pp cmbdt dua ve mot bien_kinhhoa.pdf

Phng php a v mt bin trong bi ton bt ng thc.

.K _ Xc - 1 -

A. l DO CHN TITrang b nhng tri thc c bn ,cn thit ,tin tin nht c bit l

nhng tri thc phng php v pht trin tr tu cho hc sinh l cc mctiu c t ln hng u trong cc mc tiu dy hc mn ton.

Bt ng thc l mt vn c gio vin v hc sinh thm nhpvi mt lng thi gian kh nhiu v y l vn c th pht trin khnng t duy ton hc cho hc sinh.

Th nhng quavic tm hiu vn ny trong qu trnh dy hc ti thy mc d

c rt nhiu phng php gi i cho nhng bi ton bt ng thc in hnhc th c nhiu dng. C nhng bi ton bt ng thc kh khi bi dnghc sinh kh gii vic s dng nhng phng php c gp nhiu khkhn, v th vi hng suy ngh khc phc nhng hn ch v phng phpgii c trc ti tm kim thm c mt phng php tin li gii quyt nhng bi ton kh v cng khi dy tr tm ti ca hc sinhv gio vin trong qu trnh t hc, khi dy lng say m tm kim nhngci mi.

V nhng l do .Di y ti xin c trao i vi qu ngnghip mt phng php gii cho nhng bi ton bt ng thc ( Thngl nhng bi bt ng thc kh, xy ra trong cc k thi hc si nh gii, thii hc). V trong mt s bi ti khai thc su thm bng nhng hotng tr tu nh tng qut, phn tch, so snh, c bit ha. ..

Ni dung ti gm ba phn :Phn I: mt bin l n ph t=h(x,y,z,...)Phn II: Mt bin l x(y hoc z)Phn III: Khai thc phng php trong lng gic

b.ni dung ti*/ Bi ton: Xt bi ton : vi iu kin R (nu c) . Chng minh rngp=f(x,y,z,...) A (hoc A)

phng php gii: Chng minh p )(tg vi Dt Chng minh Atg )( vi t D Vn t ra l nh gi biu thc p a v biu thc mt bin g(t) v chngminh Atg )(

- Vic chng minh Atg )( y ti ch s dng cch bin i ( d ondu bng xy ra),ngoi ra i vi hoc sinh lp 12 c th lm mt cch nhanhchng hn bng cch s dng o hm lp bng bin thin gii.

- Cn nh gi p ni chung l phong ph ty thuc tng bi ton lachn cch nh gi thch hp (dng cch bin i , s dn g bt ng thc c inbunhiacopki,csi,....) .

http://kinhhoa.violet.vn


.K _ Xc - 2 -

*/ kin thc b sung1.Bt ng thc c bn :a.Bt ng thc csi:cho )2(,...,, 21 nxxx n s khng m khi :

nnn xxxnxxx ...... 2121 ng thc xy ra khi v ch khi nxxx ...21

b. Bt ng thc bunhiacopxki:2

221122

22

122

22

1 )...()...)(...( nnnn yxyxyxyyyxxx ng thc xy ra khi v ch khi

n

n

yx

yx

yx ...

2

2

1

1

c. Bt ng thc svac-x(h qu ca bt ng thc bunhiacopxki ) :vi )2(,...,, 21 nyyy n l s dng:

n

n

n

n

yyyxxx

yx

yx

yx

...

)...(...

21

221

2

2

22

1

21

ng thc xy ra khi v ch khi :n

n

yx

yx

yx ...

2

2

1

1

2.Tnh cht:a. Nu p c gi tri khng i khi ta hon v vng quanh cc bin x,y,z..chng hn p=f(x,y,z)=f(y,z,x)=f(z,x,y) .khi khng mt tnh tng qut ta c th gi s x=max(x,y,z,...) hocx=min(x,y,z,...)b. Nu p c gi tr khng i khi ta hon v mt cch bt k cc bin x,y,z...chng hn p=f(x,y,z)=f(x,z,y)=f(y,x,z)=f(y,z,x)=f(z,x,y)=f(z,y,x) .khi khng mt tnh tng qut ta c th sp xp cc bin theo mt th t

.... zyxI. mt bin l n ph t=h(x,y,z,...). Sau y l mt s v d m uBi ton 1:Vi x,y l s dng chng minh rng:

2233 yxxyyx (1)Gii:V x l s dng nn:

(1) x

yx

yx

y

23

1 . tx

y =t th t>0(1) tr thnh t 3 -t 2 -t+10 (t-1) 2 (t+1)0 (ng vi mi t>0) pcm

Tng qut ta c bi ton sau:Cho x,y l s dng. Cmr: ),2(11 Nnnyxxyyx nnnn

Chng minh hon hon tng t!Bi ton 2: Vi x,y khc khng chng minh rng:


.K _ Xc - 3 -

)2(22

2

2

2

4

4

4

4

x

y

yx

x

y

y

x

x

y

y

x

Gii:t t=

x

yyx th 2

x

yyx

x

yyx

t (p dng bt csi)khi (2) tr thnh:

