
X l (*) v (**) bng tnh cht dy t s bng nhau c22 4 3 13
1, 2, 12 8 3 13
m n pm n p= = = = = =
VD3.Cho 2 5, 0, 5a b a b> + = .Tm Min 3 6M a b= +
Gi s du bng xy ra ti 2 5, 5(*)a m b n m n= = + = .Ta CS nh sau3
2
3 23 3 3 2 2
6 56 6 6 6 6 6 5 56 5
3( )
3 ( ) 2 2
66 ( )( )
5 5
m ma a
a a m m a
m nb b b b b m n bb b
- + + +
+ + + + + - +
=> dng c GT th2 53 6
(**)2 5
m n=
T (*) v (**) ta c th rt th c 2 520 25
,9 9
m n= =
VD4. Cho 2 3, 0, 2011a b a b> = Tm Min 7 5M a b= +
Ta khoan quan tm n du bng m hy em CS m s 7a v n s 5b nh sau:7 7
7 5 5 5 7 5
... ... ( )m n
m nm n
a a a b b b a bm n
m m m n n n m n++ + + + + + + + => dng c gii thit th
7 2
5 3
m
n= nn ta
chn m=10 v n=21. Du bng xy ra khi7 5
10 21
a b= v 2 3 2011a b = (lun c a v b !)
VD5. Cho 2 2 216
425
a b c ab+ + + = Tm Max M ab bc ca= + +
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Ging nh VD1 du bng xy ra ti a=b. Ta CS nh sau:=>Ta lm xong qu
trnh m t cc binnhng th l cha v cc bin cnmang theo mnh h s. Ln lt
nhn ccBT vi x,y,z,t ta s c h thc
Gith s cn c( )
1 (nhn o GT)1 1 1 16 / 25
16/ 252 2 2 ( ook at M)
x y x z m y z t y zv
tx t my mz L
+ + +== = = =
= + = =
2 9/5011 1
41/5016
2 2 50/4125
x y my x
y
x my m
+ == = =
+ = =
C m ri th du bng cn chi l kh khi 2 2 250 16
, 441 25
a b c a b c ab= = + + + = .Thi t gii tip nh.
II.Tnh khkiu tch
Xt phn sA
B
Kh ni phn s: ta phn tch B thnh nhn t v tch A lm cc tha s sao cho
sau CSc c iu mnh mun
?1 21 2
1 2 1 2
? . ......
. ... . ...
A AA A
B B B B
+ +
Kh mrng: phn tch B thnh nhn t ri cng c phn s vi cc nhn t ca B
saocho CS c c iu mnh mun ?1 2
1 2
... ?. ...A B B A
B B+ +
Nhn xt:-Vic nhn nhn
1 2, ...B B kh d dng
-im ri gip ta chn h scho cc bin em CS sau cho ph hp
VD1.Chox,y,z>0 tho mn x y z 3+ + = CMR:+ +
+ + -+ + +
x y z x y z
y z x
3 3 3 2 2 2
3 3 3
4 ( )
9 278 8 8.
Vai tr x,y,z nh nhau nn du bng khi chng bng nhau v bng 1.
Ta phn tch bi ton tm bin CS nh :
=+ + - +
x x
y y y y
3 3
3 28 ( 2)( 2 4)
nn ta s chn thm 2 biu thc + - + y y y 2( 2), ( 2 4) thc hin.
Nhng =+
x
y
3
3
1
98, (y+2)=3, - +y y2( 2 4) =3 nn ta s lm nh sau
x y y y x
y
3 2
3
2 2 4
27 27 38
+ - ++ +
+
Lm tng t c VTx y z x y z
2 2 210( ) ( ) 18
27
+ + - + + -=
x y z2 2 2
12 ( )
27
- + +(noy!)
VD2.CMR4 3 2
22
a b ca b
b c ca a+ + + vi , , 0a b c >
2 2
2 2
2 2
2( )
2 ( )
2 ( )
( )
a b ab
a mc m ac
b mc m bc
ab ab
+
+
+
2 2
2 2
2 2
( ) 2 ( )
( ) 2 ( )
( ) 2 ( )
( )
x a b x ab
y a mc my ac
b mc mz bc
tab t ab
+
+
+
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Nhn vai tr a,b,c khng h nh nhau nhng hy c th cho a=b=c ta c du
bng.Nu c th th cng th liu xem
4
2
3 4 3 2
2
2
4
3 2
2
ab b c a
b c
b a b cc a b a b
ca b c ca a
ca c
a
+ + +
+ + + + +
+
.Bi ny xoy tht !
