$x = \frac{1}{\sqrt{3} + \sqrt{2}}$ 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) $x^4 + \frac{1}{x^4}$

1. 問題の内容

$x = \frac{1}{\sqrt{3} + \sqrt{2}}$ 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) $x^4 + \frac{1}{x^4}$

2. 解き方の手順

まず、$x$ を簡単にする。$x$ の分母を有利化する。

$x = \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$

次に、$\frac{1}{x}$ を求める。

$\frac{1}{x} = \frac{1}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2}$

(1) $x + \frac{1}{x} = (\sqrt{3} - \sqrt{2}) + (\sqrt{3} + \sqrt{2}) = 2\sqrt{3}$

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

$x^2 + \frac{1}{x^2} = (2\sqrt{3})^2 - 2 = 12 - 2 = 10$

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

$x^3 + \frac{1}{x^3} = (2\sqrt{3})^3 - 3(2\sqrt{3}) = 8 \cdot 3\sqrt{3} - 6\sqrt{3} = 24\sqrt{3} - 6\sqrt{3} = 18\sqrt{3}$

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

$x^4 + \frac{1}{x^4} = (10)^2 - 2 = 100 - 2 = 98$

3. 最終的な答え

(1) $x + \frac{1}{x} = 2\sqrt{3}$

(2) $x^2 + \frac{1}{x^2} = 10$

(3) $x^3 + \frac{1}{x^3} = 18\sqrt{3}$

(4) $x^4 + \frac{1}{x^4} = 98$