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

1. 問題の内容

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

(1) $x+y$

(2) $xy$

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

2. 解き方の手順

まず、$x$と$y$の分母を有理化する。

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

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

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

3. 最終的な答え

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

(2) $xy = 2$

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