由 y = e ^ln(1+2^1/2) + e^-ln(1+2^1/2)
點樣計到 y = 2^3/2
(「e」 係 natural exponential function, 「ln 」係 natural log,)
後面嗰嚿有負號
我用計數機計過係一樣，但中間啲步驟係點樣？
maths question (extract)
2009-12-06 2:49 am
回答 (2)
2009-12-06 2:58 am
✔ 最佳答案
Note that alnb = ln(b^a), where a and b are real, b > 0So, y
= e^ln(1 + 2^1/2) + e^-ln(1 + 2^1/2)
= e^ln(1 + 2^1/2) + e^ln(1 + 2^1/2)^-1
= (1 + 2^1/2) + (1 + 2^1/2)^-1
= [(1 + 2^1/2)^2 + 1] / (1 + 2^1/2)
= [1 + 2(2^1/2) + (2^1/2)^2 + 1] / (1 + 2^1/2)
= (4 + 2^3/2) / (1 + 2^1/2)
= (4 + 2^3/2)(1 - 2^1/2) / [(1 + 2^1/2)(1 - 2^1/2)]
= [4 - 4(2^1/2) + 2(2^1/2) - (2^3/2)(2^1/2)] / [1 - (2^1/2)^2]
= (4 - 2(2^1/2) - 4) / (-1)
= 2(2^1/2)
= 2^3/2
參考: Physics king
2009-12-06 6:59 am
唔駛咁長
(1 + √2) + 1/(1 + √2)
= 1 + √2 + (1 - √2)/(1 - 2)
= 1 + √2 - 1 + √2
= 2√2
= 2^(3/2)
(1 + √2) + 1/(1 + √2)
= 1 + √2 + (1 - √2)/(1 - 2)
= 1 + √2 - 1 + √2
= 2√2
= 2^(3/2)
收錄日期: 2021-04-19 20:46:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091205000051KK01163