$x = 2 - \sqrt{3}$ のとき、以下の式の値を求める問題です。 (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

$x = 2 - \sqrt{3}$ のとき、以下の式の値を求める問題です。

(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、$x$ の逆数 $\frac{1}{x}$ を求めます。

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

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

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

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

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

$x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = 4^2 - 2 = 16 - 2 = 14$

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

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

$x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = 4^3 - 3(4) = 64 - 12 = 52$

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

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

$x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = 14^2 - 2 = 196 - 2 = 194$

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

$(x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) = x^5 + \frac{x^2}{x^3} + \frac{x^3}{x^2} + \frac{1}{x^5} = x^5 + \frac{1}{x} + x + \frac{1}{x^5} = x^5 + \frac{1}{x^5} + (x + \frac{1}{x})$

$x^5 + \frac{1}{x^5} = (x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x}) = (14)(52) - 4 = 728 - 4 = 724$

3. 最終的な答え

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

(2) $x^2 + \frac{1}{x^2} = 14$

(3) $x^3 + \frac{1}{x^3} = 52$

(4) $x^4 + \frac{1}{x^4} = 194$

(5) $x^5 + \frac{1}{x^5} = 724$