can somebody prove (1 - E^(-2/x))/(1 - E^(-1/x)) equals to E^(-1/x)+1 where E is exponential?
回答 (3)
2016-06-21 2:55 pm
{1 - E^(-2/x)} / [1 - E^(-1/x)]
= {1^2 - [E^(-1/x)]^2} / [1 - E^(-1/x)]
= [1 + E^(-1/x)] [1 - E^(-1/x)] / [1 - E^(-1/x)]
= 1 + E^(-1/x)
= {1^2 - [E^(-1/x)]^2} / [1 - E^(-1/x)]
= [1 + E^(-1/x)] [1 - E^(-1/x)] / [1 - E^(-1/x)]
= 1 + E^(-1/x)
2016-06-21 5:40 pm
{[1 - (e^(-2/x))]/[1 - (e^(-1/x))]} * {[1 + (e^(-1/x))]/[1 + (e^(-1/x))]}
[1 - (e^(-2/x))] * [1 + (e^(-1/x))]/[1 - (e^(-2/x))]; from rule (e^a)(e^b) = e^(a + b)
[1 + (e^(-1/x))]
[1 - (e^(-2/x))] * [1 + (e^(-1/x))]/[1 - (e^(-2/x))]; from rule (e^a)(e^b) = e^(a + b)
[1 + (e^(-1/x))]
2016-06-21 5:01 pm
Proof:
[1-e^(-2/x)]/[1-e^(-1/x)]=
[1-1/e^(2/x)]/[1-1/e^(1/x)]=
[e^(2/x)-1][e^(1/x)]/
[e^(2/x)][e^(1/x)-1]=
[e^(1/x)+1][e^(1/x)-1]/
[e^(1/x)][e^(1/x)-1]=
[e^(1/x)+1]/e^(1/x)=
1+e^(-1/x)
[1-e^(-2/x)]/[1-e^(-1/x)]=
[1-1/e^(2/x)]/[1-1/e^(1/x)]=
[e^(2/x)-1][e^(1/x)]/
[e^(2/x)][e^(1/x)-1]=
[e^(1/x)+1][e^(1/x)-1]/
[e^(1/x)][e^(1/x)-1]=
[e^(1/x)+1]/e^(1/x)=
1+e^(-1/x)
收錄日期: 2021-04-18 15:09:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160621065206AAgiuyt