二次方程2條

2009-10-18 9:37 pm

2^2x-3(2^x)-4=0



3^2x+3^x-2=0

回答 (1)

2009-10-18 10:16 pm
✔ 最佳答案
1.
2^2x - 3(2^x) - 4 = 0
(2^x)^2 - 3(2^x) - 4 = 0

Put y = 2^x
y^2 - 3y - 4 = 0
(y - 4)(y + 1) = 0
y = 4 or y = -1
2^x = 2^2 or 2^x = -1 (rejected)
x = 2


3^2x + 3^x - 2 = 0
(3^x)^2 + 3^x - 2 = 0

Put y = 3^x
y^2 + y - 2 = 0
(y - 1)(y + 2) = 0
y = 1 or y = -2
3^x = 1 or 3^x = -2 (rejected)
3^x = 3^o
x = 0


收錄日期: 2021-04-30 13:02:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091018000051KK00792

檢視 Wayback Machine 備份