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V MT CCH TM GI TR LN NHT , NH NHTCA BIU THC CHA HAI BIN S
B Ch - Thi Bnh tng www.mathvn.com
Bi vit ny chng ti xin trao i v mt phng php tm gi tr ln nht (GTLN), gi trnh nht (GTNN) ca biu thc cha hai bin s nh tp gi tr, trong hai bin b rngbuc bi mt iu kin cho trc .Bi ton : Cho cc s thc x , y tho mn iu kin : G(x ; y) = 0 .
Tm GTLN , GTNN (nu c) ca biu thc P = F(x ; y).Phng php gii :Gi T l tp gi tr ca P. Khi , m l mt gi tr ca T khi v ch khi h sau c nghim (x;y):
G(x;y) 0F(x;y) m
Sau tm cc gi tr ca tham s m h trn c nghim . T suy ra tp gi tr T ca P ,ri suy ra GTLN , GTNN (nu c) ca P.Sau y l cc bi ton minh ho .
Bi ton 1 : Cho hai s thc x , y tho mn iu kin : 3 3 3 3 3x( x 1) y y 1 xy Tm GTLN , GTNN ca biu thc 3 3 3F x y xy .
Li gii : Gi T1 l tp gi tr ca F . Ta c 1m T h sau c nghim: 3 3 3 3 33 3 3
x( x 1) y y 1 xyx y xy m
(I)
t3 3
3
S x yP xy
th x, y S,P : 2S 4P
H (I) tr thnh2 2S S 3P 0 S 2S 3m 0 (II)S P m P m S
Ta c :
22 2 24(S S)S 4P S S 4S 0 0 S 43
T , h (I) c nghim h (II) c nghim ( S ; P ) tho mn 2S 4P phng trnh
2S 2S 3m 0 c nghim S : 0 S 4 , iu ny xy ra khi v ch khiS
1
2
1 3m 0 1m
30 S 1 1 3m 41 1 3m 50 S 1 1 3m 4
0 m 8 . Do : 1T 0;8
Vy : minF = 0 , maxF = 8.Bi ton 2 : Cho cc s thc x, y tho mn : 32 2x - xy + y
Tm GTLN , GTNN ca biu thc 2 2G = x + xy - 2yLi gii : Gi T2 l tp gi tr ca G . Ta c 2m T h sau c nghim:
3 2 2
2 2x - xy + yx + xy - 2y = m (III)
Trang 2
Nu y = 0 th h (III) tr thnh 2
23x
x m
33
xm
Nu y 0 th t x = ty ta c h :2
2 2 2
2 2 2
2
3yy (t t 1) 3 t t 1y (t t 2) m 3(t t 2) mt t 1
2
2
3y t t 1(m 3)t (m 3)t m 6 0
(IV)
Trng hp ny h (III) c nghim h (IV) c nghim y 0 phng trnh2(m 3)t (m 3)t m 6 0 (2) c nghim :
Nu m = 3 th (2) c nghim t = 32
Nu m 3 th (2) c nghim 2t 3m 6m 81 0 1 2 7 m 1 2 7 (m 3 )
Kt hp cc trng hp trn ta c cc gi tr ca m h (III) c nghim l :1 2 7 m 1 2 7 . Do : 2T 1 2 7 ; 1 2 7
Vy : minG = 1 2 7 , maxG = 1 2 7 Bi ton 3 : (Tuyn sinh i hc khi A nm 2006 )
Cho hai s thc thay i x 0 , y 0 tho mn : 2 2(x y)xy x y xy
Tm gi tr ln nht ca biu thc 3 31 1Ax y
Li gii : Gi T3 l tp gi tr ca A . Ta c 3m T h sau c nghim x 0 , y 0 :2 2 2 22 2
2 2 2
3 3 3 3
(x y)xy x y xy (x y)xy x y xy(x y)xy x y xy1 1 (x y)(x y xy) xy(x y)
m m mx y (xy) (xy)
2
2
(x y)xy (x y) 3xyx y( ) mxy
(V)
tS x yP xy
( 2S 4P ) , ta c h :2
2
SP S 3PS( ) mP
(VI)
