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  • 計算:"3y/x+y"+"y/y-x"+"2x/x^2-y^2" "a+2/a^2-4a+4"-"1/2a+4"-"a-1/2a^2-8"

    計算:"3y/x+y"+"y/y-x"+"2x/x^2-y^2" "a+2/a^2-4a+4"-"1/2a+4"-"a-1/2a^2-8"
    a^2+a+1-"a^3/a-1" "1/a-2"-"12/a^2-4"
    一共4題。
    數(shù)學(xué)人氣:364 ℃時間:2020-06-30 20:30:18
    優(yōu)質(zhì)解答
    "3y/x+y"+"y/y-x"+"2x/x^2-y^2"
    =[3y*(x-y)-y(x+y)]/[(x+y)(x-Y)]
    =(2xy-4y^2)/[(x-y)(x+y)]
    =2xy/(x+y)
    a+2/a^2-4a+4"-"1/2a+4"-"a-1/2a^2-8
    =(a+2)/(a-2)^2-1/[2(a+2)]-(a-1)/[2(a+2)(a-2)]
    =(a+2)/(a-2)^2-(a-2)/[2(a+2)(a-2)]-(a-1)/[2(a+2)(a-2)]
    =(a+2)/(a-2)^2-[(2a-3)/[2(a+2)(a-2)]
    =[2*(a+2)^2-(2a-3)(a-2)]/[2(a+2)(a-2)^2]
    =(15a+2)/2(a-2)^2(a+2)
    3.a^2+a+1-"a^3/a-1
    =[(a^2+a+1)(a-1)-a^3]/(a-1)
    =-1/(a-1)
    4.1/a-2"-"12/a^2-4
    =1/(a-2)-12/(a+2)(a-2)
    =[(a+2)-12]/(a+2)(a-2)
    =(a-10)/(a+2)(a-2)
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