问题
解答题
化简求值：(
|
答案
∵
+|y-1|=0，x+2
∴x=-2，y=1，
则原式=
•y(x+2y)(x-2y) (x+2y)2 4xy+x(x-2y) x-2y
=
•y(x+2y)(x-2y) (x+2y)2 x(x+2y) x-2y
=xy=-2．
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化简求值：(
|
∵
+|y-1|=0，x+2
∴x=-2，y=1，
则原式=
•y(x+2y)(x-2y) (x+2y)2 4xy+x(x-2y) x-2y
=
•y(x+2y)(x-2y) (x+2y)2 x(x+2y) x-2y
=xy=-2．