1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2004)(x+2005)

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/05/19 00:36:57

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+.+1/(x+2004)(x+2005)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+-----+(1/(x+2004)-1/(x+2005)
=1/x-1/(x+2005)
=2005/(x^2+2005x)
用裂项相消法
1/（x+n)(x+n+1)=1/(x+n)-1/(x+n+1)

x^4+x^3+x^2+x+1=0,x^2006+x^2005+x^2004+x^2003+x^2002 x-1)(X-2)(x-3)...(x-50)+x(x-2)(X-3)...(X-50)+...+x(x-1)(x-2) x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4 x^5+x^4 = (x^3-x)(x^2+x+1)+x^2+x 1+x+x^2+x^3+.+x^99+x^100 |X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10| {2X+1}+{X} x^2-|x|+1 ｜x-1｜+｜x-2｜ x(x+1)^3+x(x+1)^2+x(x+1)+x+1因式分解 x(2-1/x)+x/(x^2-2x)÷(3-x)/(x^2-4) 解方程x/(x-2)=2x/(x-3)+(1-x)/(x-5x+6)