(1+x)^q-(1-x)^q-(2x)^q=0
q>1 is a constant, solve for x.
(1+x)^q-(1-x)^q-(2x)^q=0
2010-04-26 1:03 am
回答 (3)
2010-04-27 5:58 am
✔ 最佳答案
=========================================================圖片參考：http://img175.imageshack.us/img175/5598/28703174.png
2010-04-26 21:58:42 補充：
http://img175.imageshack.us/img175/5598/28703174.png
2010-04-26 2:31 am
why x=1 is the only solution?
2010-04-27 01:00:49 補充：
What it q is not an integer?
2010-04-27 01:00:49 補充：
What it q is not an integer?
2010-04-26 2:14 am
put x=1,
(1+1)^q-(1-1)^q-(2*1)^q=2^q-0^q-2^q=0
(1+1)^q-(1-1)^q-(2*1)^q=2^q-0^q-2^q=0
收錄日期: 2021-04-23 23:22:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100425000051KK01043