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计算：[(x+y/x-y)^2·(2y-2x/3x+3y)]-[(x^2/x^2-y^2)/(x/y)]

分类: 作业答案 • 2021-12-28 15:50:14
问题描述：

计算：[(x+y/x-y)^2·(2y-2x/3x+3y)]-[(x^2/x^2-y^2)/(x/y)]
今天止

答

[(x+y/x-y)^2*(2y-2x/3x+3y)]-[(x^2/x^2-y^2)/(x/y)]
={[(x+y)/(x-y)]^2*(2y-2x)/(3x+3y)}-[x^2/(x^2-y^2)*y/x]
=-{[(x+y)/(x-y)]^2*2(x-y)/3(x+y)}-[x^2/(x^2-y^2)*y/x]
=-(x+y)/(x-y)-[x/(x^2-y^2)*y]
=-(x+y)/(x-y)-[xy/(x^2-y^2)]
=-(x+y)^2/(x^2-y^2)-[xy/(x^2-y^2)]
=-[(x+y)^2+xy]/(x^2-y^2)]
=-(x^2+y^2+2xy+xy)/(x^2-y^2)
=-(x^2+y^2+3xy)/(x^2-y^2)第3步：-{[(x+y)/(x-y)]^2*2(x-y)/3(x+y)}-[x^2/(x^2-y^2)*y/x]应该是化简为：-[2(x+y)/3(x-y)]-[x/(x^2-y^2)*y]吧...是？你的这个答案是正确的吗？、请稍等

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