請幫忙解答 thanks
find x
(x^2-3x)^2-2x^2+6x+1=0
http://img338.imageshack.us/img338/7296/mathsqe6.png
A-Maths
2007-07-27 4:36 am
回答 (2)
2007-07-27 4:47 am
✔ 最佳答案
(x^2-3x)^2-2x^2+6x+1=0(x^2-3x)(x^2-3x-2)+1=0
.'.x^2-3x=1
x^2-3x-2= -1
x^2-3x-1=0
x= 3正負√13
2007-07-27 4:54 am
(x^2-3x)^2-2(x^2-3x)+1=0
(x^2-3x-1)^2 =0
x^2-3x-1 =0
x =( 3正負 開方根13)除2
2007-07-26 20:55:14 補充：
補充一下(x^2-3x)^2-2(x^2-3x) 1=0let u be x^2-3x條式就變左u^2-2u 1=0 (u-1)^2 =0
(x^2-3x-1)^2 =0
x^2-3x-1 =0
x =( 3正負 開方根13)除2
2007-07-26 20:55:14 補充：
補充一下(x^2-3x)^2-2(x^2-3x) 1=0let u be x^2-3x條式就變左u^2-2u 1=0 (u-1)^2 =0
收錄日期: 2021-04-13 00:52:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070726000051KK04732