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  • 初二幾道分式計(jì)算

    初二幾道分式計(jì)算
    1.b/(a-b)+a/(a+b)-2ab/(b^2-a^2)
    2.(2a+2)/(a+1)-(a^2-1)/(a^2-2a+1)
    3.(x^2-2x)/x^2-1/[x-1-(2x-1/x+1)
    數(shù)學(xué)人氣:756 ℃時(shí)間:2020-05-13 00:58:01
    優(yōu)質(zhì)解答
    1.=b(a+b)/(a-b)(a+b)+a(a-b)/(a+b)(a-b)-2ab/(b^2-a^2)
    =(ab+b^2+a^2-ab)/(a^2-b^2)+2ab/b^2-a^2
    =a^2+b^2/(a^2-b^2)-2ab/a^2-b^2
    =(a-b)^2/(a+b)(a-b)
    =a-b/a+b
    2.=2(a+1)/(a+1)-(a^2-1^2)/(a-1)^2
    =2-(a+1)(a-1)/(a-1)^2
    =2-(a+1)/(a-1)
    3 =x(x-2)/(x^2-1^2)-1/{(x+1)(x-1)/(x+1)++(2x-1)/(x+1)
    =x(x-2)/(x^2-1^2)-1/(x^2-2x)/(x+1)
    =x(x-2)/(x^2-1^2)-(x+1)/x(x-2)
    =x+1/(x+1)(x-1)
    =1/(x-1)
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