已知xy=-2,x-y=3,求(x+y)(x-y)-y²+(x-y)²-(6x²y-2xy²)÷(2y)的值.
问题描述：
已知xy=-2,x-y=3,求(x+y)(x-y)-y²+(x-y)²-(6x²y-2xy²)÷(2y)的值.
答
第二个等式变为y=x-3代人一式得，x²-3x+2=0。原式=3（x+y)-y²+9-(3x²-xy)=3(x+y)-y²+7-3x²=3(2x-3)-(x-3)²+7-3x²=-4(x²-3x+2)-3=-3
答
(x+y)(x-y)-y²+(x-y)²-(6x²y-2xy²)÷(2y)
=x²-y²-y²+x²-2xy+y²-（6x²y÷2y-2xy²÷2y）
=x²-y²-y²+x²-2xy+y²-（3x²-xy）
=-x²-y²-xy
=-x²-y²+2xy-3xy
=-(x²+y²-2xy）-3xy
=-(x-y)²-3xy
=-3²+6
=-3
答
(x+y)(x-y)-y²+(x-y)²-(6x²y-2xy²)÷(2y)
=x²-y²-y²+x²-2xy+y²-3x²+xy
=-x²-y²-xy
=-x²-y²+2xy-3xy
=-(x-y)²-3xy
=-3²+6
=-3
答
(x+y)(x-y)-y2+(x-y)2-(6x2y-2xy2)÷(2y)
=x^2-y^2-y^2+x^2-2xy+y^2-3x^2+xy
=-2x^2-y^2-xy