02)2(2)2( 222 ttt (t+2)(t 3 -2t 2 -t+3)0(2')+) Vi t2: ta c t 3 -2t 2 -t+3=(t-2)(t 2 -1)+1>0nn bt ng thc (2') ng+) Vi t -2: ta c t 3 -2t 2 -t+3=(t+2)[(t-2) 2 +3] - 11 > 0v t+20 nn bt ng thc (2') ngvy bt ng thc (2) ng du bng xy ra khi t= -2 hay x=-y pcmBi ton 3:( chn i tuyn d thi HSG ton QG 2006-2007)x,y,z l s thc tha mn 2222 zyx .Tm gi tr ln nht, nh nht ca biuthc xyzzyxP 3333 -Nhn xt : D on du gi tr LN,NN t c kh i x=y=z hoc ti cc imbin.Th vo ta c phn on 2222 PGii: T ng thc 2222 )()(2 zyxzxyzxyzyx

))((3 222333 zxyzxyzyxzyxxyzzyx v iu kin ta c:)

22)(2)(())((

2222 zyxzyxzxyzxyzyxzyxp

t 60 tzyxt2222)22()2(

213

2)

222( 2

32

ttttttpdu bng xy ra khi v ch khi 2tvy Pmin= 22 khi x= 2 ,y=z=0 hoc hon v Pmax= 22 khi x= 2 ,y=z=0 hoc hon vSau y ta xt mt s v d m phi nh gi biu thc P mi thy c nphBi ton 4: Cho

23

0,,

zyx

zyx Cmr:

215111

zyxzyx

Gii: p dng bt ng thc csi ta c:zyx

zyxxyz

zyxzyx

zyx 913111 3

t230 tzyxt


.K _ Xc - 4 -

Vy:2

15

23

.4

2749

.2427

499111

tt

ttt

tt

zyxzyx

du bng xy ra khi v ch khi x=y=z=21

pcmTng qut ta c bi ton: Cho )2(,...,, 21 nxxx n l s dng ;

)(... *21 Rkkxxx n 22;0 bnakb .Cmr :

kakbn

xxxbxxxa

n

n

22

2121 )

1...

11()...( (*)S lc li gii:

kakbn

kbn

akk

bnk

bnat

kt

tbn

t

bnat

xxx

bnxxxa

xxxbxxxa

n

n

n

n

22

2

22

2

2

22

221

2

2121

21

)(1.2.)()1(

...

)...()1...11()...(

Nhn xt1:- T bi ton (*) ta c bit ha1.Vi a=1; b=4 ; n=3 ; k=

23 ta c bi ton :

Cho

23

0,,

zyx

zyx Cmr:

251)111(4

zyxzyx

kt hp bt ng thc bunhiacopxki ta cbi ton 2':(olimpic-ton s cp -i Hc Vinh)

Cho

23

0,,

zyx

zyx C mr: 2 2 22 2 21 1 1 173. 2x y zy z x

Tht vy : p dng bt ng thc bunhacopxki ta c)4(

17114)41)(1( 222222 yxyxyxyx

tng t sau cng li kt hp bi ton trn ta suy ra iu phi chng minhVi a=1;b=9;n=3;k=1 kt hp bt ng thc bunhiacopxki ta c bi ton

Cho

10,,

zyxzyx CMR : 82111 222222 zzyyxx

( thi i hc ,cao ng nm 2003 -2004)2.vi a= -1; b=1 ; n=2 ; k= 2 ta c bi ton :Cho

20,

yxyx Cmr: 2)(11 yx

yx

bng cch thay i gi thit , t n ph ta c bi ton 2'':


.K _ Xc - 5 -

cho

10,

yxyx Cmr: 2

11 y

yx

x

Tht vy: bng cch t: a= x1 ; b= y1 v kt hp bt ng thcbunhacopxki v bi ton trn ta suy ra iu phi chng minhTng qut: (tp ch crux )

)2(,...,, 21 nxxx n l s dng v mxxx n ...21 ,m>0:Cmr::

1...

2

2

1

1

nmn

xm

x

xm

x

xm

x

n

n

Chng minh hon ton tng t !- Nu i chiu ca bt ng thc iu kin (bi ton (*))th bi ton thay inh th no?Tr li cu hi ny ta c bi ton mi : Cho )2(,...,, 21 nxxx n l s dng;

)(... *21 Rkkxxx n 22;0 bnakb .Cmr :

kakbn

xxxbxxxa

n

n

22

2121 )

1...

