[Tailieuluyenthi.com]Chuyen de Bat Dang Thuc Hay
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Transcript of [Tailieuluyenthi.com]Chuyen de Bat Dang Thuc Hay
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
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:
EMBED Equation.DSMT4
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
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
EMBED Equation.DSMT4
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
EMBED Equation.DSMT4
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.
4. S dng phng php nh ngha v php bin i tng ng
5. S dng phng php nh ngha v php bin i tng ng
Tng t: .cng li c iu phi chng minh
6. S dng phng php nh ngha v php bin i tng ng
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
25. S dng phng php lm tri
a)
b)
c) v
26. S dng phng php lm tri
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
EMBED Equation.DSMT4
Sinh vin: Nguyn Mnh Hng Lp: CSP Ton Tin K48 Trang 6
_1269083020.unknown
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