(1/a+1) + (1/a-1) ,,點通份呀??
+ 個負號落去呀??
(1/a+1 + 1/a-1) ,,點通份呀??
2006-11-10 3:55 am
回答 (4)
2006-11-10 4:07 am
✔ 最佳答案
(1/a+1) + (1/a-1)= [(a - 1) / (a + 1)(a - 1)] + [(a + 1) / (a + 1)(a - 1) ]
{所有數乘到分母 = (a + 1)(a - 1) (所有數分母既L.C.M.) 黎通分母}
= [(a - 1) + (a + 1)] / [(a + 1)(a - 1)]
= (2a) / (a + 1)(a - 1)
2006-11-10 4:08 am
1/(a+1) + 1/(a-1)
=(a-1)/[(a-1)(a+1)] + (a+1)/[(a-1)(a+1)]
= [(a-1) + (a+1)] / (a^2-1)
= (a-1+a-1) /(a^2-1)
= 2a / (a^2-1)
=(a-1)/[(a-1)(a+1)] + (a+1)/[(a-1)(a+1)]
= [(a-1) + (a+1)] / (a^2-1)
= (a-1+a-1) /(a^2-1)
= 2a / (a^2-1)
2006-11-10 4:02 am
我想......應該......
(1/a+1) + (1/a-1)
=1/a+1 + 1/a-1
=1/a+ 1/a
=2/a
(1/a+1) + (1/a-1)
=1/a+1 + 1/a-1
=1/a+ 1/a
=2/a
參考: 個人知識
2006-11-10 3:59 am
(1/a+1) + (1/a-1)
=(a-1+a+1)/(a+1)(a-1)
=(2a)/(a^2-1)
=(a-1+a+1)/(a+1)(a-1)
=(2a)/(a^2-1)
收錄日期: 2021-04-12 18:10:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061109000051KK03608