问题 解答题
知x2-xy-2y2=0,且x≠0,y≠0,求代数式
x2-2xy-5y2
x2+2xy+5y2
的值.
答案

∵x2-xy-2y2=0,

∴(x-2y)(x+y)=0,

∴x-2y=0或x+y=0.

∴x=2y或x=-y.

当x=2y时,

x2-2xy-5y2
x2+2xy-5y2
=
(2y)2-2•2y•y-5y2
(2y)2+2•2y•y+5y2
=
-5y2
13y2
=-
5
13
;

当x=-y时,

x2-2xy-5y2
x2+2xy+5y2
=
(-y)2-2•(-y)•y-5y2
(-y)2+2•(-y)•y+5y2
=
-2y2
4y2
=-
1
2
.

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