(急速)分式运算: (1)(x+y)·x^2-y^2分之x^2 +x-y分之y^2 -(2x^2y-2xy^2)÷(x^2-2xy+y^2)

2025-05-20 19:36:16
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回答1:

(1)(x+y)·x^2-y^2分之x^2 +x-y分之y^2 -(2x^2y-2xy^2)÷(x^2-2xy+y^2)
=(x+y)*x²/(x+y)(x-y)+y²/(x-y)-2xy(x-y)/(x-y)²
=x²/(x-y)+y²/(x-y)-2xy/(x-y)
=(x²+y²-2xy)/(x-y)
=(x-y)²/(x-y)
=x-y