分解因式:x-1-x(x-1)+x(x-1)的平方+……-x(x-1)的2003次方+想(x-1)的2004次方

2个回答

  • x-1-x(x-1)+x(x-1)^2-x(x-1)^3+……-x(x-1)^2003+x(x-1)^2004

    =x-1-[x(x-1)+x(x-1)^3+..+x(x-1)^2003]+[x(x-1)^2+x(x-1)^4+...+x(x-1)^2004]

    =x-1-x[(x-1)+(x-1)^3+..+(x-1)^2003]+x[(x-1)^2+(x-1)^4+...+(x-1)^2004]

    =x-1-x{(x-1)[1-(x-1)^1002]/[1-(x-1)]}+x{(x-1)^2[1-(x-1)^1002]/[1-(x-1)]}

    =(x-1)+x(x-1)[1-(x-1)^1002][(x-1)-1]/(2-x)

    =(x-1)-x(x-1)[1-(x-1)^1002]

    =(x-1){1-x[1-(x-1)^1002}

    =(x-1)[1-x+x(x-1)^1002]

    =(x-1)[-(x-1)+x(x-1)^1002]

    =(x-1)(x-1)[x(x-1)^1001-1]

    =(x-1)^2[x(x-1)^1001-1]

    正确答案,欢迎采纳!