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
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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！
=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！
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