(1) $\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} + \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}}$ を簡単にせよ。 (2) $\frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + 2} + \frac{1}{2 + \sqrt{5}}$ を簡単にせよ。

1. 問題の内容

(1) $\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} + \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}}$ を簡単にせよ。

(2) $\frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + 2} + \frac{1}{2 + \sqrt{5}}$ を簡単にせよ。

2. 解き方の手順

(1) 各項の分母を有理化する。

$\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} = \frac{3\sqrt{2}(\sqrt{6} - \sqrt{3})}{(\sqrt{6} + \sqrt{3})(\sqrt{6} - \sqrt{3})} = \frac{3\sqrt{2}(\sqrt{6} - \sqrt{3})}{6 - 3} = \frac{3\sqrt{12} - 3\sqrt{6}}{3} = \sqrt{12} - \sqrt{6} = 2\sqrt{3} - \sqrt{6}$

$\frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} = \frac{4\sqrt{3}(\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})} = \frac{4\sqrt{18} - 4\sqrt{6}}{6 - 2} = \frac{4(3\sqrt{2}) - 4\sqrt{6}}{4} = 3\sqrt{2} - \sqrt{6}$

$\frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{6}(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{\sqrt{18} - \sqrt{12}}{3 - 2} = 3\sqrt{2} - 2\sqrt{3}$

よって、

$2\sqrt{3} - \sqrt{6} - (3\sqrt{2} - \sqrt{6}) + 3\sqrt{2} - 2\sqrt{3} = 2\sqrt{3} - \sqrt{6} - 3\sqrt{2} + \sqrt{6} + 3\sqrt{2} - 2\sqrt{3} = 0$

(2) 各項の分母を有理化する。

$\frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \frac{\sqrt{2} - \sqrt{3}}{-1} = \sqrt{3} - \sqrt{2}$

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

$\frac{1}{2 + \sqrt{5}} = \frac{2 - \sqrt{5}}{(2 + \sqrt{5})(2 - \sqrt{5})} = \frac{2 - \sqrt{5}}{4 - 5} = \frac{2 - \sqrt{5}}{-1} = \sqrt{5} - 2$

よって、

$\sqrt{3} - \sqrt{2} + 2 - \sqrt{3} + \sqrt{5} - 2 = - \sqrt{2} + \sqrt{5} = \sqrt{5} - \sqrt{2}$

3. 最終的な答え

(1) 0

(2) $\sqrt{5} - \sqrt{2}$