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Dng 1 Da vo nh ngha v cc php bin i tng ng

chuyn bt ng thc - SSC v thc hnh gii ton

Li ni uTrong b mn Ton trng ph thng th chuyn Bt ng thc c xem l mt trong nhng chuyn kh, nhiu hc sinh kh thm ch gii cn lo ngi trnh n bi v hc sinh cha hnh thnh c nhng phng php gii hc sinh ng dng vo vic chng minh Bt ng thc.Qua ni dung v Bi tp ln em xin trnh by chuyn : mt s phng php chng minh Bt ng thc v ng dng ca n trong vic chng minh v gii quyt cc bi ton c lin quan. Cc bi tp y vi kh c nng dn ln nhm gip hc sinh bt lng tng khi gp cc bi ton v chng minh bt ng thc, gip hc sinh c th t nh hng c phng php chng minh v hng th hn khi hc v bt ng thc.

Ni dung ca chuyn bao gm:Phn I - Kin thc c bn cn nm: y l phn tm tt mt s kin thc l thuyt c bn m hc sinh cn nm s dng trong qu trnh chng minh Bt ng thc.Phn II - Cc phng php chng minh bt ng thc: Tng hp cc phng php chng minh Bt ng thc thng dng cho hc sinh THCS. Vi mi phng php c cc kin thc cn nm, cc v d minh ho, bi tp p dng hc sinh t mnh hnh thnh c t duy cm nhn v phng php .Phn III - ng dng ca vic chng minh bt ng thc: Trnh by nhng ng dng ph bin ca chng minh Bt ng thc.Phn IV - Hng dn, gii cc BT p dng: y l phn gii chi tit ca cc BT p dng cho tng phng php chng minh Bt ng thc trn.Phn V - Bi tp tng hp t gii: Bao gm cc bi tp tng hp cho tt c cc dng phng php chng minh Bt ng thc. C s l lun Thc tin

Cc phng php chng minh Bt ng thc th rt phong ph nhng cho hc sinh hnh thnh c phng php chng minh cng nh ng dng Bt ng thc trong Ton hc th cha c. S hc sinh hiu v c im kh ca phn ny rt thp thm ch khng c, a s cc em ch c im Trung Bnh hoc Yu. Ngoi ra, s lng thi gian nghin cu chuyn su phn Bt ng thc trong kin thc ca chng trnh THCS rt t nn hc sinh t thi gian n cc kin thc m gio vin ging trong phn ny. Do hc sinh khng c hng th khi hc sinh bt gp dng ton Bt ng thc ny. Do thi gian nghin cu lm bi ti ngn nn ti khng th a ra c s liu iu tra c th c nhng ti mong rng qua ti ny ti hi vng n s l cng c hu ch cho nhng em c hng th hc tp b mn Ton ni chung v chuyn Bt ng ni ring.Phn I - kin thc c bnI Mt s bt ng thc cn nh:

Bt ng thc C sy:

Vi

du bng xy ra khi

Bt ng thc Bunhiacopski:

Du ng thc xy ra

Bt ng thc Tr- b-sp:

Nu

EMBED Equation.3

Nu

EMBED Equation.3

Du bng xy ra khi

II - Mt s bt ng thc ph c chng minh l ng.

du( = ) khi x = y = 0

III Cc bt ng thc trong tam gicIV Cc hm lng gic thng dngV Cc tnh cht c bn

Tnh cht 1: a > b b < a

Tnh cht 2: a > b v b > c => a > c

Tnh cht 3: a > b a + c > b + c

H qu : a > b a - c > b c

a + c > b a > b c

Tnh cht 4 : a > c v b > d => a + c > b + d

a > b v c < d => a - c > b d

Tnh cht 5 : a > b v c > 0 => ac > bd

a > b v c < 0 => ac < bd

Tnh cht 6 : a > b > 0 ; c > d > 0 => ac > bd

a > b > 0 => an > bna > b an > bn vi n l .

VI Cc hng ng thc ng nh

VII Cc kin thc v to vec t

VIII Cc kin thc v tnh cht ca t l thc:

Phn II Cc phng php chng minh Bt ng thc

Cc phng php chng minh Bt ng thc v cng a dng y ti xin trnh by nhng dng phng php thng dng nht nh sau:

Dng 1 Da vo nh ngha v cc php bin i tng ng

Dng 2 S dng bt ng thc Bunhiacopxky v cc bt ng thc ph.Dng 3 S dng Bt ng thc Cauchy

Dng 4 Chng minh bng phn chng

Dng 5 Phng php lng gic

Dng 6 Phng php chng minh qui npDng 7 Phng php p dng cc tnh cht ca cc dy t s bng nhau

Dng 8 Phng php dng tam thc bc hai

Dng 9 Phng php dng tnh cht bc cu

Dng 10 - Phng php dng cc bt ng thc trong tam gic

Dng 11 Phng php i bin s

Dng 12 Phng php lm tri (chng minh bt ng thc c n s hng)Ngoi cc phng php chng minh Bt ng thc nu trn th cn rt nhiu cc phng php khc nh: Phng php to vect, bt ng thc cha du gi tr tuyt i, s dng cc tr, Nhng do cc kin thc l thuyt cc em cha c nn ti ch xin trnh by mt s phng php nh trn.

Dng 1- Da vo nh ngha v cc php bin i tng tng ng

y l phng php c bn nht, da vo cc tnh cht c bn ca bt ng thc n gin bin i cc bt ng thc phc tp ca ra thnh cc bt ng thc n gin v ng hoc cc bt ng thc c chng minh l ng. phn ny cc bn ch n cc hng ng thc:

Phng php:

Khi bin i tng ng ta c gng lm xut hin cc iu kin cho trong gi thit nhm p dng c iu kin ca gi thit chng minh c bt ng thc l ng. Chuyn v chng minh bt ng thc (

Chuyn v cc tha s v dng hng ng thc d chng minh

Lm xut hin cc tch cc tha s c cha cc yu t ca bi ta xt du cc tha s

Chia nh tng v chng minh sau cng v theo v cc bt ng thc con c iu phi chng minh.

