Re: [中學] 求解!三次根號的題目
: 試過配方 拆開比較係數 或是移項再立方但還是解不出來,希望各位大師能幫忙解惑!
有夠難打的 提供參考
令x=2^(1/3) ==>x^3 =2
令原式y=(x-1) ^ (1/3)
(x+1)^3 = x^3+ 3x^2 +3x +1 = 3(x^2+x+1)
^^^^
=2
左右2邊 *(x-1)
(x-1) (x+1)^3 = (x-1) 3(x^2+x+1) = 3 (x^3 -1) = 3
^^^
=2
3
(x-1) = --------------
(x+1)^3
左右開3次方
3^(1/3)
(x-1) ^(1/3) = y = ---------------
x+1
3^(1/3) (x^2-x+1)
y = ------------ * ---------------
(x+1) (x^2-x+1)
x=2^(1/3) 代入
3^(1/3) *{4^(1/3) -2^(1/3)+1}
= ------------------------------------
x^3 +1
^^
=2
3^(1/3) *{4^(1/3) -2^(1/3)+1}
= --------------------------------- {把 3^(1/3) 除到分母}
3
{4^(1/3) -2^(1/3)+1}
= ----------------------------- = (4/9)^(1/3) -(2/9)^(1/3)+(1/9)^(1/3)
9^(1/3)
a+b+c= 4/9 -2/9 +1/9 =1/3
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 111.251.196.151 (臺灣)
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1568509202.A.63E.html
※ 編輯: suker (111.251.196.151 臺灣), 09/15/2019 09:02:30
推
09/15 12:17,
4年前
, 1F
09/15 12:17, 1F
→
09/16 00:07,
4年前
, 2F
09/16 00:07, 2F
→
09/16 10:50,
4年前
, 3F
09/16 10:50, 3F
推
09/17 02:14,
4年前
, 4F
09/17 02:14, 4F
→
09/17 02:14,
4年前
, 5F
09/17 02:14, 5F
討論串 (同標題文章)