$x = \frac{2}{\sqrt{5} + \sqrt{3}}$、$y = \frac{2}{\sqrt{5} - \sqrt{3}}$ のとき、次の式の値を求めよ。 (1) $x+y$ (2) $xy$ (3) $x^2 + y^2$ (4) $\frac{y}{x} + \frac{x}{y}$

1. 問題の内容

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

(1) $x+y$

(2) $xy$

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

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

2. 解き方の手順

まず、$x$ と $y$ の分母を有理化します。

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

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

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

(2) $xy = (\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = 5 - 3 = 2$

(3) $x^2 + y^2 = (x+y)^2 - 2xy = (2\sqrt{5})^2 - 2(2) = 4 \times 5 - 4 = 20 - 4 = 16$

(4) $\frac{y}{x} + \frac{x}{y} = \frac{y^2 + x^2}{xy} = \frac{x^2 + y^2}{xy} = \frac{16}{2} = 8$

3. 最終的な答え

(1) $x+y = 2\sqrt{5}$

(2) $xy = 2$

(3) $x^2 + y^2 = 16$

(4) $\frac{y}{x} + \frac{x}{y} = 8$