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  • (1)計(jì)算:(5x)/(3y)*(-9y)/(10x^2) (2)化簡(jiǎn):(x+1)/(x)÷(1)/(x^2-x)

    (1)計(jì)算:(5x)/(3y)*(-9y)/(10x^2) (2)化簡(jiǎn):(x+1)/(x)÷(1)/(x^2-x)
    (3)計(jì)算:(1+m)/(1-m)*(m^2-1)的結(jié)果
    (4)化簡(jiǎn):x^2/y-x-y^2/y-x的結(jié)果
    (5)計(jì)算:(2-x)/(x^2-4)+(x+3)/(2+x)=
    數(shù)學(xué)人氣:427 ℃時(shí)間:2020-05-17 13:27:56
    優(yōu)質(zhì)解答
    (1)計(jì)算:(5x)/(3y)*(-9y)/(10x^2)
    =(5x)(-9y)/(3y)(10x^2)
    = -3/2x
    (2)化簡(jiǎn):(x+1)/(x)÷(1)/(x^2-x)
    =(x+1)/(x)÷(1)/(x)(x-1)
    =(x^2-1)/(x)(x-1)÷(1)/(x)(x-1)
    =x^2-1
    (3)計(jì)算:(1+m)/(1-m)*(m^2-1)
    =(1+m)/(1-m)*(m+1)(m-1)
    = -(1+m)(m+1)
    = -(m+1)^2
    (4)化簡(jiǎn):x^2/y-x-y^2/y-x
    =(x^2-y^2)/(y-x)
    =(x+y)(x-y)/(y-x)
    = -x-y
    (5)計(jì)算:(2-x)/(x^2-4)+(x+3)/(2+x)
    =(2-x)/(x+2)(x-2)+(x+3)/(2+x)
    = -1/(x+2)+(x+3)/(2+x)
    = (-1+x+3)/(x+2)
    = 1
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