(x+y)(x-y)+(x-y)²-(6x²y-2xy²)÷2y=(x²-y²)+(x²-2xy+y²)-[(2y)(3x²-xy)]÷2y=x²-y²+x²-2xy+y²-3x²+xy=-xy-3x²=-x(y+3x)=-(-2)×[3分之1+3...