Maths problem: how to do 29, 30, thanks?
回答 (1)
2020-08-02 8:33 pm
✔ 最佳答案
29.2^x 恆正. 故: 令 y = 2^x,
2^x(4^x+2^x-2) = 0
iff. (2^x)^2 + 2^x - 2 = 0
iff. y^2 + y - 2 = 0
iff. (y-1)(y+2) = 0
iff. y = 1
iff. x = 0
所以恰有一解 x = 0.
30.
(1-ai)/(1+ai) = (1-ai)^2/(1+a^2)
= [(1-a^2)-2ai]/(1+a^2)
虛部是
-2a/(1+a^2) = -3/5
即
10a = 3(1+a^2)
或即
(3a-1)(a-3) = 0
所以 a = 1/3 或 a = 3.
收錄日期: 2021-04-24 07:54:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200802110036AAmcApi