Mt s v d:V d 1:

Chng t rng vi th:

Gii

Bt ng thc lun ng v .

V d 2:

Cho Chng minh rng:

Gii

EMBED Equation.DSMT4

V .

Vy

V d 3:

Vi chng minh:

Gii

Hin nhin ng.

Vy .V d 4: Chng minh rng mi a,b,c,d th :

Gii

EMBED Equation.DSMT4 Vy :

V d 5: Chng minh rng nu: th (1)

Gii

Suy ra iu phi chng minh.

V:

EMBED Equation.DSMT4 Bi tp p dung:

Bi 1: Cho a + b = 2. Chng minh rng:

Bi 2:Chng minh rng vi mi s nguyn dng n ta c:

Bi 3: Chng minh (m,n,p,q ta u c

m+ n+ p+ q+1( m(n + p + q +1) Bi 4: Chng minh rng:

Bi 5: Chng minh bt ng thc : Trong : a > 0 , b > 0

Bi 6: Chng minh rng: Vi mi s dng a, b, c, d ta c:

Dng 2 S dng bt ng thc Bunhiacpsky v cc bt ng thc phy l phng php ph bin nht trong vic chng minh Bt ng thc. Chng ta da vo iu kin cho bi ta la chn phng php cho thch hp. Ngoi ra, ta cn phi ch n du ca BT c th s dng bt ng thc no chng minh. Khi p dng cc BT c chng minh l ng th bn nn tch nh BT cn chng minh ra thnh cc v nh sau cng v theo v c BT cn chng minh.Mt s v d:

V d 1:

Chng minh rng vi mi s thc dng x,y,z ta c:

Gii

Do ta c:

Du = xy ra khi x=y=z

V d 2: Chng minh rng:

Gii

Theo bt ng thc Becnuli ta c:

V:

V d 3: Cho Chng minh rng:

Gii

p dng bt ng thc Bunhiacopxki cho 4s 1,1,a,b ta c:

p dng bt ng thc Bunhiacopxki cho 4s 1,1,a2,b2 ta c:

V d 4: Cho a,b,c>0 Chng minh rng:

Gii

Ta c:

V :

Nn:

V d 5: Cho 4 s dng a,b,c,d chng minh rng:

Gii

p dng bt ng thc ph:

Ta c:

Tng t:

Cng v theo v ta c:

Ta chng minh:

Bi tp p dng:Bi 1: Cho x,y,z tho mn

Chng minh rng:

Bi 2: Cho a>b>c>0 v .Chng minh rng

Bi 3: Cho x , y l 2 s thc tho mn x2 + y2 =

Chng minh rng : 3x + 4y 5

Bi 4: Cho a, b, c 0 ; a + b + c = 1 . Chng minh rng:

Bi 5:Cho a, b, c l di 3 cnh mt tam gic, p l na chu vi. Chng minh rng:

Bi 6: Cho a, b,c l 3 s khc 0. Chng minh rng:

Bi 7 Cho ba s .Tho mn

Chng minh rng:

(*)Dng 3 s dng bt ng thc Cauchyy l phng php chng minh BT m hc sinh THCS d nhn dng chng minh l s dng Bt ng thc Cauchy . Ta cn phi ch n du ca BT c th s dng bt ng thc no chng minh. Khi p dng cc BT c chng minh l ng th bn nn tch nh BT cn chng minh ra thnh cc v nh sau cng v theo v c BT cn chng minh.V d 1: Cho 3 s dng a,b,c chng minh rng:

Gii

Vn dng bt ng thc Csi, ta c:

Cng v theo v (1) (2) v (3) ta c:

Vy:

V d 2: Cho a,b,c >0 tho mn

Chng minh rng:

Gii

Ta c:

p dng bt ng thc Csi:

Nhn li ta c:

V d 3: Gi s a,b,c d, l 4 s dng tho mn:

EMBED Equation.DSMT4 Chng minh rng:

Gii

T gi thit ta c:

p dng bt ng thc Csi ta c:

Bi tp p dng:Bi 1: Chng minh rng: ( a+ b + c ) ( + + ) 9 vi a,b,c > 0

Bi 2: Cho a,b,c l di 3 cnh ca mt tam gic vi chu vi 2p

Chng minh rng:

a)

b)

Bi 3: Cho a, b, c 0 ; a + b + c = 1 . Chng minh rng:

Bi 4:Cho a, b, c l di 3 cnh v 2p l chu vi ca mt tam gic.

Chng minh rng:

Dng 4 Chng minh bng phn chngy l phng php chng minh BT da vo cc phng php chng minh phn chng trong Ton hc. chng minh mnh A ng th ta gi s mnh A sai v chng minh rng t mnh A sai ta suy ra mt iu mu thun kt lun A l ng.Mun chng minh bt ng thc ng, ta gi s sai, tc l ng, t chng minh nhng lp lun chnh xc ta suy ra iu mu thun t gi thit. Kt lun ng. iu v l c th l tri vi gi thit, hoc l nhng iu tri ngc nhau , t suy ra ng thc cn chng minh l ng.

