与えられた$x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$に対して、以下の値を求める問題です。 (1) $x + \frac{1}{x}$, $x^2 - \frac{1}{x^2}$, $x^2 + \frac{1}{x^2}$, $x^4 + \frac{1}{x^4}$ (2) $x$の小数部分を$a$とするとき、$a$, $a^2 + a$, $\frac{\sqrt{a+1} - \sqrt{a}}{\sqrt{a+1} + \sqrt{a}}$ (3) $\sqrt{x^2 - 6x + 9} + \sqrt{9x^2 + 6x + 1}$

1. 問題の内容

与えられた$x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$に対して、以下の値を求める問題です。

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

(2) $x$の小数部分を$a$とするとき、$a$, $a^2 + a$, $\frac{\sqrt{a+1} - \sqrt{a}}{\sqrt{a+1} + \sqrt{a}}$

(3) $\sqrt{x^2 - 6x + 9} + \sqrt{9x^2 + 6x + 1}$

2. 解き方の手順

(1)

まず、$x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} = \frac{(\sqrt{5} + 1)^2}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{5 + 2\sqrt{5} + 1}{5 - 1} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$

$\frac{1}{x} = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} - 1)^2}{(\sqrt{5} + 1)(\sqrt{5} - 1)} = \frac{5 - 2\sqrt{5} + 1}{5 - 1} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$

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

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

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

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

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

(2)

$x = \frac{3 + \sqrt{5}}{2} = \frac{3 + 2.236\dots}{2} = \frac{5.236\dots}{2} = 2.618\dots$

よって、$x$の整数部分は2であり、小数部分$a = x - 2 = \frac{3 + \sqrt{5}}{2} - 2 = \frac{3 + \sqrt{5} - 4}{2} = \frac{\sqrt{5} - 1}{2}$

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

$\frac{\sqrt{a+1} - \sqrt{a}}{\sqrt{a+1} + \sqrt{a}} = \frac{(\sqrt{a+1} - \sqrt{a})^2}{(\sqrt{a+1} + \sqrt{a})(\sqrt{a+1} - \sqrt{a})} = \frac{a+1 - 2\sqrt{a(a+1)} + a}{a+1 - a} = 2a + 1 - 2\sqrt{a(a+1)} = 2a + 1 - 2\sqrt{a^2 + a} = 2(\frac{\sqrt{5} - 1}{2}) + 1 - 2\sqrt{1} = \sqrt{5} - 1 + 1 - 2 = \sqrt{5} - 2$

(3)

$\sqrt{x^2 - 6x + 9} + \sqrt{9x^2 + 6x + 1} = \sqrt{(x-3)^2} + \sqrt{(3x+1)^2} = |x - 3| + |3x + 1|$

$x = \frac{3 + \sqrt{5}}{2} = 2.618\dots$

$x - 3 = 2.618\dots - 3 = -0.382\dots < 0$

$3x + 1 = 3(\frac{3 + \sqrt{5}}{2}) + 1 = \frac{9 + 3\sqrt{5}}{2} + \frac{2}{2} = \frac{11 + 3\sqrt{5}}{2} > 0$

$|x - 3| + |3x + 1| = -(x - 3) + (3x + 1) = -x + 3 + 3x + 1 = 2x + 4 = 2(\frac{3 + \sqrt{5}}{2}) + 4 = 3 + \sqrt{5} + 4 = 7 + \sqrt{5}$

3. 最終的な答え

(1) $x + \frac{1}{x} = 3$, $x^2 - \frac{1}{x^2} = 3\sqrt{5}$, $x^2 + \frac{1}{x^2} = 7$, $x^4 + \frac{1}{x^4} = 47$

(2) $a = \frac{\sqrt{5} - 1}{2}$, $a^2 + a = 1$, $\frac{\sqrt{a+1} - \sqrt{a}}{\sqrt{a+1} + \sqrt{a}} = \sqrt{5} - 2$

(3) $\sqrt{x^2 - 6x + 9} + \sqrt{9x^2 + 6x + 1} = 7 + \sqrt{5}$