$x = \frac{3-\sqrt{5}}{2}$ 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) $x^2 - \frac{1}{x^2}$ (4) $x^4 + \frac{1}{x^4}$ (5) $x^4 - \frac{1}{x^4}$

1. 問題の内容

$x = \frac{3-\sqrt{5}}{2}$ 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) $x^2 - \frac{1}{x^2}$

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

(5) $x^4 - \frac{1}{x^4}$

2. 解き方の手順

まず、$\frac{1}{x}$ を求める。

$\frac{1}{x} = \frac{2}{3 - \sqrt{5}} = \frac{2(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{2(3 + \sqrt{5})}{9 - 5} = \frac{2(3 + \sqrt{5})}{4} = \frac{3 + \sqrt{5}}{2}$

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

$x + \frac{1}{x} = \frac{3 - \sqrt{5}}{2} + \frac{3 + \sqrt{5}}{2} = \frac{3 - \sqrt{5} + 3 + \sqrt{5}}{2} = \frac{6}{2} = 3$

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

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

$x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = 3^2 - 2 = 9 - 2 = 7$

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

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

$x - \frac{1}{x} = \frac{3 - \sqrt{5}}{2} - \frac{3 + \sqrt{5}}{2} = \frac{3 - \sqrt{5} - 3 - \sqrt{5}}{2} = \frac{-2\sqrt{5}}{2} = -\sqrt{5}$

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

$x^2 - \frac{1}{x^2} = 3 \times (-\sqrt{5}) = -3\sqrt{5}$

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

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

$x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = 7^2 - 2 = 49 - 2 = 47$

(5) $x^4 - \frac{1}{x^4}$

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

$x^4 - \frac{1}{x^4} = 7 \times (-3\sqrt{5}) = -21\sqrt{5}$

3. 最終的な答え

(1) 3

(2) 7

(3) $-3\sqrt{5}$

(4) 47

(5) $-21\sqrt{5}$