H (V) c nghim x 0 , y 0 h (VI) c nghim ( S ; P ) tho mn 2S 4P .V 2 2 2 21 3SP x y xy (x y) y 0
2 4 vi mi x 0 , y 0 S 0
P vi mi x 0 ,
y 0T :
Nu m 0 th h (V) v nghim Nu m > 0 th t phng trnh 2S S( ) m m
P P S m.P thay vo phng
trnhu ca h (VI) c : 2 2mP mP 3P (m m)P 3 ( v SP > 0 nn P 0 )
Trang 3
c P t phng trnh ny th m m 0 m 1 ( m > 0 ) v ta c3P
m( m 1) , do
3Sm 1
. Trng hp ny h (VI) c nghim ( S ; P )tho
mn 2S 4P khi v ch khi :
23 12( )m 1 m( m 1)
24( m 1)3 3 m 4( m 1) m 4
m( m 1)
0 m 16 (m 1) Tm li cc gi tr ca m h (V) c nghim x 0 , y 0 l : 0 m 16 , m 1 Do : 3T 0;16 \ 1Vy : maxA = 16 ( ch khng tn ti minA )
Bi ton 4 : ( HSG quc gia - Bng A + B nm 2005 )Cho hai s thc x, y tho mn : x 3 x 1 3 y 2 y Hy tm gi tr ln nht v nh nht ca biu thc K x y
Li gii : KX : x 1, y 2 Gi T4 l tp gi tr ca K . Ta c 4m T h sau c nghim:
3( x 1 y 2) mx 3 x 1 3 y 2 y (VII)x y m x y m
t u x 1 v v y 2 th u, v 0 v h (VII) tr thnh :
2 2 2
mu v3(u v) m 3
u v m 3 1 muv ( m 3)
2 9
u , v l hai nghim ca phng trnh :
22 2 2m 1 mt t ( m 3) 0 18t 6mt m 9m 27 0
3 2 9 (3)
T , h (VII) c nghim ( x ; y ) sao cho x 1, y 2 khi v ch khi (3) c hai nghimkhng m v iu kin l :
2t
t
2
t
9(m 18m 54) 0m 9 3 21S 0 m 9 3 153 2m 9m 27P 0
18
. Do 49 3 21T ;9 3 15
2
Vy : minK = 9 3 212
, maxK = 9 3 15
Bnh lun: u th ca phng php trn l quy bi ton tm GTLN, GTNN v bi ton tmtham s h c nghim , v vy khng cn ch r gi tr ca bin s biu thc t GTLN,GTNN . Nu dng cc bt ng thc nh gi th nht thit phi ch r cc gi tr ca bins ti biu thc t GTLN, GTNN. Cc bn c th m rng phng php ny cho biuthc c nhiu hn hai bin s .Cui cng l cc bi tp minh ho phng php trn :Bi 1 : Cho hai s thc x , y tho mn : 2 2 2( ) 7x y x y .
Trang 4
Tm gi tr ln nht , nh nht ca biu thc 3 3P x(x 2) y(y 2)Bi 2 : Cho cc s thc x, y tho mn : 2 24x - 3xy +3y = 6 .
Tm gi tr ln nht v nh nht ca biu thc 2 2F x xy 2y Bi 3 : Cho cc s thc khng m x , y tho mn : x y 4 .
Tm gi tr ln nht v nh nht ca biu thc 1 9Q x y Bi 4 : Cho cc s dng x , y tho mn : xy x y 3
Tm GTLN ca biu thc 2 23x 3yG x yy 1 x 1
Bi 5 : (Cao ng kinh t k thut nm 2008) .Cho hai s x , y tho m n : 2 2x y 2
Tm GTLN , GTNN ca biu thc 3 3P 2(x y ) 3xy Bi 6 : (i hc Khi B nm 2008). Cho hai s thc x , y thay i v tho mn hthc: 2 2x y 1
Tm GTLN , GTNN ca biu thc2
2
2(x 6xy)P1 2xy 2y
................................Ht............................
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