Mt s hnh thc chng minh bng phn chng:

Dng mnh o. Ph nh ri suy ra iu tri vi gi thit. Ph nh ri suy ra tri vi iu ng. Ph nh ri suy ra hai iu tri ngc nhau.Mt s v d:

V d 1: Cho v

Chng minh rng t nht mt trong hai bt ng thc sau y l ng

Gii

Gi s hai bt ng thc trn u sai, c ngha ta c :

v

v

V a+b =2cd

Mu thun

Nn s c t nht mt trong hai bt ng thc cho l ng

V d 2: Cho 3 s dng a,b,c nh hn 2. Chng minh rng c t nht mt trong cc bt ng thc sau l sai:

Gii

Gi s cc bt ng thc sau u ng, nhn ba ng thc li ta c

M

Tng t ta c:

Suy ra:

Mu thun

Vy c t nht mt trong cc bt ng thc cho l sai

V d 3: Cho 6 s t nhin khc 0 nh hn 108. Chng minh rng c th chn c 3 trong 6 s , chng hn a,b,c sao cho a 0

1+

EMBED Equation.3 > + b

m 0 < a,b < 1

T ta suy ra 1+

EMBED Equation.DSMT4 Vy <

Tng t ta c:

V (

Cng cc bt ng thc ta c:

V d 4:Cho Chng minh rng:

a.

b.

Gii

a. Ta c: (1)Mt khc:

Suy ra: (2)

T (1) v (2) suy ra

b. Ta chng minh:

Ta c:

V ( )

( theo cu a).

Bi tp p dng:Bi 1:

Cho Chng minh rng:

Bi 2:

Cho tho mn Chng minh rng:

Bi 3:

Cho a, b, c l di 3 cnh ca tam gic c chu vi bng 1. Chng minh rng:

Bi 4:

Cho a, b, c l di 3 cnh ca tam gic c chu vi bng 2. Chng minh rng:

Dng 10 Phng php dng cc bt ng thc trong tam gicy l phng php s dng cc bt ng thc trong tam gic lm cc gi thit chng minh cc bt ng thc.

phng php chng minh ny cc bn nn ch mt s kin thc c bn sau:

Kin thc:

1. Cc bt ng thc trong tam gic:

Vi a, b, c l 3 cnh ca mt tam gic th

Nu

EMBED Equation.DSMT4 th s o ca 3 gc A, B, C cng ng vi bt ng thc trn.2. Cng thc lin quan n tam gic

Mt s v d:V d 1:Cho a, b, c l 3 cnh ca tam gic. Chng minh rng:

Gii

Ta c:

(1)

Kt qu (2) lun ng v trong tam gic ta lun c.

Vy bt ng thc (1) c chng minh.

V d 2:Cho a;b;cl s o ba cnh ca tam gic chng minh rng:

a. a2 + b2 + c2 < 2(ab + bc + ac)b. abc>(a + b - c).(b + c - a)( c + a - b)

Gii

V a, b, c l s o 3 cnh ca tam gic nn ta c

EMBED Equation.3 Cng tng v cc bt ng thc trn ta c:

a2+b2+c2< 2(ab+bc+ac)

Ta c a > (b-c ( ( > 0

b > (a-c ( ( > 0

c > (a-b ( (

Nhn tng v ca ng thc li ta c:

V d 3:Trong ABC c chu vi a + b +c = 2p ( a, b, c l di 3 cnh ).

CMR : + + 2 ( + + )

Gii

Ta c : p - a = > 0 ( v b + c > a bt ng thc tam gic ) Tng t : p - b > 0 ; p- c > 0.

Mt khc : p - a + p - b = 2p - a - b = c

p - b + p - c = a

p - c + p - a = b ta p dng tnh cht ca Bt ng thc: ta c:

+ =

+

+

Cng theo v ca bt ng thc ta c :

+ + 2 ( + + )

Du = xy ra khi ABC uV d 4:

Cho a, b, c , l di ba cnh ca mt tam gic

(a+b-c)(b+c-a)(c+a-b) abc

Gii

Bt ng thc v ba cnh ca tam gic :

T

(a + b c)( a b + c)( b c +a)( b + c a)( c a + b)( c + a b)

(a + b c)2( b + c a)2( c + a b)2

(a + b c)( b + c a )( c + a b) abc

V a, b, c l 3 cnh ca mt tam gic nn

v abc > 0

Bi tp p dng: Cho a, b, c l di 3 cnh ca tam gicBi 1:Chng minh rng:

Nu vi th

Bi 2:

Chng minh rng:

Bi 3: Chng minh rng: Vi mi p, q sao cho p + q = 1 th

Bi 4:

Cho a, b, c l di 3 cnh ca mt tam gic vi a < b < c. Chng minh rng:

Bi 5:

Chng minh rng: vi a, b,c l di 3 cnh ca mt tam gic th ta c:

Dng 11 Phng php i bin sKhi ta gp mt s bt ng thc c bin phc tp th ta c th dng phng php i bin s a cc bt ng thc cn chng minh v dng n gin hn, tc l ta t cc bin mi biu th c cc bin c sao cho bin mi c th gn hn hoc d chng minh hn. Sau khi i bin s ta s dng cc phng php chng minh trn chng minh bt ng thc.

Phng php lng gic cng l mt dng ca phng php i bin s.Mt s v d:

V d 1:Cho a,b,c > 0 Chng minh rng: (1)

Gii

t x=b+c ; y=c+a ;z= a+b ta c a= ; b = ; c =

Ta c (1)

EMBED Equation.3

EMBED Equation.3

(

Bt ng thc cui cng ng v ( ; ) iu phi chng minh.

V d 2:

Cho a,b,c > 0 v a+b+c < 1. Chng minh rng:

(1)

Gii

t x = ; y = ; z =

Ta c

(1) Vi x+y+z < 1 v x ,y,z > 0

Theo bt ng thc Csi ta c:

3.

3. .

EMBED Equation.3 M x+y+z < 1

Vy iu phi chng minh.V d 3:Cho a, b, c l cc s thc dng. Chng minh rng:

Gii

y l v d 1 nhng ta s dng cch i bin khc:Ta t

Bt ng thc

(ng).Vy Bt ng thc c chng minh.

Du = xy ra

V d 4:

Cho cc s thc dng a, b, c sao cho abc = 1

Chng minh rng:

Gii

Do nn ta c th t: vi .Nn bt ng thc ta c th vit li nh sau:

(Ta chng minh c)

Vy BT c chng minh. Du = xy ra

Bi tp p dng:

Bi 1:Chng minh rng ; vi mi s thc x, y ta c bt ng thc:

Bi 2:

Cho x, y, z l cc s dng tho mn: + = 4

CMR: + + 1

(i hc khi A nm 2005)

Bi 3:Cho a, b, c l cc s thc dng tho mn abc=1 .