VD3.CMR2 2 216 1
(64 )9
a b cc a b
b c c a a b+ + - -
+ + +vi , , 0a b c >
Nhn i: a,b c vai tr nh nhau nn ta st a=b=mc th du bng xy ra ti2
2 2 2( ) ( ) 16 1 2 8 2
(64 ) (32 )9 1 9
mc mc c mc mc mc m
mc c c mc mc mc m m+ + - - + = -
+ + + +
Nht vo Fx570ES s cho ra m=2.V khi ta s lm nh sau:2
2 2 2 2
2
4 4( )
9 3
4 4 16 1( ) (64 )
9 3 916
( ) 8
ab c a
b c
b a b ca c b c a b
c a b c c a a bc
a b ca b
+ + +
+ + + + - -
+ + + ++ +
+
VD4.Cho x,y>0 x+y=4 Tm Min6 10
2 3M x yx y
= + + +
Ta CS nh sau
62 6
102 10
mx mx
ny yy
+
+
=> Du bng xy ra ti2
2
6 6 104(*)
10
mx
m nny
= + =
=
V phn cn li ca M l (2 ) (3 )m x n y- + - gii quyt c cn 2 3m n- =
- hay 1n m= + (**)
T (*) v (**) ta c6 10
41m m
+ =+
nht v my tnh i s cho m=3/2 -> n=5/2
VD5.Cho 2a Tm Min2
1M a
a= +
C du bng l a=2 ri th c th m lm 32 2
3 1 3 1 9.2 3
4 8 8 4 8 4
a aM a
a= + + + + =
VD6. Tm Min 29
(1 )(1 )(1 )y
M xx y
= + + +
Ta s CS 1+x bng cch tch 1 thnh a sv x thnh b s. Khi s to ra a b
bx+ . V thng ny ch
c th kh khi CS (1 )y
x+ bng cch tch 1 ra a sv
xra b s tc to ra
b
a bb
y
x+ .Nu lm nh vy
du bng xy ra ti2
2 21 9 9(1 )(1 ) (10 )x y
y x M x xa b xb x x
= = = = + + = + +
Th ny th x ch c
th bng 3 v y s l 9. Li gii chun nh sau2
3 3 32 4 4 4
3 3 3
3 3 3 3(1 )(1 )(1 ) 4 .4 .4 256
3 3 3 3 3 3 3 (3 ) ( )
x x x y y y x yM
x x x xy y y y
= + + + + + + + + + =
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VD7.Chox,y,z>0 Tm Minz
(1 3 )( 8 )( 9 )( 6)
xyM
x x y y z z =
+ + + +
Ta s CS mi tha s ca tch nh sau
(1 3 )x+ tch 1 thnh a s3x thnh b s s to ra bin a b bx+ v du bng
ti1 3x
a b=
khc x (bc1) t th ( 8 )x y+ tch x thnh a sv 8y thnh b ss to ra
bin a ba bx y+ . Du
bng ti8x y
a b
=
Lp lun tng t c4
1 3 8 9 6 1 1 1 3 12, y , 1,
3 9 8 6 3 3.9.8.6 2 6
x x y y z z a y x z ax z
a b a b a b a b b x z y x b
= = = = = = = = = = = = =
Li gii hon chnh nh sau
43 3 8 8 9 9(1 ... )( ... )( ... )( 1 ... 1) 7 z6 6 6 6 6 6
x x y y z z x y z xy+ + + + + + + + + + + +
VD8.Tm Min2
3
3
1
x xM
x
- +=
-
Ta c 2 21 1 . 1x x x x- = - + + nn cn tm m,n sao cho2 2 2(1 ) (1
) 3 ( ) 2, 1
3
m nm x n x x x x nx n m x m n n m n m n
+- + + + = - + = + - + + = - = = = .
