当xy≠0时,试说明代数式[(x y)(x-y)-(x y)²-2y(x-y)-2xy]÷xy的值与x,y的

如题所述

解:原式=[(x+y)(x-y)-(x+y)^2-2y(x-y)-2xy)]/xy
=[x^2-y^2-x^2-2xy-y^2-2xy+2y^2]/xy
=[-2xy-2xy]/xy
=-4
所以与xy取值无关
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