CMR :

Bi 4:Cho x, y, z >0 tho mn . CMR Bi 5:Cho x, y, z l cc s thc dng tho mn: .

CMR:

Dng 12 - Phng php lm tri (chng minh bt ng thc c n s hng)Dng cc tnh bt ng thc a mt v ca bt ng thc v dng tnh c tng hu hn hoc tch hu hn.Phng php chung tnh tng hu hn : S =

Ta c gng bin i s hng tng qut u v hiu ca hai s hng lin tip nhau

Khi : S =

Phng php chung v tnh tch hu hn:

P =

Bin i cc s hng v thng ca hai s hng lin tip nhau:

=

Khi P =

Mt s v d:

V d 1:

Chng minh rng: Vi n l s nguyn

Gii:

Ta c:

Khi cho k chy t 1 n n ta c

1 > 2

Cng tng v cc bt ng thc trn ta c:

V d 2:

Chng minh rng:

Gii:Ta c

Cho k chy t 2 n n ta c

Vy

V d 3:

Chng minh rng:

Gii:Ta c:

Do :

Vy:

V d 4:

Chng minh rng:

Gii:

Ta c:

Vy: .

Bi tp p dng:Bi 1:

Chng minh BT sau:

a.

b.

Ngoi cc phng php chng minh Bt ng thc nu trn th cn rt nhiu cc phng php khc nh: Phng php to vect, bt ng thc cha du gi tr tuyt i, s dng cc tr, Nhng do cc kin thc l thuyt cc em cha c nn ti ch xin trnh by mt s phng php nh trn. Cc bi ton v Bt ng thc rt a dng v phc tp, cc phng php chng minh nu trn ch l nhng phng php tng i thng dng cho hc sinh THCS c gng nhn nh v lm quen vi cc dng trn p dng phng php thch hp.Phn III ng dng ca Bt ng thc.Bt ng thc c ng dng rng ri nhiu trong vic tm GTLN, GTNN, gii phng trnh v h phng trnh, dng gii phng trnh nghim nguyn v rt nhiu ng dng khc na.I - Dng BT tm GTLN v GTNN

Kin thc : Nu f(x) m th f(x) c gi tr nh nht l m.

Nu f(x) M th f(x) c gi tr ln nht l M.

Ta thng hay p dng cc bt ng thc thng dng nh: Csi, Bunhiacpxki , bt ng thc cha du gi tr tuyt i, kim tra trng hp xy ra du ng thc tm cc tr. Tm cc tr ca mt biu thc c dng l a thc , ta hay s dng phng php bin i tng ng , i bin s , mt s bt ng thc Tm cc tr ca mt biu thc c cha du gi tr tuyt i , ta vn dng cc bt ng thc cha du gi tr tuyt iCh :

Xy ra du '' = '' khi AB 0

Du ''= '' xy ra khi A = 0

Mt s v d:

V d 1:

a. Tm gi tr nh nht ca biu thc

A = (x2 + x)(x2 + x - 4)

b. Tm gi tr nh nht ca biu thc :

B = - x2 - y2 + xy + 2x +2y

Gii

Ta c:

a.A = (x2 + x)(x2 + x - 4) . t : t = x2 + x 2

=> A = (t - 2)(t + 2) = t2 - 4 - 4

Du bng xy ra khi : t = 0 ( x2 + x - 2 = 0

(x - 2)(x + 2) = 0 ( x = -2 ; x = 1

=> min A = - 4 khi x = -2 ; x = 1

b. Tng t

V d 2: Tm gi tr nh nht ca biu thc

a. C =

b. D =

c. E =

Gii

p dng BT :

Du '' = ''xy ra khi AB 0

=> C =

Du '' = '' xy ra khi (2x - 3)(1 - 2x) 0 (

Vy minC = 2 khi

b, Tng t : minD = 9 khi : -3 x 2

c. Tng t: minE = 4 khi : 2 x 3

V d 3:

Cho ba s dng x , y , z tho mn : + + 2

Tm gi tr ln nht ca tch : P = xyz.

Gii

(1 - ) + ( 1 - ) = + 2

Tng t : 2

2

T suy ra : P = xyz

MaxP = khi x = y = z =

V d 4: Cho G =

Tm gi tr ln nht ca G.

Gii

Tp xc nh : x 1 ; y 2 ; z 3

Ta c : G = + +

Theo BT Csi ta c : =>

Tng t : ;

=> G

Vy MaxG = t c khi x = 2 ; y = 2 ; z = 6

Bi tp p dng:Bi 1:

Tm gi tr ln nht ca:

S = xyz.(x+y).(y+z).(z+x) vi x,y,z > 0 v x+y+z =1

Bi 2: Cho xy+yz+zx = 1. Tm gi tr nh nht ca

Bi 3:Tm gi tr nh nht ca biu thc A = x2 2xy + 3y2 2x 10y + 20Bi 4:Tm gi tr ln nht , nh nht ca phn thc D =

Bi 5; Tm s t nhin nh nht bit rng s ny khi chia cho 9 c s d l 5 v khi chia cho 31 c s d l 28.

II Dng BT gii phng trnh v h phng trnh

Nh vo cc tnh cht ca bt ng thc , cc phng php chng minh bt ng thc , ta bin i hai v ( VT , VP ) ca phng trnh sau suy lun ch ra nghim ca phng trnh. Nu VT = VP ti mt hoc mt s gi tr no ca n ( tho mn TX) => phng trnh c nghim.

