設a^2x=√(3+√8),則(a^3x+a^-3x)/(a^x-a^-x)之值=?

2019-08-25 7:54 pm

回答 (2)

2019-08-25 10:44 pm
✔ 最佳答案
∵3+√8 = 3+2√2 = 1²+2√2+(√2)² = (1+√2)²
∴a²ˣ =√(3+√8) = 1+√2. . . . . . . . . . . . . ①

a³ˣ+a⁻³ˣ  (aˣ)³ + (1/aˣ)³
---------- = -------------------
aˣ-a⁻ˣ   aˣ - 1/aˣ
     (aˣ + 1/aˣ ) [(aˣ)² - aˣ(1/ aˣ) + (1/ aˣ)²]
    = -------------------------------------------------
         aˣ - 1/aˣ
     (a²ˣ + 1) [a²ˣ - 1 + (1/a²ˣ)]
    = ------------------------------------    (#)
         a²ˣ - 1
     (2+√2){√2 + [1/(1+√2)]}
    = ------------------------------------    ( 代入① )
         √2
     √2 (√2+1){√2 + [1/(1+√2)]}
    = -----------------------------------------
         √2
    = (√2+1){√2 + [1/(1+√2)]}
    = (√2+1)√2 + 1
    = 3+√2
______________________________________________
# 公式《x³+y³ = (x+y)( x²-xy+y²)》
2019-08-25 9:21 pm
令a^x=k,則 k²=√(3+√8) ⇒ k²=1+√2 ⇒(k²-1)²=2⇒k⁴-k²=k²+1。
則(a^3x+a^-3x)/(a^x-a^-x)
= (k³ + 1/k³)/(k - 1/k)
= (k⁶ + 1)/(k⁴ - k²)
= (k⁶ + 1)/(k² + 1)
= (k⁴ - k² + 1)
= k²+2
= 3+√2


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