次の数の分母を有理化する。 (1) $\frac{1}{\sqrt{5}+1}$ (2) $\frac{2}{4-\sqrt{6}}$ (4) $\frac{3\sqrt{7}}{\sqrt{7}-2}$ (5) $\frac{1}{\sqrt{3}+\sqrt{2}}$ (7) $\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}$ (8) $\frac{\sqrt{20}}{\sqrt{7}-\sqrt{5}}$ (10) $\frac{4+\sqrt{3}}{3+\sqrt{3}}$ (11) $\frac{3\sqrt{3}-\sqrt{6}}{2\sqrt{3}-\sqrt{6}}$

1. 問題の内容

次の数の分母を有理化する。

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

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

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

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

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

(8) $\frac{\sqrt{20}}{\sqrt{7}-\sqrt{5}}$

(10) $\frac{4+\sqrt{3}}{3+\sqrt{3}}$

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

2. 解き方の手順

分母を有理化するには、分母の共役な複素数を分母と分子にかける。

(1)

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

(2)

$\frac{2}{4-\sqrt{6}} = \frac{2}{4-\sqrt{6}} \cdot \frac{4+\sqrt{6}}{4+\sqrt{6}} = \frac{2(4+\sqrt{6})}{4^2 - (\sqrt{6})^2} = \frac{2(4+\sqrt{6})}{16-6} = \frac{2(4+\sqrt{6})}{10} = \frac{4+\sqrt{6}}{5}$

(4)

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

(5)

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

(7)

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

(8)

$\frac{\sqrt{20}}{\sqrt{7}-\sqrt{5}} = \frac{\sqrt{20}}{\sqrt{7}-\sqrt{5}} \cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} = \frac{\sqrt{20}(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^2 - (\sqrt{5})^2} = \frac{2\sqrt{5}(\sqrt{7}+\sqrt{5})}{7-5} = \frac{2\sqrt{35}+2 \cdot 5}{2} = \frac{2(\sqrt{35}+5)}{2} = \sqrt{35}+5$

(10)

$\frac{4+\sqrt{3}}{3+\sqrt{3}} = \frac{4+\sqrt{3}}{3+\sqrt{3}} \cdot \frac{3-\sqrt{3}}{3-\sqrt{3}} = \frac{(4+\sqrt{3})(3-\sqrt{3})}{3^2 - (\sqrt{3})^2} = \frac{12 - 4\sqrt{3} + 3\sqrt{3} - 3}{9-3} = \frac{9-\sqrt{3}}{6} = \frac{9}{6} - \frac{\sqrt{3}}{6} = \frac{3}{2} - \frac{\sqrt{3}}{6} = \frac{9-\sqrt{3}}{6}$

(11)

$\frac{3\sqrt{3}-\sqrt{6}}{2\sqrt{3}-\sqrt{6}} = \frac{3\sqrt{3}-\sqrt{6}}{2\sqrt{3}-\sqrt{6}} \cdot \frac{2\sqrt{3}+\sqrt{6}}{2\sqrt{3}+\sqrt{6}} = \frac{(3\sqrt{3}-\sqrt{6})(2\sqrt{3}+\sqrt{6})}{(2\sqrt{3})^2-(\sqrt{6})^2} = \frac{18+3\sqrt{18}-2\sqrt{18}-6}{12-6} = \frac{12+\sqrt{18}}{6} = \frac{12+3\sqrt{2}}{6} = \frac{4+\sqrt{2}}{2}$

3. 最終的な答え

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

(2) $\frac{4+\sqrt{6}}{5}$

(4) $7+2\sqrt{7}$

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

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

(8) $\sqrt{35}+5$

(10) $\frac{9-\sqrt{3}}{6}$

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