Nu VT > VP hoc VT < VP ti mi gi tr ca n

phng trnh v nghim

Cn i vi h phng trnh ta dng bt ng thc bin i tng phng trnh ca h , suy lun v kt lun nghim. Bin i mt phng trnh ca h , sau so snh vi phng trnh cn li , lu dng cc bt ng thc quen thuc.Mt s v d:

V d 1:

Gii phng trnh : 13 + 9 = 16x

Gii

iu kin : x 1 (*)

p dng bt ng thc Csi ta c :

13 + 9

= 13.2. + 3.2. 13( x - 1 + ) + 3(x + 1 + ) = 16xDu '' = '' xy ra

( ( x = tho mn (*)

Phng trnh (1) c nghim ( du '' = '' (2) xy ra Vy (1) c nghim x = .

V d 2:

Gii phng trnh : + - x2 + 4x - 6 = 0 (*)

Gii

TX :

(*) ( + = x2 - 4x + 6

VP = (x - 2)2 + 2 2 , du '' = '' xy ra khi x = 2=> vi x = 2 ( tho mn TX ) th VT = VP = 2

=> phng trnh (*) c nghim x = 2

V d 3:

Gii phng trnh : + = x2 - 6x + 13Gii

TX : -2 x 6

VP = (x - 3)2 + 4 4 . Du '' = '' xy ra khi x = 3

VT2 = (.1 + .1)2 (6 - x + x + 2)(1 + 1) = 16

=> VT 4 , du '' = '' xy ra khi = ( x = 2

=> khng c gi tr no ca x VT = VP => Phng trnh v nghim

V d 4:

Gii h phng trnh:

Gii

(1) ( x3 = - 1 - 2(y - 1)2 ( x3 - 1 ( x - 1 . (*)

(2) ( x2 1 ( v 1 + y2 2y) ( -1 x 1 (**)T (*) v (**) => x = -1 . Thay x = -1 vo (2) ta c : y = 1

=> H phng trnh c nghim duy nht : x = -1 ; y = 1V d 5:Gii h phng trnh :

Gii

p dng : BT : du '' = '' xy ra khi a = bTa c :

Mt khc:

T (*) v (**) => x4 + y4

EMBED Equation.DSMT4 Du '' = '' xy ra khi : x = y = z m x + y + z = 1 nn : x = y = z =

Vy h phng trnh c nghim : x = y = z =

Bi tp p dng:Bi 1:Gii h phng trnh: (vi x, y, z > 0)

Bi 2: Gii phng trnh sau:

Bi 3: Gii phng trnh:

Bi 4:Gii h phng trnh sau:

Bi 5: Gii h phng trnh sau:

III Dng BT gii phng trnh nghim nguynMt s v d:

V d 1: Tm cc s nguyn x,y,z tho mn

Gii

V x,y,z l cc s nguyn nn

(*)

M

Cc s x,y,z phi tm l

V d 2:Tm cc cp s nguyn tho mn phng trnh:

(*)

Gii

(*) Vi x < 0 , y < 0 th phng trnh khng c ngha

(*) Vi x > 0 , y > 0

Ta c

t (k nguyn dng v x nguyn dng )

Ta c

Nhng

M gia k v k+1 l hai s nguyn dng lin tip khng tn ti mt s nguyn dng no c nn khng c cp s nguyn dng no tho mn phng trnh

Vy phng trnh c nghim duy nht l :

Bi tp p dng:Bi 1: Tm nghim nguyn dng ca phng trnh:

Phn IV - gii v hng dn gii BT p dngDng 1 Da vo nh ngha v cc php bin i tng ng

Bi 1:

Gii

Ta c:

Tng t:

Vy:

iu phi chng minh.

Bi 2:

Gii

Ta c vi mi s nguyn dng k:

=

v

Do :

iu phi chng minh.

Bi 3:

Gii

(1)

Du bng xy ra khi

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3 Bi 4:

Gii

EMBED Equation.3

EMBED Equation.3

a2b2(a2-b2)(a6-b6) 0 a2b2(a2-b2)2(a4+ a2b2+b4) 0

Bt ng thc cui ng vy ta c iu phi chng minh.

Bi 5:

Gii

Vi a > 0 , b > 0 => a + b > 0

Ta c :

4a2 - 4ab + 4b2 a2 + 2ab + b2

3(a2 - 2ab + b2 ) 0

3(a - b)2 0 . Bt ng thc ny ng

=> Du '' = '' xy ra khi a = b.

Bi 6:

Gii

Ta c:

Cng v theo v ta c:

Dng 2 S dng bt ng thc Bunhiacopxky v bt ng thc ph Bi 1:

Gii

p dng bt ng thc Bunhiacpxki, ta c:

Theo gi thuyt ta c:

V theo (1)

t Ta gii phng trnh bc 2

Bi 2:

Gii

Do a,b,c i xng ,gi s abc

p dng BT Tr- b-sp ta c:

==

Vy .Du bng xy ra khi a=b=c=

Bi 3:

Gii

p dng bt ng thc Bunhiacpxki ta c:

(x2 + y2)2 = ()2 ( ; )

(x2 + y2)(1 - y2 + 1 - x2) => x2 + y2 1

Ta li c : (3x + 4y)2 (32 + 42)(x2 + y2) 25

=> 3x + 4y 5

ng thc xy ra

iu kin : .

Bi 4:

Gii

p dng bt dng thc Bunhiacpxki vi 2 b 3 s ta c:

=>

=>

Du '' = '' xy ra khi : a = b = c =

Bi 5:

Gii

Ta c

Kt qu ny lun ng v v

Theo bt ng thc Bunhiacpski ta c:

Vy bt ng thc c chng minh.Bi 6:

Gii

p dng bt ng thc Bunhiacpski cho cc s ta c :

p dng bt ng thc Csi ta c:

Do :

(pcm).