11()...( (**)t bi ton (**) ta c th khai thc ta c nhng bi ton mi kh th v ...*)Nh vy khi lm mt bi ton ta c th dng hot ng tr tu khai thc subi ton_ trn c mt chu k hot ng kh hay l :bi ton c th tngqutc bit (phn tch , so snh...)bi ton mi tng qut.(ch tng qut c nhiu hng :theo hng s ,theo s bin hoc s m) Bi ton 5:(THTT/ T4/352/2007) Vi x,y,z l s dng v xyz 1Cmr:

23

xyzz

xzy

y

yzx

x (5)

Gii:t a= x , b= y , c= zBi ton tr thnh : a,b,cl s dng v abc 1 cmr

23

2

2

2

2

2

2

abc

c

acbb

bcaa (4')

p dng bt ng thc svac-x ta c:VT 2 (5')

abcacbbcacba

222

2)( 2 = 2222

4)(abcacbbca

cba

]3)[(3)(

)](3)[(3)(

)(3)(

2

4

2

4

222

4

cbacba

cabcabcbacba

cabcabcbacba

{v ab+bc+ca 3 2)(3 abc 3}t t=(a+b+c) 2 th t9 { v a+b+c 33 abc 3}


.K _ Xc - 6 -

ta c )3(32

tt =

33

.

1232

12159.3

33

123

12153

t

t

t

tt =29

VT 2 (5') 29 VT(4')

23

du bng xy ra khi x=y=z=1 pcm

Tng qut ta c bi ton sau:vi )2(,...,, 21 nxxx n dng v 1...21 nxxx

Cmr:2

...

...

..... 1211432

2

321

1 n

xxxx

x

xxxxx

x

xxxx

x

nn

n

nn

Bi ton 6: Cho

10,,

zyxzyx Cmr:

109

111 222 z

z

yy

x

xP

Nhn xt: Ta ngh n p dng bt svac-x nhng y chiu ca bt ngthc li ngc.Mt ngh ny sinh l bin i P lm i chiu bt ng thc ?Gii : Ta c :

333

2222

3

4

3

4

3

4

2

3

2

3

2

3

2

2

2

2

2

2

1)(1

)111

(1)1

1()1

1()1

1(

zyxzyxzyx

zz

z

yyy

xx

x

z

z

yy

x

x

z

zz

yyy

x

xxP

t 222 zyxt t k31 t

p dng bt ng thc bunhiacopxki v csi ta c:

321

23)

3(3)](1[

21

3))((3

222222222

222333

ttt

zyxzyxzyx

xyzzxyzxyzyxzyxzyx

Vy

109

109

3103

)957)(31(

109

109

31033103

313

21

3231

21 222

2

22

tt

tt

tt

ttt

tt

t

ttt

tP

du bng xy khi v ch khi x=y=z=31

pcmKhi gp bi ton c iu kin phc tp kh s dng th phi x l iu kin . Taxt bi ton sau:Bi ton 7:(Tp ch ton hc tui th)Cho

)1)(1)(1)(1()1;0(,,

zyxxyzzyx Cmr: x 2 +y 2 +z 2

43

Gii:


.K _ Xc - 7 -

(1) 1-(x+y+z)+xy+yz+zx=2xyz x 2 +y 2 +z 2 =2-2(x+y+z)+(x+y+z) 2 -4xyzp dng bt Csi ta c : xyzzyx

3

3 nn

x 2 +y 2 +z 2 2-2(x+y+z)+(x+y+z) 2 -43

3

zyx

t t=x+y+z th 30 t .Khi :x 2 +y 2 +z 2 3 2 24 1 15 3 32 2 (2 3) ( )

27 27 4 4 4t t t t t

du bng xy ra khi t=23 hay x=y=z=

21

pcmNhn xt 2 : T tng phng php gii trn ta c th sng to cc btng thc :chng hn -T bt ng thc c si1.C ho x,y l s dng.Cmr: xyyxyx 888)( 22322 2.(THTT-248 - 1998):Cho x,y,z l s dng khng ln hn 1. Cmr:a.

3)1)(1)(1(

311 zyx

zyx

b. )1)(1)(1(311

zyxzyx

T ta c bi ton tng qut : (ch : cu b cht hn cu a)Cho )2(,...,, 21 nxxx n l s dng khng ln hn : Cmr:

))...()((...

2121

1

n

n

n

n

xaxaxan

a

xxx

a

Hd: p dng bt ng thc csi ta c:n

nn

n

xxxnaxaxaxa

)...())...()(( 2121

bt ng thc tr thnh:1 1 1

1( ) 0(*)

nn n n n n

n

a a na t na t n a t na t

t n n n tn

p dng bt ng thc csi ta c:

nnnnn

n

antnatann

nantnatnatnatn 11)()1()1())...()(()1(

kt hp iu kin bi ton nn bt ng thc (*) ngngoi ra t cch chng minh ta c bt ng thc cht hn sau:

Cho )2(,...,, 21 nxxx n l s dng khng ln hn a .Cmr:))...()((1

...

