次の計算問題を解きます。 $\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

1. 問題の内容

次の計算問題を解きます。

$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

2. 解き方の手順

まず、それぞれの分数を有理化します。

$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{2}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} = \frac{\sqrt{6}+2}{3-2} = \sqrt{6}+2$

$\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{2}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \frac{\sqrt{6}-2}{3-2} = \sqrt{6}-2$

したがって、

$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}} = (\sqrt{6}+2) - (\sqrt{6}-2) = \sqrt{6}+2-\sqrt{6}+2 = 4$

3. 最終的な答え

4