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  • 分式加減計算題

    分式加減計算題
    X+2 X^2-2X+1 2X-6
    ---X-------- - ----
    X-1 X^2-X-6 X^2-9
    X^2-1 1
    ----- + X(1+ - )
    X-1 X 其中X=(根號2) - 1
    數(shù)學人氣:719 ℃時間:2020-03-23 01:13:49
    優(yōu)質(zhì)解答
    [(x+2)/(x-1)]*[(x^2-2x+1)/(x^2-x-6)]-(2x-6)/(x^2-9)
    =[(x+2)/(x-1)]*[(x-1)^2/(x-3)(x+2)]-2(x-3)/(x+3)(x-3)
    =(x+2)(x-1)^2/[(x-1)(x-3)(x+2)]-2/(x+3)
    =(x-1)/(x-3)-2/(x+3)
    =[(x-1)(x+3)-2(x-3)]/(x+3)(x-3)
    =(x^2+2x-3-2x+6)/(x+3)(x-3)
    =(x^2+3)/(x+3)(x-3)
    (x^2-1)/(x-1)+x(1+1/x)
    =(x+1)(x-1)/(x-1)+x+1
    =x+1+x+1
    =2x+2
    =2根號2-2+2
    =2根號2
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