121

1

21

11

n

nn

n

nn

xaxaxan

a

n

n

xxx

a

n

n


.K _ Xc - 8 -

chng minh hon ton tng t !3.Cho

30,,

222 zyxzyx Cmr: 3027 xyzzyx

4.Cho

20,,zyxxyz

zyx Cmr: 6 zyx- T bt ng thc bunhiacsxki, svac -x v ng thc

2222 )()(2 zyxzxyzxyzyx 1. Cho x,y,z nm trong on [1;2] .Cmr : 6)(0 zyxzxyzxy2. Cho

1

0,,xyz

zyx Cmr:3

101222 zxyzxyzyx

3. Cho

0,,1

zyxxyz Cmr: 43

222 zyxx

z

z

y

yx

4. Cho

21

0,,

zyx

zyx Cmr:5

108111222 zzyyxx

5.(THTT- 346/2006) Cho

0,,

1zyx

zyx Cmr:))((8 222222222 xzzyyxzyxzxyzxy

6. Cho

]2;1[,, zyxzyxzxyzxy

Cmr:43

)(4)(4)(4 22

2

2

2

2

yx

z

xz

y

zy

x

- Hay t bt ng thc schur :2)(9)(40))(())(())(( zyx

zyx

xyzzxyzxyyzxzzxyzyyzxyxx

1: Cho xyz l s khng m . Cmr: )(212 222 zxyzxyzyxxyz s lc li gii: Bt ng thc ca bi ton tng ng vi

12)()(4)(412)( 22 xyzzyxzxyzxyhayzxyzxyxyzzyxkt hp bt ng thc trn v bt ng thc csi ta cn chng minh:

127

)29( 2 tt vi zyxt 29

, t { cn29t hin nhin ng}

Bng cch thm bt cc biu thc vo ta c nhiu bi ton khc nhauChng hn:

zyxtctbxyzt

axyzctbxyzzyxzxyzxya ;9])()(4[ 2

ta c: Chn a,b sao cho:

cbacba

20,, th:


.K _ Xc - 9 -

)(233)()( 222 zxyzxyacbazyxcbxyzzyxa vi a=3 b=5 c=1 ta c bi ton:2.Cho x,y,z l s khng m . chng minh rng :

)(61)(5)(3 222 zxyzxyzyxxyzzyx Bng cch tng t ta c bi ton:3.Cho x,y,z l s dng chng minh rng

)(58)(2 222 zyxzyxxyz (THTT-s 356)4.Cho x,y,z l s dng chng minh rng

)1)(1)(1(32222 zyxxyzzyx

5.Cho

]34

;0[,,3

zyx

zxyzxy Cmr: 13)(4 zyxxyz

T ng thc ,bt ng thc c bn,n gin ta c th to v s bi ton! kt thc phn I ti xin a ra thm mt s bi ton lm theo phng phpny:

*--------------Mt s bi ton----------*I1.Chng minh rng: 4444 2

2721

1227

21

yx

yx vi mi x,y thuc RHD: yxt I2.Cho

)2;0(,,3

zyxzyx Cmr:

222222 4

1

4

1

4

1

)2)(2)(2(27

zyxzyx

HD: t = 2)( zyx :I3.Cho

1

0,,zyx

zyx Cmr :121)()()( 444 yxzxzyzyx

HD: Gi s 0 zyx t )( zyxt ta chng minh c)31()()()( 444 ttyxzxzyzyx

I4 . Cho

0,,4222

zyxxyzzyx Cmr: 3 zyx

I5. Cho

]1;0(,, zyxzyxzxyzxy Cmr:

3)()()( 2

2

2

2

2

2

zyx

z

yxz

y

xzy

x

I6. Cho ),2(,...,, 21 Nnnxxx n l s dng v )0(...21 knkxxx n .Chng minh rng: )(

1...

11 322

222

11 knkn

xxxxxx nn


.K _ Xc - 10 -

I7 . Vi )2(,...,, 21 nxxx n dng v 1...21 nxxx . Cmr:

n

n

xxxxxx

n

n

1...

1...

2

2121

Nhn xt 3:- Nu chng minh g(t) 0 bng cch bin i nh trn th trc tin phi d onc du bng xy ra ti u nh gi hay tch nhm hp l .-Khi t n ph th phi tm iu kin st ca n ph c bit l chng minhg(t)0 bng phng php o hm.II. Mt bin l x(y hoc z): v d trn th chng ta phi lm xut hin n ph.sau y ta xt mt lp

bi ton m n ph chnh l x hoc y hoc z1.a v mt bin nh iu kin :Bi ton 8: Cho

0,,

1zyx

zyx Cmr:278 xyzzxyzxy

Gii:T k bi ton ta thy 0110 zzp dng bt csi ta c:xy+yz+zx-xyz=z(x+y)+xy(1-z)z(x+y)+

2

2

yx (1-z)

xy+yz+zx-xyz=z(1-z)+2

21

z (1-z)=

4123 zzz =

278

278)

35()

31(

41 2 zz vi mi z, 10 z

du bng xy ra khi x=y=z=31 pcm

Bi ton s 9: Cho

0,,

3zyx

zyx Cmr: )9()(25 zxyzxyxyz Gii: Khng mt tnh tng qut gi s z=min(x,y,z)T iu kin d thy 10 z

04

)2()1(04

230)3(2)2()2

3(5

0)(2)2()2

(50)(2)2(5)9(23

2

2

zzzz

zzzz

yxzzyxyxzzxy

ng vi ]1;0[z . Du bng xy ra khi x=y=z=1pcmNhn xt4:- Nu ly iu kin 30 z th bt ng thc nh gi biu thc trn l khngng. y chng ta s dng tnh cht 1 lm hn ch iu kin ca bin c th nh gi c biu thc .


