(x+1)^2+P(×+1)≡x^2+Q (要step)
(x+1)^2+P(×+1)≡x^2+Q
求P及Q
左方
=(x+1)2+P(×+1)
=x2+2x+1+Px+1
-------------------↑ -------
上面點解“1”唔係“P”既?
(x+1)^2+P(×+1)≡x^2+Q (要step)
2007-12-24 11:10 pm
回答 (2)
2007-12-24 11:31 pm
✔ 最佳答案
(x+1)^2+P(×+1)≡x^2+Qx^2+2x+1+Px+P≡x^2+Q
x^2(P+2)x+P+1≡x^2+Q
比較係數
P+2=0
P=-2
Q=P+1=-2+1=-1
∴P=-2 , Q=-1
參考: me
收錄日期: 2021-04-19 22:55:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071224000051KK02541