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  • 計算:-x(x^2+2x-3)-x^2(x+1)+3x-1 2ab(3/2·a^n-1+1/2b^n-1)-3ab(2/3·a^n-1-b^n-1) [x(x^4-y^4)-3xy(-

    計算:-x(x^2+2x-3)-x^2(x+1)+3x-1 2ab(3/2·a^n-1+1/2b^n-1)-3ab(2/3·a^n-1-b^n-1) [x(x^4-y^4)-3xy(-
    -x(x^2+2x-3)-x^2(x+1)+3x-1
    2ab(3/2·a^n-1+1/2b^n-1)-3ab(2/3·a^n-1-b^n-1)
    [x(x^4-y^4)-3xy(-y)^3·x^2·y
    數(shù)學(xué)人氣:267 ℃時間:2020-04-09 02:22:06
    優(yōu)質(zhì)解答
    -x(x^2+2x-3)-x^2(x+1)+3x-1
    =-x³-2x²+3x-x³-x²+3x-1
    =-2x³-3x²+6x-1
    2ab(3/2·a^n-1+1/2b^n-1)-3ab(2/3·a^n-1-b^n-1)
    =3a^nb+ab^n-2a^nb+3ab^n
    =a^nb+4ab^n
    [x(x^4-y^4)-3xy(-y)^3·x^2·y
    =x^5-xy^4+3x^3y^5
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