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  • 計(jì)算:(x^3+x^2y) / (x^2-y^3) - x(x-y)/ (xy+x^2)- (2xy)/(x^2-y^2)

    計(jì)算:(x^3+x^2y) / (x^2-y^3) - x(x-y)/ (xy+x^2)- (2xy)/(x^2-y^2)
    數(shù)學(xué)人氣:168 ℃時(shí)間:2019-08-22 12:47:58
    優(yōu)質(zhì)解答
    [(x^3+x^2y)/(x^2y-y^3)]-{[x(x-y)]/(xy+x^2)}-[(2xy)/(x^2-y^2)] 原式=[x^2(x+y)/y(x+y)(x-y))]-[x(x-y)]/[x(y+x)]-[2xy/(x+y)(x-y)]=[x^2/y(x-y)]-(x-y)/(x+y)-2xy/(x+y)(x-y)=[x^2(x+y)-y(x-y)^2-2xy^2]/y(x+y)(x...
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