Bi 7:

Gii

T gi thit ra

Ta c:

EMBED Equation.3

p dng bt ng thc Bunhicopsky ta c:

Do ta c:

Cng v theo v ta c:

Dng 3 s dng Bt ng thc CauchyBi 1:

Gii

VT = + + ++ ++ 3p dng BT csi cho cc bi s:

v ; v; v

Ta c : + 2=2 ; + 2 ; + 2

+ + ++ +

EMBED Equation.DSMT4 6 + + ++ ++39

VT 9 ( pcm ).

Bi 2:

Gii

a)Theo bt ng thc Csi ta c:

Nhn theo v ta c:

b)Ta p dng vi x,y>0 ta lun c bt ng thc ng:

Thay cc gi tr vo ta c:

Tng t:

Cng theo v ta c:

Bi 3:

Gii

p dng bt ng thc Csi , ta c:

Tng t : ;

Cng tng v ca 3 bt ng thc trn ta c:

Du ng thc xy ra khi a = b = c =0 tri vi gi thit : a + b + c = 1

Vy :

Bi 4:

Gii

Theo bt ng thc Csi ta c:


EMBED Equation.DSMT4 . iu phi chng minh.

Dng 4 chng minh bng phn chngBi 1:Gii:

Gi s c 3 bt ng thc trn u ng, c ngha:

Ta c

Tng t ta c:

Bi 2:

Gii:Chng minh bng phn chng. Gi s trong 25 s t hin cho, khng c hai s no bng nhau. Khng mt tnh tng qut, gi s

Suy ra

Th th

EMBED Equation.DSMT4 (1)

Ta li c

EMBED Equation.DSMT4 (2)

T (1) v (2) suy ra < 9, tri vi gi thit. Vy tn ti hai s bng nhau trong 25 s .

Dng 5 - Phng php lng gicBi 1:

Gii

Bin i iu kin: a2 + b2 - 2a - 4b + 4 = 0( (a-1)2 + (b-2)2 = 1

t

A (pcm)

Bi 2:

Gii:

Bin i bt ng thc: a2 + b2 + 2(b-a) ( - 1 ( (a-1)2 + (b + 1)2 ( 1

t vi R ( 0 (

Ta c:

(

T ( (a-1)2 + (b+1)2 = R2 ( 1 ( a2 + b2 + 2(b - a) ( - 1 (pcm)

Bi 3:

Gii:

T k 1 - a2 ( 0 ( |a| ( 1 nn

t a = cos( vi 0 ( ( ( ( ( = sin(. Khi ta c:

A=

=

EMBED Equation.3 (pcm)

Bi 4:

Gii:

Do |a| ( 1 nn :

t a = vi ((( . Khi :

A = (pcm)

Bi 5:

Gii:

Do |a| ( 1 nn:

t a = vi ((( . Khi :

A = = (5-12tg()cos2( = 5cos2(-12sin(cos(=

=

( - 4 = (pcm)

Bi 6:

Gii:

t a = tg(, b = tg(, c = tg(. Khi bt ng thc ((

(

( (sin((-()(+(sin((-()( ( (sin((-()(. Bin i biu thc v phi ta c:

(sin((-()(= (sin[((-()+((-()]( = (sin((-()cos((-()+sin((-()cos((-()( ((sin((-()cos((-()(+(sin((-()cos((-()(=(sin((-()((cos((-()(+(sin((-()((cos((-()(( (sin((-()(.1 + (sin((-()(.1 = (sin((-()( + (sin((-()( ( (pcm)

Bi 7:

Gii:

(1) (

t tg2(=, tg2(= vi (,( ( ( Bin i bt ng thc

(

( cos( cos( + sin( sin( = cos((-() ( 1 ng ( (pcm)

Du bng xy ra ( cos((-() = 1 ( (=( ( .

Bi 8:

Gii:

t a = tg(, b = tg(. Khi

=

=

(pcm).

Dng 6 Phng php qui np

Bi 1:

Gii:

Vi n =2 ta c (ng)

Gi s BT (1) ng vi n =k ta phi chng minh

BT (1) ng vi n = k+1

Tht vy khi n =k+1 th:

(1)

Theo gi thit quy np:

EMBED Equation.3

k2+2k o th:

,

Do ta c:


Cng tng v bn ng thc trn ta c:

(pcm).

Bi 2:

Gii:Khng mt tnh tng qut ta gi s : T :

v a + b = c + d

Nu: b th

EMBED Equation.3

EMBED Equation.3 999

Nu: b = 998 th a = 1

EMBED Equation.3 = t gi tr ln nht khi d = 1; c = 999

Vy gi tr ln nht ca = 999 + khi a = d =1; c = b = 999.Bi 3:

Gii:

T 2, v l. Vy 0 < a < 1.

Tng t ta c: 0 < b < 1 v 0 < c < 1.

Ta c: suy ra

v (1)

M

(2)

T (1) v (2) ta suy ra:

Dng 10 - Phng php dng cc bt ng thc trong tam gic

Bi 1:

Gii

Ta c:

EMBED Equation.DSMT4 Ta chng t

Kt qu trn lun ng v v v .

Vy

Bi 2:

Gii

V a, b, c l di 3 cnh ca tam gic nn:

Theo bt ng thc Cosi ta c:

iu phi chng minh.

Bi 3:

Gii

t v p + q = 1

L mt tam thc bc hai theo p c bit s

Trong mt tam gic ta c:


EMBED Equation.DSMT4 ab =

Ta c d thy vi mi a, b th : -

M : (a - b)2 =

(a + b)2 =

Suy ra : - ab .

Bi 2:

Gii

Ta c : =

( + ) ( + + + )

Tng t:

( + + + )

( + + + )

Cng theo v 3 BT trn:

+ + . 4 ( + )

M + = 4

Vy + + 1

Du = xy ra khi x = y = z = .

Bi 3:

Gii

Ta t vi v do nn

Nn BT

Mt khc theo BT Cauchy- Schwarz ta c:

Vy BT c chng minh.

