1/(x+1)(x=2)+1/(x+2)(x+3)+%+...+1/(x+2003)（x+2004)=1/3x+6012

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：2024/04/29 05:55:35

1/(x+1)(x=2)+1/(x+2)(x+3)+%+...+1/(x+2003)（x+2004)=1/3x+6012 注意是解方程!

1/(x+1)(x+2)+1/(x+2)(x+3)+%+...+1/(x+2003)（x+2004)=1/3x+6012
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+.1/(x+2003)-1/(x+2004)=1/3(x+2004)
1/(x+1)-1/(x+2004)=1/3(x+2004)
3(x+2004)-3(x+1)=(x+1)
3x+6012-3x-3=x+1
x=6008
这里采用的是拆项法,不理解,

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