問題 (3), (4), (5) をそれぞれ解きます。 (3) $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{8}} + \frac{1}{\sqrt{32}}$ (4) $\frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{2+\sqrt{3}}{2-\sqrt{3}}$ (5) $\frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+2}$

1. 問題の内容

問題 (3), (4), (5) をそれぞれ解きます。

(3) $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{8}} + \frac{1}{\sqrt{32}}$

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

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

2. 解き方の手順

(3) まず、各項を整理します。

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$

$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

したがって、

$\frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}} + \frac{1}{4\sqrt{2}} = \frac{4}{4\sqrt{2}} - \frac{2}{4\sqrt{2}} + \frac{1}{4\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{4 \times 2} = \frac{3\sqrt{2}}{8}$

(4) 各項を有理化します。

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

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

したがって、

$7-4\sqrt{3} + 7+4\sqrt{3} = 14$

(5) 各項を有理化します。

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

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

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

したがって、

$\sqrt{2}-1 + \sqrt{3}-\sqrt{2} + 2-\sqrt{3} = 1$

3. 最終的な答え

(3) $\frac{3\sqrt{2}}{8}$

(4) $14$

(5) $1$