(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×(-2)]
=2×(3分之1-6)
=2×(3分之1-3分之18)
=2×(-3分之17)
=-3分之34
=-11又3分之1
以上回答你满意么?