计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/10/02 20:33:13
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
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
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