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

1. 問題の内容

$x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, $y = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ のとき、次の式の値を求めよ。

(1) $x^2+y^2$

(2) $x^3y+xy^3$

(3) $\frac{x}{y}+\frac{y}{x}$

2. 解き方の手順

まず、$x$と$y$の分母を有理化する。

$x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} = \frac{(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{5-2\sqrt{15}+3}{5-3} = \frac{8-2\sqrt{15}}{2} = 4-\sqrt{15}$

$y = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{5+2\sqrt{15}+3}{5-3} = \frac{8+2\sqrt{15}}{2} = 4+\sqrt{15}$

$x+y = (4-\sqrt{15}) + (4+\sqrt{15}) = 8$

$xy = (4-\sqrt{15})(4+\sqrt{15}) = 16 - 15 = 1$

(1) $x^2+y^2 = (x+y)^2 - 2xy = 8^2 - 2(1) = 64 - 2 = 62$

(2) $x^3y+xy^3 = xy(x^2+y^2) = xy((x+y)^2 - 2xy) = 1(8^2 - 2(1)) = 64 - 2 = 62$

(3) $\frac{x}{y}+\frac{y}{x} = \frac{x^2+y^2}{xy} = \frac{62}{1} = 62$

3. 最終的な答え

(1) 62

(2) 62

(3) 62