Du = xy ra

Bi 4: Gii

T gi thit ta c th t: vi a,b,c >0

Nn BT CM

(ng)

Du = xy ra

Bi 5:

Gii

T

Ta t vi

Nn BT cn CM CM BT

Mt khc ta c:

Nn

Vy BT lun ng

Du = xy ra .Dng 12 Phng php lm tri (chng minh bt ng thc c n s hng).Hng dn:

Bi 1:

a. Ta c:

Cho n chy t 1 n k .Sau cng li ta c

b. Ta c:

Cc BT ng dng ca Bt ng thcI - Tm GTLN GTNN

Gii

V x,y,z > 0 ,p dng BT Csi ta c:

x+ y + z

p dng bt ng thc Csi cho x+y ; y+z ; x+z ta c:

Du bng xy ra khi x=y=z=

Vy S

Vy S c gi tr ln nht l khi x=y=z=.

Bi 2:

Gii

p dng BT Bunhiacpski cho 6 s (x,y,z) ;(x,y,z)

Ta c

p dng BT Bunhiacpski cho () v (1,1,1)

Ta c:

T (1) v (2)

Vy c gi tr nh nht l khi x=y=z=

Bi 3:

Gii

A = x2 2xy + 3y2 2x 10y + 20

= (x- y -1 )2 + 2(y- 3)2 + 1

m (x- y -1 )2 0 v (y- 3)2 0

vy x2 2xy + 3y2 2x 10y + 20 1

du = xy ra


Vy Amin =1 khi x= 4 , y = 3Bi 4:

Gii

C D = xc nh vi mi x v lun dng

Ta c: A(x2+1) = x2 +x +1

(A-1)x2 - x +A -1 =0

+ Nu A=1 th x =0 v ngc li

+ Nu A1 mun tn ti x th iu kin cn v : 1- 4(A-1)2 0

4A2 -8A + 3 0 A

Vy Amax= Amin=

(Khi x=1) (khi x=-1)Bi 5:

Gii

Gi a l s t nhin cn tm

Ta c a= 29 q +5 =31 q =28 (q, q, )

l s l; q-q,

A nh nht khi q, nh nht. Lc

2q, =29(q-q,) -23 cng nh nht

Hay q-q, nh nht

q-q, =1 v 2q, = 29-23 =6

Vy: q, =3 v a =121 l s cn tmII - Gii phng trnh v h phng trnh:

Bi 1:

Gii

p dng : Nu a, b > 0 th :

(2) (

( 6

Mt khc : v x, y, z > nn 6

V ;

Nn

Du '' = '' xy ra khi x = y = z , thay vo (1) ta c:

x + x2 + x3 = 14 (x - 2)(x2 + 3x + 7) = 0

x - 2 = 0 x = 2

Vy h phng trnh c nghim duy nht : x = y = z = 2

Bi 2:

Gii

Ta c


Vy

Du ( = ) xy ra khi x+1 = 0 x = -1

Vy khi x = -1

Vy phng trnh c nghim duy nht x = -1

Bi 3:

Gii

p dng BT BunhiaCpski ta c:

Du (=) xy ra khi x = 1

Mt khc

Du (=) xy ra khi y = -

Vy khi x =1 v y =-

Vy nghim ca phng trnh l .

Bi 4:

Gii

p dng BT Cosi ta c:

V x+y+z = 1 Nn

Du (=) xy ra khi x = y = z =

Vy c nghim x = y = z =

Bi 5:

Gii

T phng trnh (1) hay

T phng trnh (2)

Nu x = th y = 2, nu x = - th y = -2

Vy h phng trnh c nghim v .

III Tm nghim nguyn ca phng trnh:

Gii

Khng mt tnh tng qut ta gi s

Ta c

M z nguyn dng vy z = 1

Thay z = 1 vo phng trnh ta c

Theo gi s xy nn 1 =

EMBED Equation.DSMT4 m y nguyn dng

Nn y = 1 hoc y = 2

Vi y = 1 khng thch hp

Vi y = 2 ta c x = 2

Vy (2 ,2,1) l mt nghim ca phng trnh

Hon v cc s trn ta c cc nghim ca phng trnh l (2,2,1) ; (2,1,2) ; (1,2,2).

Phn V - Bi tp t gii phn trn ti gii thiu cc phng php cng nh bi tp v v d c th cn phn Bi tp t gii ny ti s cung cp mt s bi tp tng hp cho hc sinh luyn tp, gip hc sinh t t duy v tm ra phng php gii thch hp. phn BT ny ti khng gii chi tit m ch hng dn, gi gip hc sinh t vn dng cc phng php chng minh trn la chn cch chng minh cho bi ton .1. Chng minh cc bt ng thc sau:

a)

b)

c)

d)

e)

2. Cho v

a)

b)

EMBED Equation.DSMT4 p dng: Vi x,y >0

3Cho Tho Chng minh:

4. Cho a,b,c >0 tho .Chng minh:

5.Cho . Chng minh rng:

.

6. Chng minh rng:

a)

b) Vi

7. Cho 0 < a < b < c. Chng minh rng:

a)

b)

8. Cho a,b,c l di 3 cnh ca tam gic.Chng minh rng:

a)

b)

c)

d)

e)

9. Cho Chng minh rng:

10. Cho a1, a2 >0 v .Chng minh rng:

(

11. Cho a, b, c >0. Chng minh:

12. a,b,c l 3 cnh ca mt tam gic. Chng minh:

13. Cho a, b, c, d >0 v abcd=1. Chng minh rng:

14. Cho tam gic ABC. Ly im M trong tam gic, cc ng thng AM, BM, CM ct cc cnh ca tam gic ln lt A1, B1, C1. Chng minh rng:

15. Cho a,b>0 v x+y=1. Chng minh rng:

16. Chng minh iu kin cn v cc s a, b, c cng du l ab+bc+ca>0 v

17. Chng minh rng khng tn ti 3 s x,y,z ng thi tho mn ba bt ng thc sau:

18. Cho 00 . Chng minh

Khng l s nguyn.

