Re: [中學] 高一數學

[Tiderus (修煉人生)]

※ 引述《alansh987 (祺瑞)》之銘言：
: http://i.imgur.com/SJYJJ5H.jpg

x + √(x^2+1) >0

y = f(x)= log[x + √(x^2+1)]

(10^y -x)^2 = x^2 + 1

x = (10^(2y)-1) / (2*10^y) = [10^y -10^(-y)] /2

: http://i.imgur.com/CKIVN0Q.jpg
: 拜託了 謝謝

2^a + 4^b = 2^(a-1) + 2^(a-1) + 4^b
≧ 3*[2^(a-1) * 2^(a-1) * 4^b]^(1/3) = 3*[2^(2a+2b-2)]^(1/3)= 3*2^2 = 12

2^(a-1) = 4^b = 4
=> a=3, b=1

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迷途中唯一的導航 是對自己誠實

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※ 編輯: Tiderus (123.240.91.95), 01/08/2016 02:28:09
→ alansh987   : 感謝!!!