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的值无关.我现在就要

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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的值无关.我现在就要当xy≠0时,试说明代数式[(x+y)(x-y)-(x+y)&s

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的值无关.我现在就要
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的值无关.我现在就要

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的值无关.我现在就要
[(x+y)(x-y)-(x+y)²-2y(x-y)-2xy]/xy
=[x²-y²-x²-2xy-y²-2xy+2y²-2xy]/xy
=-6xy/xy
=-6

将括号里的化开得
x2-y2-x2-2xy-y2-2xy+2y2-2xy=-4xy
所以-6xy/xy=-6
所以与xy的取值无关