$x = \sqrt{2} - 1$ のとき、次の式の値を求めよ。 (1) $x + \frac{1}{x}$ (2) $x^2 + \frac{1}{x^2}$ (3) $x^3 + \frac{1}{x^3}$ (4) $x^4 + \frac{1}{x^4}$ (5) $x^5 + \frac{1}{x^5}$

1. 問題の内容

$x = \sqrt{2} - 1$ のとき、次の式の値を求めよ。

(1) $x + \frac{1}{x}$

(2) $x^2 + \frac{1}{x^2}$

(3) $x^3 + \frac{1}{x^3}$

(4) $x^4 + \frac{1}{x^4}$

(5) $x^5 + \frac{1}{x^5}$

2. 解き方の手順

まず、$\frac{1}{x}$ を求める。

$\frac{1}{x} = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1$

(1) $x + \frac{1}{x} = (\sqrt{2} - 1) + (\sqrt{2} + 1) = 2\sqrt{2}$

(2) $x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = (2\sqrt{2})^2 - 2 = 8 - 2 = 6$

(3) $x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = (2\sqrt{2})^3 - 3(2\sqrt{2}) = 16\sqrt{2} - 6\sqrt{2} = 10\sqrt{2}$

(4) $x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = (6)^2 - 2 = 36 - 2 = 34$

(5) $x^5 + \frac{1}{x^5} = (x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x}) = (6)(10\sqrt{2}) - 2\sqrt{2} = 60\sqrt{2} - 2\sqrt{2} = 58\sqrt{2}$

3. 最終的な答え

(1) $2\sqrt{2}$

(2) $6$

(3) $10\sqrt{2}$

(4) $34$

(5) $58\sqrt{2}$