23. Cho a ,b ,c l di ca 3 cnh ca mt tam gic . Chng minh rng:

24.Cho n l s nguyn dng, chng minh:

25. Chng minh rng:

a) Vi b) Vi

c)

26. Cho Chng minh rng:

27. Cho a>b>0. Chng minh rng:

28. Cho a,b,c >0, a+b+c=1. Chng minh rng:

29. Cho C tng bng 1. Chng minh rng:

30. Cho hnh thang ABCD c AD//BC c din tch S. Gi E l giao im ca hai ng cho. Chng minh rng 31. Cho a,b,c l 3 cnh ca tam gic c chu vi 2p. Chng minh rng:

32. Cho 3x-4y =7 .Chng minh rng: Hng dn Gi BT t gii S dng phng php nh ngha v php bin i tng ng

1. a)

b)

c)

d)

e)

EMBED Equation.DSMT4 2. S dng phng php nh ngha v php bin i tng ng

a)

b)

3. S dng phng php nh ngha v php bin i tng ng

iu phi chng minh.



Tng t: .cng li c iu phi chng minh


a)ng

b)

7 S dng phng php nh ngha v php bin i tng ng v cc

a) v ();

b)

8. S dng phng php nh ngha v php bin i tng ng v cc

Bt ng thc trong tam gic.

a)cng li c iu phi chng minh;

b)

Nhn cc v bt ng thc trn ta c iu phi chng minh

c)

d)

e)

9.p dng Bt ng thc ph

p dng v

10. p dng

11. .p dng Bt ng thc ph Chia ra 3 bt ng thc nh sau :

Cng v theo v v p dng

12. p dng phng php i bin s

t v p dng vi

13. Ta c nn

V

14. Gi ln lt l din tch cc tm gic ABC, MBC, MAC, MAB. Ta c:

Tng t cho v Sau p dng (x > 0).

15.

p dng v

16. S dng phng php chng minh phn chng

Xt hai trng hp

TH1: th a > 0, b > 0, c > 0

TH2: abc < 0 th a < 0, b < 0, c < 0.

17. Bnh phng hai v ca bt ng thc cho, chuyn v v tri ri nhn li

18.

19. V l

20. Gi s

Cng li ta c: , v l !

21. S dng cc tnh cht c bit ca t l thc.

22. S dng cc tnh cht c bit ca t l thc tng t cho cc BT khc ri cng li ta c: 1 < A < 2

23. Tng t bi 22:

24. S dng phng php lm tri


a)

b)

c) v


27. S dng BT Cosi

28. S dng BT Cosi

29. S dng BT Cosi

30. S dng BT Cosi

Ch cn chng minh v

vi x = BC, y = AD

31. S dng BT Bunhiacopsky

32. S dng BT Bunhiacopsky

. Ti liu tham kho(-------- ( ----------(1. ton nng cao v cc chuyn i s 8

NXB gio dc: 2000 Tc gi: Nguyn Ngc m Nguyn Vit Hi V Dng Thy 2. ton nng cao i s 279 bi ton chn lc nh xut bn tr 2005 V i Mau3. Cc bi ton pht trin bi dng hc sinh gii i s 9

Tc gi: V i Mau4. Chuyn bt ng thc v ng dng

Nh xut bn tr 1998 Tc gi: V i Mau5. Chuyn Bi dng hc sinh gii Trung hc c sNh xut bn gio dc - 2005 Tc gi: Nguyn V ThanhMc lc1Li ni u

2Phn I - kin thc c bn

2I Mt s bt ng thc cn nh:

3II - Mt s bt ng thc ph c chng minh l ng.

3III Cc bt ng thc trong tam gic

3IV Cc hm lng gic thng dng

3V Cc tnh cht c bn

3VI Cc hng ng thc ng nh

3VII Cc kin thc v to vec t

3VIII Cc kin thc v tnh cht ca t l thc:

4Phn II Cc phng php chng minh Bt ng thc

5Dng 1- Da vo nh ngha v cc php bin i tng tng ng

7Dng 2 S dng bt ng thc Bunhiacpsky v cc bt ng thc ph

11Dng 3 s dng bt ng thc Cauchy

13Dng 4 Chng minh bng phn chng

16Dng 5 Phng php lng gic

20Dng 6 Phng php chng minh qui np

22Dng 7 - Phng php p dng cc tnh cht ca cc dy t s bng nhau

24Dng 8 Phng php dng tam thc bc hai

25Dng 9 Phng php dng tnh cht bc cu

27Dng 10 Phng php dng cc bt ng thc trong tam gic

30Dng 11 Phng php i bin s

33Dng 12 - Phng php lm tri (chng minh bt ng thc c n s hng)

37Phn III ng dng ca Bt ng thc.

37I - Dng BT tm GTLN v GTNN

39II Dng BT gii phng trnh v h phng trnh

41III Dng BT gii phng trnh nghim nguyn

44Phn IV - gii v hng dn gii BT p dng

44Dng 1 Da vo nh ngha v cc php bin i tng ng

46Dng 2 S dng bt ng thc Bunhiacopxky v bt ng thc ph

52Dng 5 - Phng php lng gic

54Dng 6 Phng php qui np

56Dng 7 - Phng php p dng cc tnh cht ca cc dy t s bng nhau

57Dng 8 Phng php dng tam thc bc hai

58Dng 9 Phng php dng tnh cht bc cu

59Dng 10 - Phng php dng cc bt ng thc trong tam gic

61Dng 11 Phng php i bin s

64Dng 12 Phng php lm tri (bt ng thc c n s hng)

64Cc BT ng dng ca Bt ng thc

69Phn V - Bi tp t gii

72Hng dn Gi BT t gii

76Ti liu tham kho

77Mc lc


Sinh vin: Nguyn Mnh Hng Lp: CSP Ton Tin K48 Trang 6

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Documents

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