解立方根問題(如檔案中的題目)

2008-08-29 6:55 am

回答 (5)

2008-08-30 4:07 am
✔ 最佳答案
2^(1/3) -1

=1/[4^(1/3)+2^(1/3)+1]

=3/[3*4^(1/3)+3*2^(1/3)+3]

=3/[2+3*4^(1/3)+3*2^(1/3)+1]

=3/[2^(1/3)+1]^3



[2^(1/3) -1]^(1/3)

= [3^(1/3)]/[2^(1/3)+1]

=[3^(1/3)]*[4^(1/3)-2^(1/3)+1] / 3

=(12/27)^(1/3)+(- 6/27)^(1/3)+(3/27)^(1/3)

=a^(1/3)+b^(1/3)+c^(1/3)

所以a=12/27,b= - 6/27,c= 3/27

a+b+c = 9/27 =1/3...........(解答)



同樣是參考老王的作法,不過有略作修改.....................
2008-09-05 7:20 am
這明明就是 Ramanujan 筆記本內的恆等式,舉例如下:

{(2)^{1/3} – 1)}^{1/3} = (1/9)^{1/3} – (2/9)^{1/3} + (4/9)^{1/3}
{5^{1/3} – 4^{1/3}}^{1/2} = (1/3){ 2^{1/3} + 20^{1/3} – 25^{1/3}}
{7(20)^{1/3} - 19}^{1/6} = (5/3)^{1/3} - (2/3)^{1/3}
{(3+2(5)^{1/4})/(3-2(5)^{1/4})}^{1/4} = (5^{1/4} + 1)/( 5^{1/4} - 1)
2008-08-29 2:12 pm
哇! 這樣解出來的話, 會不會被認為是抄襲的呀?
2008-08-29 11:00 am
是有理化嘛...?!

看到就懶了~"~
2008-08-29 9:59 am
http://tw.knowledge.yahoo.com/question/question?qid=1106100804250

偷看了老王的答案。

a+b+c=1/3

歹勢!

2008-08-29 10:35:53 補充:
不會啦!

本來知識就是這樣,總不能說我能抄襲老師的知識.

老實講就算有過程,有答案.

我也還是看不太懂.

說不定思瑜可以找到更簡單的方法.


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