题目内容
先化简:(xy-2y)(xy+2y)-(x-y)2-x2y2,并求当x=
、y=2时的代数式的值.
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原式=x2y2-4y2-(x2-2xy+y2)-x2y2
=x2y2-4y2-x2+2xy-y2-x2y2
=-5y2-x2+2xy,
当x=
,y=2时,原式=-20-
+2=-18
.
=x2y2-4y2-x2+2xy-y2-x2y2
=-5y2-x2+2xy,
当x=
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