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

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/16 01:53:29

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

[(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)
再问：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...
再答：是
再问：？你的这个答案是正确的吗？、
再答：请稍等

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