.K _ Xc - 11 -

- Ta c bi ton Tng qut ca bi 9 sau:

bi ton 9' cho

34

0;00,,3

ba

bazyxzyx

Cmr: 0)3()( babxyzzxyzxya

HD: Khng mt tnh tng qut gi s z=min(x,y,z) t iu kin d thy 043;010

ba

zbzaz ta c:

0)43()1(41)3()3(

)(4

)3()3()()()3()(

2

2

ba

zzbbazaz

bzazbayxazbzaxybabxyzzxyzxya

Ch : Nu 3ba th vic chng minh bi ton tng qut khng cn s dng

tnh cht 1Thay i hnh thc bi ton:- S dng ng thc 2222 )()(2 zyxzxyzxyzyx ta c th a bi

ton trn v bi ton tng ng nhng hnh thc khc :chng hn bi 9 c th pht biu di dng tng ng :Cho

0,,

3zyx

zyx (THTT-2006) Cmr: 4222 xyzzyx hay s dng ng thc

)](3))[((3 2333 zxyzxyzyxzyxxyzzyx bi ton 9 c th c pht biu di dng :Cho

0,,

3zyx

zyx Cmr: 93)(2 333 xyzzyx

- t n ph : a=mx;b=my;c=mz hoc a=x

1 ;b=y1 ;c=

z

1 ..v..v..chng hn: bi ton 9 c th pht biu di dng tng ng

Cho

0,,

1zyx

zyx Cmr: )(18275 zxyzxyxyz

Cho

0,, zyxzxyzxyxyz Cmr: )(18275 222222 zyxxyzzyx

-S dng tnh cht bc cu v bt ng thc c:chng hn bi 9: T bt ng thc csi: xyzzyx 3333 ta c bi ton

0,,

3zyx

zyx Cmr: )(615333 zxyzxyzyx *)T cch chng minh bi ton tng qut trn ta c bi ton Tng t


.K _ Xc - 12 -

bi ton9'' Cho

32

0;00,,

3

ba

bazyx

zyx

chng minh rng:

0)3()( babxyzzxyzxyaCh : chng minh : s dng tnh cht 1 vi z=max(x,y,z)

c bit ha ta c bi ton: Vi a=1; b=-2 : Cho

0,,

3zyx

zyx Cmr: 12 xyzzxyzxySau y ta xt tip mt s bi ton s dng tnh cht ny lm hn ch

phm vi ca bin:Bi ton 10: Cho

3]2;0[,,

zyxzyx Cmr: 9333 zyx

Gii:Khng mt tnh tng qut , gi s z=max(x,y,z)T iu kin 21 z . Ta c:

333 zyx x 3+y 3+3xy(x+y) +z 3=(x+y) 3+z 3=(3-z) 3+z 3==9z 3 -27z+27=9(z-1)(z-2)+99 vi mi z,1 z2du bng xy ra khi (x,y,z)=(0,1,2) v hon v ca npcmBi ton 11 : ( thi ton quc gia _bng B_1996;USAMO_2001)

Cho

40,,xyzzxyzxy

zyx Cmr: x+y+zxy+yz+zx(11)Gii: Gi s z=min(x,y,z) , t iu kin ta c :

100)2)(1(34 223 zzzzzxyzzxyzxy (11')xy+yz+zx+xyz=4 (x+y)z=4-xyz-xyx+y=

z

xyxyz 4 (11'')Mt khc : 0=xy(1+z)+z(x+y)-4xy(1+z)+2 xy .z-4

12

0)2)(1

2( zxyxyzxy (11''')(11) 0)1)(( xyzzyxT (11'),(11''),(11''') ta c :


.K _ Xc - 13 -

0)1()1(1

2)(1

244

)(44)1(4)1)((

2

22

22

2

22

z

zz

z

zz

zzz

z

z

zxyzzxyzxyzz

z

xyxyzxyzzyx

du bng xy ra khi x=y=z=1pcmBi ton 12: Cho

0,,

3zyx

zyx Cmr: 7)(2 222222 zyxzyxGii:Ta c 0)]1)(1)(1[()]1)(1)][(1)(1)][(1)(1[( 2 zyxxzzyyxDo trong ba s )1)(1();1)(1();1)(1( xzzyyx c t nht mt s khngm . Gi s 10)1)(1( yxxyyxTa c:

77)22()1(9674)2(2)3()1(2)()(2

22234

22222222222222

zzzzzzz

zzzzzyxzyxzyxzyx

du bng xy ra khi v ch khi

10)22()1(

0)1)(1(3

22

zyxzzz

yxzyx

pcm Bng cch s dng tnh cht trn ta c th to ra cc bi ton michng hn: cho

]4;21[,,1

zyx

xyz Cmr:

