帮帮帮帮帮帮帮帮手.解方程:1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/2x+4012

来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 11:16:26
帮帮帮帮帮帮帮帮手.解方程:1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/2x+4012

帮帮帮帮帮帮帮帮手.解方程:1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/2x+4012
帮帮帮帮帮帮帮帮手.
解方程:
1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/2x+4012

帮帮帮帮帮帮帮帮手.解方程:1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/2x+4012
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
1/(x+2)(x+3)=1/(x+2)-1/(x+3)
.
...
1/(x+2005)(x+2006)=1/(x+2005)-1/(x+2006)
不难得出 1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2005)(x+2006)=1/(x+1)-1/(x+2006)=1/2x+4012
再解方程...

因为 1/（x+1）（x+2）=1/x+1 - 1/x+2
。。。
所以
原式恒等于 1/x+1 - 1/x+2006 =1/2x+4012
下面会作了么？

拆项，如
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
每项都拆后可以抵消
左边为1/(x+1)-1/(x+2006)

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