作业帮计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/08 12:16:45
计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020
计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020
计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020
应该是1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/（2x+4020 ）吧
1/(x+1)-1/(x+2)+1/(x+2）-1/(x+3)+1/(x+3)-1/(x+4)+...+1/(x+2009)-1/(x+2010)=1/（2x+4020）
1/(x+1)-1/(x+2010)=1/[2（x+2010）]
2009/(x+1)(x+2010)=1/[2(x+2010)]
2009/(x+1)=1/2
x=4017
原方程可以写成; 1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4).....+1/(x+2009)-1/(x+2010)=1/2x+4020
1/(x+1)-1/(x+2010)=1/2x+4020
剩下的自己算 如果还有其他问题请。。。
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