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  • 分解因式:4/(x^2-y^2)+(x+y)/(xy^2-x^2y)÷(x^2+xy-2y^2)/(x^2+2xy)

    分解因式:4/(x^2-y^2)+(x+y)/(xy^2-x^2y)÷(x^2+xy-2y^2)/(x^2+2xy)
    不好意思題目打錯了,應該是:[4/(x^2-y^2)+(x+y)/(xy^2-x^2y)]÷(x^2+xy-2y^2)/(x^2y+2xy^2),答案是-1/x+y,求過程
    數(shù)學人氣:215 ℃時間:2020-06-04 14:32:37
    優(yōu)質(zhì)解答
    原式=[4/(x-y)(x+y)+(x+y)/xy(y-x)]xy(x+2y)/(x+2y)(x-y)=[4/(x-y)(x+y)+(x+y)/xy(y-x)]xy/(x-y)=-(x+y)/(x-y)^2+4xy/(x-y)^2(x+y)=[(x+y)^2-4xy]/-(x-y)^2(x+y)=(x-y)^2/-(x-y)^2(x+y)=-1/(x+y)
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