与えられた複数の数式を計算して、簡単にします。

1. 問題の内容

与えられた複数の数式を計算して、簡単にします。

2. 解き方の手順

問題を一つずつ解いていきます。

(7) $\frac{1}{\sqrt{2}+1}$

分母を有理化します。

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

(9) $\frac{2\sqrt{2}}{3-\sqrt{5}}$

分母を有理化します。

$\frac{2\sqrt{2}}{3-\sqrt{5}} = \frac{2\sqrt{2}}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}} = \frac{2\sqrt{2}(3+\sqrt{5})}{9-5} = \frac{6\sqrt{2}+2\sqrt{10}}{4} = \frac{3\sqrt{2}+\sqrt{10}}{2}$

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

分母を有理化します。

$\frac{2\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{2\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(2\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{3-2} = 2(3) - 2\sqrt{6} - \sqrt{6} + 2 = 6 - 3\sqrt{6} + 2 = 8 - 3\sqrt{6}$

(10) $\frac{3-\sqrt{7}}{3+\sqrt{7}}$

分母を有理化します。

$\frac{3-\sqrt{7}}{3+\sqrt{7}} = \frac{3-\sqrt{7}}{3+\sqrt{7}} \times \frac{3-\sqrt{7}}{3-\sqrt{7}} = \frac{(3-\sqrt{7})^2}{9-7} = \frac{9 - 6\sqrt{7} + 7}{2} = \frac{16 - 6\sqrt{7}}{2} = 8 - 3\sqrt{7}$

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

分母を有理化します。

$\frac{2\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} = \frac{2\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} = \frac{(2\sqrt{2}-\sqrt{3})(\sqrt{2}-\sqrt{3})}{2-3} = \frac{4 - 2\sqrt{6} - \sqrt{6} + 3}{-1} = \frac{7 - 3\sqrt{6}}{-1} = -7 + 3\sqrt{6}$

(13) $\frac{3}{\sqrt{2}} - \frac{\sqrt{50}}{4} = \frac{3}{\sqrt{2}} - \frac{5\sqrt{2}}{4} = \frac{3\sqrt{2}}{2} - \frac{5\sqrt{2}}{4} = \frac{6\sqrt{2} - 5\sqrt{2}}{4} = \frac{\sqrt{2}}{4}$

(14) $\sqrt{18} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{8}} = 3\sqrt{2} - \frac{\sqrt{2}}{2} + \frac{1}{2\sqrt{2}} = 3\sqrt{2} - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{4} = \frac{12\sqrt{2} - 2\sqrt{2} + \sqrt{2}}{4} = \frac{11\sqrt{2}}{4}$

(15) $\frac{1}{\sqrt{5}} - \frac{1}{\sqrt{20}} - \frac{1}{\sqrt{125}} = \frac{1}{\sqrt{5}} - \frac{1}{2\sqrt{5}} - \frac{1}{5\sqrt{5}} = \frac{1}{\sqrt{5}}(1 - \frac{1}{2} - \frac{1}{5}) = \frac{1}{\sqrt{5}}(\frac{10-5-2}{10}) = \frac{3}{10\sqrt{5}} = \frac{3\sqrt{5}}{50}$

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

(17) $\frac{3\sqrt{5}-5\sqrt{3}}{\sqrt{5}+\sqrt{3}} + \frac{3\sqrt{5}+4\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{(3\sqrt{5}-5\sqrt{3})(\sqrt{5}-\sqrt{3}) + (3\sqrt{5}+4\sqrt{3})(\sqrt{5}+\sqrt{3})}{5-3} = \frac{(15 - 3\sqrt{15} - 5\sqrt{15} + 15) + (15 + 3\sqrt{15} + 4\sqrt{15} + 12)}{2} = \frac{(30 - 8\sqrt{15})+(27 + 7\sqrt{15})}{2} = \frac{57 - \sqrt{15}}{2}$

(18) $\frac{\sqrt{2}+1}{\sqrt{2}-1} + \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} + \frac{2-\sqrt{3}}{\sqrt{3}+\sqrt{2}} = \frac{(\sqrt{2}+1)^2}{2-1} + \frac{(\sqrt{3}+\sqrt{2})^2}{3-2} + \frac{(2-\sqrt{3})(\sqrt{3}-\sqrt{2})}{3-2} = (2+2\sqrt{2}+1) + (3+2\sqrt{6}+2) + (2\sqrt{3}-2\sqrt{2}-3+\sqrt{6}) = 3+2\sqrt{2} + 5+2\sqrt{6} + 2\sqrt{3}-2\sqrt{2}-3+\sqrt{6} = 5+2\sqrt{3}+3\sqrt{6}$

3. 最終的な答え

(7) $\sqrt{2}-1$

(9) $\frac{3\sqrt{2}+\sqrt{10}}{2}$

(11) $8 - 3\sqrt{6}$

(10) $8 - 3\sqrt{7}$

(12) $-7 + 3\sqrt{6}$

(13) $\frac{\sqrt{2}}{4}$

(14) $\frac{11\sqrt{2}}{4}$

(15) $\frac{3\sqrt{5}}{50}$

(16) $-1$

(17) $\frac{57-\sqrt{15}}{2}$

(18) $5+2\sqrt{3}+3\sqrt{6}$