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  • 1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy

    1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy
    4.(1-1-x分之1)÷1-x分之1+x
    分式的加減法
    數(shù)學(xué)人氣:867 ℃時(shí)間:2020-06-13 16:45:21
    優(yōu)質(zhì)解答
    解1題:
    原式=[1/(a+1)]-[2/(1-a)]
    =[1/(a+1)]+[2/(a-1)]
    ={ (a-1)/[(a+1)(a-1)] }+{2(a+1)/[(a+1)(a-1)] }
    =[(a-1)+2(a+1)]/[(a+1)(a-1)]
    =(a-1+2a+2)/(a²-1)
    =(3a+1)/(a²-1)
    解2題:
    原式=[x/(x-y)]-[y/(x+y)]
    ={ x(x+y)/[(x+y)(x-y)] }-{ y(x-y)/[(x+y)(x-y)] }
    =[x(x+y)-y(x-y)]/[(x+y)(x-y)]
    =(x²+xy-xy+y²)/(x²-y²)
    =(x²+y²)/(x²-y²)
    解3題:
    原式=[x²/(x+y)-y²/(x+y)]×[xy/(x-y)]
    =[(x²-y²)/(x+y)]×[xy/(x-y)]
    =[(x+y)(x-y)/(x+y)]×[xy/(x-y)]
    =xy
    解4題:
    原式=[1-1/(1-x)]÷[(1+x)/(1-x)]
    ={ [(1-x)-1]/(1-x) }×[(1-x)/(1+x)]
    =[-x/(1-x)]×[(1-x)/(1+x)]
    =-x/(1+x)
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