化简：1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+...+1/(x+99)(x+101)
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解决时间 2021-05-06 05:45
- 提问者网友：骨子里的高雅
- 2021-05-05 22:43
化简：1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+...+1/(x+99)(x+101)
最佳答案
- 五星知识达人网友：污到你湿
- 2021-05-05 23:37
=【1/（x+1）-1/（x+3）】/2+【1/（x+3）-1/（x+5）】/2……………………………………
都消了
【1/（x+1）-1/(x+101)】/2=50/（x+1）（x+101)
全部回答
- 1楼网友：人類模型
- 2021-05-06 03:04
1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+...+1/(x+99)(x+101)
=(2/(X+1)(X+3)+2/(X+3)(X+5)+2/(X+5)(X+7)+...+2/(X+99)(X+101))/2
=(1/(X+1)-1/(X+3)+1/(X+3)-1/(X+5)+1/(X+5)-1/(X+7)+...+1/(X+99)-1/(X+101))/2
=(1/(X+1)-1/(X+101))/2
=((X+101-X-1)/(X+1)(X+101))/2
=(100/(X+1)(X+101))/2
=50/((X+1)(X+101))
- 2楼网友：蕴藏春秋
- 2021-05-06 02:21
这是要用的公式
原式=1/2[1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+1/(x+5)-1/(x+7)+...+1/(x+99)-1/(x+101)]
=1/2[1/(x+1)-1/(x+101)
=50/(x+1)(x+101)
望采纳
- 3楼网友：罪歌
- 2021-05-06 00:49
1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+...+1/(x+99)(x+101)=2[1/(x+1)-1/(x+3)+……+1/(x+99)-1/(x+101)]=2[1/(x+1)-1/(x+101)]=200/[(x+1)(x+101)]
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