$x = \frac{2}{\sqrt{7}+\sqrt{3}}$、$y = \frac{2}{\sqrt{7}-\sqrt{3}}$のとき、$x+y$と$xy$の値を求めよ。

1. 問題の内容

$x = \frac{2}{\sqrt{7}+\sqrt{3}}$、$y = \frac{2}{\sqrt{7}-\sqrt{3}}$のとき、$x+y$と$xy$の値を求めよ。

2. 解き方の手順

まず、$x$と$y$の分母を有理化する。

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

$y = \frac{2}{\sqrt{7}-\sqrt{3}} = \frac{2(\sqrt{7}+\sqrt{3})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})} = \frac{2(\sqrt{7}+\sqrt{3})}{7-3} = \frac{2(\sqrt{7}+\sqrt{3})}{4} = \frac{\sqrt{7}+\sqrt{3}}{2}$

次に、$x+y$と$xy$を計算する。

$x+y = \frac{\sqrt{7}-\sqrt{3}}{2} + \frac{\sqrt{7}+\sqrt{3}}{2} = \frac{\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}}{2} = \frac{2\sqrt{7}}{2} = \sqrt{7}$

$xy = \frac{\sqrt{7}-\sqrt{3}}{2} \cdot \frac{\sqrt{7}+\sqrt{3}}{2} = \frac{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}{4} = \frac{7-3}{4} = \frac{4}{4} = 1$

3. 最終的な答え

$x+y = \sqrt{7}$

$xy = 1$