与えられた6つの根号を含む分数を計算し、できる限り簡約化し、必要に応じて分母の有理化を行う問題です。 (1) $\frac{\sqrt{6}}{\sqrt{10}}$ (2) $\frac{\sqrt{18}}{\sqrt{12}}$ (3) $\frac{\sqrt{15}}{\sqrt{21}}$ (4) $\frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$ (5) $\frac{\sqrt{8}}{5\sqrt{24}}$ (6) $\frac{2\sqrt{27}}{3\sqrt{45}}$

1. 問題の内容

与えられた6つの根号を含む分数を計算し、できる限り簡約化し、必要に応じて分母の有理化を行う問題です。

(1) $\frac{\sqrt{6}}{\sqrt{10}}$

(2) $\frac{\sqrt{18}}{\sqrt{12}}$

(3) $\frac{\sqrt{15}}{\sqrt{21}}$

(4) $\frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$

(5) $\frac{\sqrt{8}}{5\sqrt{24}}$

(6) $\frac{2\sqrt{27}}{3\sqrt{45}}$

2. 解き方の手順

(1) $\frac{\sqrt{6}}{\sqrt{10}} = \sqrt{\frac{6}{10}} = \sqrt{\frac{3}{5}} = \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{\sqrt{15}}{5}$

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

(3) $\frac{\sqrt{15}}{\sqrt{21}} = \sqrt{\frac{15}{21}} = \sqrt{\frac{5}{7}} = \frac{\sqrt{5}}{\sqrt{7}} = \frac{\sqrt{5} \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{35}}{7}$

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

(5) $\frac{\sqrt{8}}{5\sqrt{24}} = \frac{\sqrt{8}}{5\sqrt{8 \times 3}} = \frac{\sqrt{8}}{5\sqrt{8}\sqrt{3}} = \frac{1}{5\sqrt{3}} = \frac{\sqrt{3}}{5\sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{5 \times 3} = \frac{\sqrt{3}}{15}$

(6) $\frac{2\sqrt{27}}{3\sqrt{45}} = \frac{2\sqrt{9 \times 3}}{3\sqrt{9 \times 5}} = \frac{2 \times 3\sqrt{3}}{3 \times 3\sqrt{5}} = \frac{6\sqrt{3}}{9\sqrt{5}} = \frac{2\sqrt{3}}{3\sqrt{5}} = \frac{2\sqrt{3} \times \sqrt{5}}{3\sqrt{5} \times \sqrt{5}} = \frac{2\sqrt{15}}{3 \times 5} = \frac{2\sqrt{15}}{15}$

3. 最終的な答え

(1) $\frac{\sqrt{15}}{5}$

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

(3) $\frac{\sqrt{35}}{7}$

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

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

(6) $\frac{2\sqrt{15}}{15}$