請解方程式： 2^2x + 2 – 5(2^x) + 1= 0?
回答 (2)
2016-06-28 4:03 am
2^(2x + 2) – 5(2^x) + 1= 0
2^x=1 or 2^x=1/4
x=0 或 x=-2
2^x=1 or 2^x=1/4
x=0 或 x=-2
2016-06-12 9:46 am
2^(2x + 2) - 5(2^x) + 1 = 0
2^(2x) 2^2 - 5(2^x) + 1 = 0
4(2^x)² - 5(2^x) + 1 = 0
[4(2^x) - 1](2^x - 1) = 0
2^x = 1/4 or 2^x = 1
x = -2 or 0
2^(2x) 2^2 - 5(2^x) + 1 = 0
4(2^x)² - 5(2^x) + 1 = 0
[4(2^x) - 1](2^x - 1) = 0
2^x = 1/4 or 2^x = 1
x = -2 or 0
收錄日期: 2021-04-18 15:07:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160612003110AA9zKAT