Bi ton c gii nh sau2 2
3 3
3 2(1 ) (1 )2 2
1 1
x x x x xM
x x
- + - + + += =
- -
VD9.Cho 34
z3
x xy xy+ + = CMR 1M x y z = + +
Ta si cm BT sau 33
( z)4
x y z x xy xy+ + + + . CS ta cn tch x lm 3 s, y lm 2 snh sau
33( ) 1 1
2 ( ) 3 ( z)
x x y x x x
x y z z xy xya b d c e a bd ce
+ + = + + + + + + + .Ta s c c h sau
3
2
3
4
31 1 1 32 3 ( oo VP )
1 34 ( )81 1 1 1 1
1 ( ch tc )1 1
( )4
a
l ka bd ce
bdc h
a b c d ece
=
= = = =
+ + = + = =
v
1 11
1 1 1
4
d e
b c
+ =
+ =
Rt th lung tung ta s c32(3 / 8) 1
11/ 4 1/ 4
cc
+ =
- .NhFx570ES cho c=16 ri th l
16
3b = ,e=4,
4
3d=
Li gii sau khi c du bng 31 1 4
.4 .4 .16z ( )2 4 3
x x y x y x y z + + + +
III.Tnh khkiu tngTa thng tch t mA thnh m bin A ri dng im ri chn
h s ph hp cho cc bin
VD1.Cho [0;2]x Tm Max 3 5(2 )M x x= -
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khc ht x ta phn tch M ra cc tha s x bc 1 v chn h s cho chng nh
sau
3 5(2 )3(2 ) (2 ) (2 ) (2 ) (2 ) 3 5
. .5
m nx
x mx x x x x x x x m nMnm m m n n n n n m n m n
+-
+ =- - - - -= =
=+
VD2. Cho [0;1]x Tm Max 2 4 2 413 9M x x x x= - + +
Ta lm nh sau2 2 2 2
2 2 2 213 9 13 (1 ) 9 (1 )(1 ) (1 ) . .2 2
mx x nx xmx x nx xm n m n
+ - + +- + + +
du bng xy ra v kh ht bin x th
Li nhFx cho ta1 3
,2 2
m n= =
IV.Bi ton kh
1.Dng cc php thn gin tm du bng
VD1:Cho , , 0, 2 3 20a b c a b c> + + CMR3 9 1
52
M a b c a b c= + + + + + (a=2,b=3,c=4)
VD2:Cho , , 0, 12, 8a b c ab bc> CMR1 1 1 8 121
2( )12
M a b cab bc ca abc
= + + + + + + (a=3,b=4,c=2)
VD3:CMR3 3 32 32
22 12 24a b c
a b cbc ca ab
+ + - - vi , , 0a b c > (a=2b=4c)
VD4:Cho 2 2 2( 2)( 1) ( 3 ) 64a a b c c+ + + + = Tm Max 3 4 5M a
b c= (a=b=c=1)
VD5:Cho { }, ,a max a b c= Tm Min 32 1 3 1a b c
Mb c a
= + + + + (a=b=c)
2.Gi thit la
VD1:Cho 0 2x< Tm Min 19952011
1x
x+
l la th c lm bnh thng m tch 1995x ra 2011 s cn2011
1
xra 1995 s
VD2:Cho4, 2, 1995
2011
a b c
a b c
+ + =Tm Max M abc=
Bi ton ny th khng n gin. Nu nhn vo bi ton ta s thy c 3 bin khc
bit nhau rt nhiu.Tuy nhin hy xem n 1995 dng nhc bit hn c 4 v 2. Ta
c th d rng du bng ti a=b=8
v c=1995. Khi [ ] 3
8.2011 1987( )8 8(2011 ) (1995 )(1995 ) 15960
3
a bM a b b a
+ + = - - - -
V.Vi cm nhn nh
*1.Cho [0;1]x CMR 2 22 3 4 6 2 11t t t t - + + + + +
2 2 2
2 2
2 2 2 2
2 2
2
2 2
11 1
13(1 ) 9( 1) 13(1 ) 9( 3)13(1 ) 9( 1)
2
1
m x xm n
m n m mm n
m n m mm n
n x x
= - + = -
- + - + = = - +
= + = +
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Du bng khi t=1 nhng nu ta CS s c2 2 2 2
2 2
2
2 3 4 6 2 2 3 11 4 62 3 4 6 2. .1 11. .1 ( 1) ( 1)
2 11 2 2 2 11
2 11 2 11 2( )
4 22 11 2
t t t t t t t t t t t t
t
- + + + - + + +- + + + + = + + + +
= + + -
2 11
2t
++
Nhng tam thc bc 2 max ti t=-1 ?????
.. =>Qua bi ton ny ti mu
n ni rng khng ph
i bt c
khi no c du b
ng c
ng c th
lmc. Bi ton trn bnh phng ln ri xt hiu s c p n !
