题目内容

已知x,y满足(x+2y)(x-2y)=-5(y2-
6
5
),2x(y-1)+4(
1
2
x-1)=0

求(1)(x-y)2;(2)x4+y4-x2y2
分析:先化简(x+2y)(x-2y)=-5(y2-
6
5
),得x2+y2=6,再化简2x(y-1)+4(
1
2
x-1)=0得xy=2,再代入即可求出(1)(x-y)2;(2)x4+y4-x2y2
解答:解:∵(x+2y)(x-2y)=-5(y2-
6
5
),
∴x2-4y2=-5y2+6,
∴x2+y2=6,
∵2x(y-1)+4(
1
2
x-1)=0,
∴2xy-2x+2x-4=0,
∴xy=2,
(1)(x-y)2=x2+y2-2xy=6-4=2;
(2)x4+y4-x2y2=(x2+y22-2x2y2-x2y2
=(x2+y22-3x2y2
=36-3×4
=24.
点评:本题考查了整式的混合运算,根据所给出的等式得出结论,再将它代入即可.
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