417 zxyzxy

*..............Mt s bi ton..................*II11. Cho

0,,

1zyx

zyx Cmr:a. xyzzy 16b. xyzzxyzxy 9c. )(419 zxyzxyxyz II12. Cho

3

0,,zxyzxy

zyx Cmr: 10)(3 xyzzyx

II13. Cho ]22

;0[, yx . Cmr:3

2211 22

xy

yx

HD: Gi s 022 yx ta i chng minh: 222 1

211 x

x

x

yy

x


.K _ Xc - 14 -

II14. Cho

0,,

1zyx

zyx Cmr:

a.27

11

11

11

2

2

2

2

2

2

x

z

z

yyx (bi T5 - THTT - 10/2004)

b.27

11

11

11

n

n

n

n

n

n

x

z

z

yyx

HD:Gi s x=max(x,y,z)

1

142

1

1)(3

1

13

1

11

11

1

1

1

1

1

1

2

2

2

2

2

22

2

22

2

2

2

2

2

2

xxx

xzy

xzy

x

zy

x

z

z

y

y

x

Cu b tng t!II15. Cho

3]2;0[,,

zyxzyx Cmr : 12 nnnn zyx

(Tng qut bi 8: chng minh tng t!)2. a dn v mt bin:T biu thc p c n bin ta nh gi a v (n -1) bin .... v cui cng a v 1bin. sau y ta xt mt s v d c trng th hin phng php ny:Bi ton 13: Cho x,y,z nm trong on [1;2] Chng minh rng : xyzzyx 5333 Gii:t ),,( zyxf xyzzyx 5333 Khng mt tnh tng qut gi s : 12 zyx

0)51)(1()51(5)1,,(),,( 23 xyzzzxyxyzzyxfzyxfV : 013)1(4415151;01 22222 zzzzzzzxyzzzMt khc : 0)51)(1()51(5)1,1,()1,,( 23 xyyyxxyyxfyxfV 01)2)(1(145151;01 222 yyyyyyyyxyyyVy 21,0)2)1)[(2(25)1,1,(),,( 23 xxxxxxxfzyxfdu bng bt ng thc xy ra khi v ch khi (x,y,z)=(2,1,1) v hon v ca(2,1,1) pcmBi ton14:(y l bi ton s 9) Cho

0,,

3zyx

zyx

Chng minh rng: )(25 zxyzxyxyz Giit xyzzxyzxyzyxp )(2),,(Ta cn chng minh 5),,( zyxf . Do vai tr ca x,y,z trong f nh nhau nn theotnh cht 2 ta gi s zyx 0 kt hp iu kin ta d dng suy ra 10 xXt


.K _ Xc - 15 -

423)

23

,

23

,()2

,

2,(),,(0))(2(

41

4)()

24)(

2(2)(2)

2,

2,(),,(

32

22

xxxx

xfzyzyxfzyxfzyx

zyx

zyx

zyzyxxyzzxyzxyzyzyxfzyxf

10;54

)2()1(5554

23),,(23

xxxxxxzyxf

du bng xy ra khi v ch khi 11

0))(2( 2

zyxx

zyx

pcmBi ton15: (Bt ng thc csi): Cho x,y,z l s dng

Chng minh rng: xyzzyx 3333 Gii: Khng mt tnh tng qut gi s 0 xyzt ),,( zyxf xyzzyx 3333 Tac:

0)2)(()(3)(),,(),,( 233 xyxyzzxyzzxyxyxyzxyyxfzyxf v xyz Mt khc: t ),( yxg 333 )(2),,( xyyxxyyxf 0))((2),(),( 2336333 xyxxyxyxxgyxgVy 0),(),(),,(),,( xxgyxgxyyxfzyxfdu bng xy ra khi v ch khi zyx

yxxyz

pcm

Bi s 16:(Bt ng thc nesbit) Cho x,y,z l s dng .Chng minh rng :

23 yx

z

xz

yzy

x

Gii:t ),,( zyxf

yxz

xz

yzy

x

. Gi s z=min(x,y,z)

Ta c : ),,(),,( xyyxfzyxfyx

xyxxy

yxyy

x

yxz

xz

yzy

x

0)(

.

))((11

)()(1

1))(())((

1))((

))(()(

))(()(

))(()(

2

xyyx

yxyxyxxy

yxyx

yx

yxxxyxyy

xyyxyx

yxxxyxzy

xyyzyx

zxyyxxyz

xxyxzzxyy

xyyzyzxyx


.K _ Xc - 16 -

Ta c :

111),,(

x

yx

y

x

yx

y

x

yx

yyxxy

xxyy

yxyx

xyyxf

23

23

)1(2221

111

2

22

2

2

2 tt

ttt

t

t

t

t

tt vi )0( t

x

yt

du bng xy ra khi v ch khi : x=y=z=1pcm

Nhn xt :- Khi a biu thc 3 bin v 2 bin hay 1 bin thng xt hiu biu thc ca btng thc v biu thc vi x(hoc y hoc z) thay bi trung bnh nhn hoctrung bnh cng .- Thng ta phi s dng tnh cht 2 mi c nh gi c

