已知xy=2,x-y=3,求（x+y）（x-y）-y^2+（x-y）^2-（6x^2y-2xy^2)

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（x+y)（x-y)-y^2+（x-y)^2-（6x^2y-2xy^2)÷2y
=（x+y)（x-y)-y^2+（x-y)^2-2y（3x^2-xy)÷2y
=（x+y)（x-y)-y^2+（x-y)^2-（3x^2-xy)
=（x+y)（x-y)+（x-y)^2-y^2-（3x^2-xy)
=[（x+y)+（x-y)]（x-y)-y^2-（3x^2-xy)
=2x(x-y)-y^2-（3x^2-xy)
=2x^2-2xy-y^2-3x^2+xy
=-x^2-xy-y^2
=-(x^2+y^2+xy)
=-(x^2-2xy+y^2+3xy)
=-[(x-y)^2+3xy]
=-[3^2+3*(-2)]
=-(9-6)
=-3======以下答案可供参考======
供参考答案1:
已知xy=2，x-y=3,求(x+y)(x-y)-y²+(x-y)²-(6x²y-2xy²)/2y的值。
∵xy=2，x-y=3，∴(x-y)²=x²-2xy+y²=x²+y²-4=9，故x²+y²=13
(x+y)(x-y)-y²+(x-y)²-(6x²y-2xy²)/2y=x²-y²-y²+9-(3x²-xy)=-2x²-2y²+9+2
=-2(x²+y²)+11=-26+11=-15