已知x/y=1/2,求2x/(x²-2xy+y² )* (x²-y&#

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[2x/(x²-2xy+y² )]* [(x²-y²)/(x+y)]+2y/(x-y) 【电脑上录入表达式,尽可能多用括号,否则多歧义】
=[2x/(x-y)²]* [(x+y)(x-y)/(x+y)]+2y/(x-y)
=2x/(x-y)+2y/(x-y)
=2(x+y)/(x-y) 【分子分母都除以y】
=2(x/y+1)/(x/y-1)
=2(1/2+1)/(1/2-1)
=-6======以下答案可供参考======
供参考答案1:
问题不全？供参考答案2:
∵ x/y=1/2
∴y=2x∴2x/(x²-2xy+y² )* (x²-y²/x+y)+2y/(x-y)
=2x(x+y)(x-y)/[x-y)²(x+y)+2y/(x-y)
=2x/(x-y)+2y/(x-y)
=2(x+y)/(x-y)
=6x/x=6