* ............Mt s bi ton............*II21. Cho

0,,1

zyxxyz Cmr : )1(4))()(( zyxxzzyyx

II22. Cho

0,,96

zyxxyzzxyzxy Cmr: 63 xyzzyx

II23. Cho

30,,

222 zyxzyx Cmr:

a. xyzzxyzxy 9b. )(419 zxyzxyxyz II24. Cho ]2

2;0[, yx chng minh rng:

322

11 22 x

yy

x

II25. Cho ]3;31[,, zyx chng minh rng:

57 xz

z

zyy

yxx (THTT-s 357)

II26. Cho x,y,z l s dng chng minh rng:)(58)(2 222 zyxzyxxyz (THTT-s 356)

II27. Cho

30,,

zxyzxyzyx Cmr: 10)(3 xyzzyx

II28. Cho

30,,

222 zyxzyx Cmr: 222222 xzzyyxzyx

II29. Cho

30,,

222 zyxzyx Cmr: xyzzxyzxy 912)(7

II20 Chng minh rng :


.K _ Xc - 17 -

2

221

yxz

x

zyz

yx (OLIMPIC 30-4)

HD: Khng mt tnh tng qut ta gi s : 121 xyz

t : z=ax ; y=bx 121 ba sau nh gi tip ta a v 1bin l b .

III. khai thc phng php trong lng gic: trn l nhng bt ng thc trong i s . vy trong lng gic liu c th

nh gi c khng? sau y ta xt mt s v d trong l ng gicBi ton17:Chng minh rng trong mi tam gic ABC ta u c:sinA+sinB+ 3 sinC 6

34 (17)

Gii:CCCBABACBA sin3

2cos2sin3

2cos

2sin2sin3sinsin)17(

t 012

cos ttC

Ta c: ))1(31(2)2

sin31(2

cos2sin32

cos2 2ttCCCC p dng bt csi :

)'17(6346

34)623()

36(

63)]131(

231[2))1(

312.

231(2))1(31(2

2

3222

tt

tttttttt

(17') ng vi mi t>0 ; v vy: sinA+sinB+ 3 sinC 634

du bng xy ra khi v ch khi

2cos36

2cos

12

cos

CBA

C

BA

pcmBi ton18: Cho tam gic ABC chng minh rng:(1-cosA)(1-cosB)(1-cosC)cosAcosBcosC (18)Gii:+) Nu tam gic c gc vung hoc gc t th bt lun ng+) Nu tam gic l nhn ,ta c:

1coscos

coscos)cos(cos1.

cos

cos1)18( CB

sCBCBA

A


.K _ Xc - 18 -

A

AA

A

A

AAVT

CBCB

CBCB

AA

cos

2sin2

2sin42

1)cos1(

21

2sin21

cos

cos1)'18(

)'18(11)cos()[cos(

21

2cos

2cos21

cos

cos1

2

Ta c: 0cos

)2

sin21(0

cos2

sin42

sin411

cos2

sin22

sin42 222

A

A

A

AA

A

AA

(18'')V tam gic nhn nn (18'') lun ng.Do (18) ng

du bng xy ra khi v ch khi CBAACB

02

sin1

12

cos

pcm Trong tam gic ABC ta c iu kin l A+B+C=180 nn gi cho chng ta s dng tnh cht 1 lm hn ch phm vi bin t c th nh gi c biu thc ,

Sau y l mt s v d:Bi ton 19: Chng minh rng trong mi tam gic ABC ta u c

3(cosA+cosB+cosC) 2(sinAsinB+sinBsinC+sinCsinA) (19)Gii:Khi hon v (A,B,C) th bt (19) khng thay i do khng mt tnh tng qutGi s A=min(A,B,C. V A+B+C=180 3A nn 1

2cos

23600 AA

)cos(]2

cossin42

sin6[2

coscos2

2cos

2sin2.sin2)cos()cos()

2cos

2cos2(cos3

)sinsinsinsinsin(sin2)coscos(cos3

CBAAACBA

CBCBACBCBCBCBA

ACCBBACBAT

Ta c: v 0)2

cos43(2

sin22

cossin42

sin612

cos23

,02

sin 2 AAAAAAA

Mt khc 1)cos(,12

cos CBCB nn

00)12()12(1248

1)2

sin1(2

sin82

sin6)2

sin21(212

cossin42

sin6cos2223

22

tttttt

AAAAAAAAT


.K _ Xc - 19 -

trong 2

sin At .V vy (cosA+cosB+cosC) 2(sinAsinB+sinBsinC+sinCsinA)

du bng xy ra khi v ch khi CBAA

CB

CB

12

sin

1)cos(1

2cos

pcmBi ton20: Chng minh rng trong mi tam gic ABC ta u c

1+cosAcosBcosC 3 sinAsinBsinC (20)Gii: Khi hon v (A,B,C) th bt (20) khng thay i do khng mt tnh tngqut . ta gi s A=max(A,B,C) Khi 60 (20 ')A . Xt