*2.Cho x,y>0 v 1x y+ = Tm Min 2 23 1 2 2 40 9S x y= + + +
Ta s dng B.C.S nh sau 2 2 2 2 2 2( )(1 2 ) ( )(40 9 ) 2 2 10 3S
a b x c d y a bx c dy= + + + + + + + +
Vic chn a,b,c,d tha mn
2 2 2 2
2 2
2 2 2 2
19; 4
2 / 31 2 40 9;
2 101 2
aa b c d
bx y
a b c d cx y d
= + = + =
= = =
= + = =
..=> Nh vy khi nm c nguyn l hot ng ca pp ny ta c th dng n vi
mi loi BTch khng ring g CS. Sau y s l mt vi v dPP phn phi li
Cho a,b,c>0 CMR3 3 3 3 3 3
2 2 2
5 5 5
3 3 3
b a c b a ca b c
ab b bc c ca a
- - -+ + + +
+ + +
Khng th ly3 3
2
5
4 3
a b b a
ab b
+ --
+nn ta nghn vic tm m,n
3 3
2
50
3
b ama nb
ab b
-+ -
+
Cng c ngha l 2 3 3 3 2 2 3( )( 3 ) (5 ) (3 ) (3 5) 0ma nb ab b b
a a ma b m n ab n b+ + - - = + + + + -
Kin thc btr: Cho3 2
1
( )
( ) 0
f x ax bx cx d
f x
= + + +
=th 2
1 1( ) 0 3 2 0f x x ax bx c " + + =
Chia cho 3b v t 3 2( ) (3 ) (3 5) 0ax f x x mx m n x nb
= = + + + + - .Du bng li xy ra ti 1x = nn
ta c h(1) 1 (3 ) (3 5) 0 1
3 2 (3 ) 0 2
f m m n n m
m m n n
= + + + + - = = -
+ + + = =
Trli bi ton ta c3 3
2
5( 2 ) 0
3
b aa b
ab b
-- + -
+3 2 2( ) 1 ( 1)( 1) 0
af x x x x x x x
b
= - - + = - + =
Ch :y l mt kthut kh mnh trong chng minh cc bi ton phn thc c bc
va phi.Kinthc bsung chnh l o hm bc nht ca cc hm a thc tuy nhin khng
cn hiu g vn bnchcn nhl vi hm a thc bc 4 ta si xt 3 24 3 2 0ax ax bx
c+ + + = .Cc bc nh hn th coinhh sbc ln bng 0.
Sau khi c xong bi ton trn chng ta sc ci nhn tng qut hn kh nhiu
cho cc bi boHnh trnh i n bi ton tng qut (THTT s371 thng 8/2008) v
Mt phng php chngminh btng thc(THTT s403 thng 1/1011)
Bi tp cho pp im ri
1. Cho , , 0 4712
a b c
a b c
>
+ + =
Tm Min 2 2 23 4 5M a b c= + +
2. Cho a+b+c=2011. Tm Max 4 2 1995M a b c=
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3. Cho 0 4, 7, 2 3b a a b c d < + Tm Min2 2
1 22c d
c dMa b
+ + +=
+
4. Cho a,b,c>0 v 2 2 2 1a b b c c a+ + = .Tm Min 3 3 32( )M a
b c= + + 5. CMR Vi a,b,c>0 th 2 2 24 3a b c a b
b c a+ + +
6. Cho a,b,c,d>0 v c+d=4, 2 24 2a b+ = . Tm Max M=2ac+bd+cd7.
CMR Vi a,b,c>0 th
3 3 2
( ) ( ) 2
a b c bab c a c a b b c+ + ++ + +
8. Tm Max ( )( )M x z y t = + + bit 2 2 2 22 2 1x y z t + + + =
9. Tm Min
2
11 74 1
2M x
x x
= + + +
10.Tm Max 22 1 4M x x x= + - - 11.Bit , 0, 60, 100x y z x y z +
+ = . Tm Max zM xy= 12.Cho , , 0x y z v 1x y z + + = Tm Min
z
x yM
xy
+=
13.Tm cc tr 2(99 101 )M x x= + - 14.Cho , , 0x y z v 2 2 23 4 5
zx y z xy+ + = Tm Min 3 2M x y z = + + 15.Cho
, 0
11
a b
ab
>
+
Tm Min 32 1999a b
Mb a
= +
16.Cho 2 2 23 2 1x y z + + .Tm Tm Min 3 4 5a b cMbc ca ab
= + +
Bi tp rn luyn knng sdng C-siLT3:Cho 0 3, 0 4x y .Tm max (3 )(4
)(2 3 )M x y x y= - - +
LT4:Cho ba s dng a, b, c tho a2 + b2 + c2 = 1. Chng minh:
3 32 2 2 2 2 2 2
a b cb c c a a b
+ + + + +
LT5:Cho [0;1]x Tm Min2010 2011
1M
a a= +
-
LT6: Tm Minab a b
Ma b ab
+= +
+
LT7: Cho a,b,c>0 v 2 2 2a b c + Tm Min2 2
2
2 2 2
1 1( )
b cM a
a b c
+= + +
LT10: Tm Min2
2
2009 2010 1 2011
1
xM
x
+ - +=
-
LT11: Min2
322
( )(2 3)M a
a b b= +
- +vi a>b>0
LT15: CMR Vi a,b,c>0 th3 3 2
( 2 ) ( 2 ) ( 2 ) 3
a b c a b c
b c a c a b a b c
+ ++ +
+ + +
LT16: Cho x,y,z>0 v xy+yz+zx=3 Tm Min1 4
z ( )( )( )M
xy x y y z z x= +
+ + +
LT17: Tm Max 2 3(1 )M x x= - vi 0
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Tc gi tham kho qu nhiu ti liu nn khng th nhr. Mi thng tin ng gp
xin gi [email protected]
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