]sin23

cos21)[cos(]cossin3[cos

211

)]cos()[cos(sin23)]cos()[cos(cos

211

sinsinsin3coscoscos1

2 AACBAAA

CBCBACBCBAT

CBACBAT

T (20') ta c: 0)60cos(sin23

cos21 AAA v 1)cos( CB nn

0)]60cos(1)[1(cos)sin3(cos21]cossin3cos

21[1 2 AAAAAAAT

V vy 1+cosAcosBcosC 3 sinAsinBsinCdu bng xy ra khi CBA

ACB

1)60cos(1)cos(

pcm

*................. Mt s bi ton.................*Chng minh rng trong mi tam gic ABC ta u c:III31.Nu tam gic ABC nhn:

33133

2cot

2cot

2cot

1tantantan

CBACBA

III32. 215

2sin

1

2sin

1

2sin

12

sin2

sin2

sin CBA

CBA

III33.1 31 c o s c o s c o s c o s c o s c o s (c o s c o s c o s )2

c o s c o s c o s

A B B C C A A B C

A B C


.K _ Xc - 20 -

III34. 233cos3cos3cos CBA

III35. 4231sin

21

sinsin 222 CBA

III36. 33 2163

2sin

2sin

2sin CBA

III37 . 2cot2cotcot CBAIII38. Nu tam gic ABC khng phi l tam gic t tha. 4)sin1)(sin1)(sin1( 222 CBb.

221

coscoscos

sinsinsin

CBACBA

Nhn xt 5: Ta c th chuyn bt ng thc c iu kin trong i s sanglng gic bng cch:*) T ng thc lng gic c bn : +) T ng thc:

)1(tantantantantantan CBACBA )2(1

2tan

2tan

2tan

2tan

2tan

2tan ACCBBA

kt hp bi ton : II12. Cho

30,,

zxyzxyzyx Cmr: 10)(3 xyzzyx

chng minh bi ny tng t bi ton 11(hoc s dng a dn v mt bin)t (1) bng cch t :

Cz

By

Ax

tan3

;tan

3;

tan3 ta c bi ton tng

ng bi ton II12 : cho tam gic ABC nhn .Cmr: CBAACCBBA tantantan

33101tantantantantantan

tng t ta c : 3302

tan2

tan2

tan)2

tan2

tan2

(tan9 CBACBA

+) T ng thc: 1coscoscos2coscoscos 222 CBACBA vBi ton 11: Cho

40,,xyzzxyzxy

zyx Cmr:x+y+z xy+yz+zxt x=2cosA ; y=2cosB ; z=2cosC ta c bi ton:Chng minh rng vi tam gic nhn ABC th:cosA+cosB+cosC 2(cosAcosB+cosBcosC+cosCcosA)y l bi ton kh p*) T bt ng thc lng gic c bn :

Ta xt bi ton 9:D thy t cch chng minh c th thay iu kin ca bi ton nh sau


.K _ Xc - 21 -

Cho

34

0;00,,3

ba

bazyxzyx

hay

34

0;00,,

3222

ba

bazyx

zyx

Cmr: 0)3()( babxyzzxyzxyac bit ha ta c bi ton :

1. a=-2;b=1 .

0,,

3zyx

zyx Cmr: )(25 zxyzxyxyz

2.a=-4;b=3 .

0,,3222

zyxzyx Cmr: )(439 zxyzxyxyz

Kt hp bt ng thc c bn trong lng gic chng hn1.

23

coscoscos CBA ta c bi ton:Cho tam gic nhn ABC . Chng minh rng:

)coscoscoscoscos(cos8coscoscos85 sACcBBAsCsBA 2.

49

sinsinsin 222 CBA ta c bi ton:Chng minh rng vi mi tam gic ABC th:

)sinsinsinsinsin(sin16sinsinsin839 ACCBBACBA tng t i vi tang,cotang v bi ton khcch : Gii bi ton i s thng qua gii bi lng gic ngi ta gi l phngphp lng gic ha. Lm ngc li gi l phng php i s ha.

C. Kt lunTrn y l mt trch dn v s vn dng phng php a v mt bin

trong vn chng minh bt ng thc. ti ny c bn thn ti v cc ng nghip cng n v th im

trn cc em c hc lc t kh tr ln. Kt qu thu c rt kh quan, cc em hctp mt cch say m hng th. Mt s em t c nhng thnh tch tt quanhng t thi hc sinh gii va qua. V tc dng tch cc trong vic bi dnghc sinh kh gii nn knh mong Hi ng khoa hc v qu thy ( c) gp bsung ti ngy mt hon thin hn, c ng dng rng hn trong qu trnhdy hc trng THPT.


.K _ Xc - 22 -

Xin chn thnh cm n!

ngy 10 thng 5 nm 2007Ngi thc hin

K_Xc

Ti liu tham kho1.Tp ch ton hc v tui tr

2.Sng to bt ng thc _pham kim hng3.Cc phng php chng minh bt ng thc _Trn tun Anh4.Cc bi ton chn lc v h thc lng trong tam gic t gic

phan huy khi_nguyn o phng5.Olimpic 30_4

pp cmbdt dua ve mot bien_kinhhoa.pdf

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