問題は、平方根の計算問題です。 1. 次の計算をしなさい。 (1) $\frac{3\sqrt{2}}{\sqrt{6}} - \frac{2}{\sqrt{3}}$ (2) $\frac{\sqrt{10}}{2} + \frac{6\sqrt{5}}{\sqrt{2}}$ (3) $\frac{3\sqrt{2}}{\sqrt{3}} - \frac{\sqrt{12}}{\sqrt{2}}$ (4) $\sqrt{8} - \frac{2}{\sqrt{2}} - \frac{2\sqrt{6}}{\sqrt{3}}$ (5) $\sqrt{7} - \frac{\sqrt{2}}{\sqrt{14}}$

はい、承知いたしました。数学の問題を解いて、指定された形式で回答します。

1. 問題の内容

問題は、平方根の計算問題です。

1. 次の計算をしなさい。

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

(2) $\frac{\sqrt{10}}{2} + \frac{6\sqrt{5}}{\sqrt{2}}$

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

(4) $\sqrt{8} - \frac{2}{\sqrt{2}} - \frac{2\sqrt{6}}{\sqrt{3}}$

(5) $\sqrt{7} - \frac{\sqrt{2}}{\sqrt{14}}$

2. 次の計算をしなさい。

(1) $(\sqrt{27} + \sqrt{90}) \div \sqrt{3}$

(2) $\sqrt{5}(\sqrt{15} - 2)$

(3) $-\sqrt{3}(\sqrt{27} + 2\sqrt{6})$

(4) $(\sqrt{3} - 2\sqrt{12}) \div 2\sqrt{6}$

2. 解き方の手順

1. (1)

$\frac{3\sqrt{2}}{\sqrt{6}} - \frac{2}{\sqrt{3}} = \frac{3\sqrt{2}\sqrt{6}}{6} - \frac{2\sqrt{3}}{3} = \frac{3\sqrt{12}}{6} - \frac{2\sqrt{3}}{3} = \frac{3 \cdot 2\sqrt{3}}{6} - \frac{2\sqrt{3}}{3} = \sqrt{3} - \frac{2\sqrt{3}}{3} = \frac{3\sqrt{3} - 2\sqrt{3}}{3} = \frac{\sqrt{3}}{3}$

2. (2)

$\frac{\sqrt{10}}{2} + \frac{6\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{10}}{2} + \frac{6\sqrt{5}\sqrt{2}}{2} = \frac{\sqrt{10}}{2} + \frac{6\sqrt{10}}{2} = \frac{7\sqrt{10}}{2}$

3. (3)

$\frac{3\sqrt{2}}{\sqrt{3}} - \frac{\sqrt{12}}{\sqrt{2}} = \frac{3\sqrt{2}\sqrt{3}}{3} - \frac{\sqrt{12}\sqrt{2}}{2} = \sqrt{6} - \frac{\sqrt{24}}{2} = \sqrt{6} - \frac{2\sqrt{6}}{2} = \sqrt{6} - \sqrt{6} = 0$

4. (4)

$\sqrt{8} - \frac{2}{\sqrt{2}} - \frac{2\sqrt{6}}{\sqrt{3}} = 2\sqrt{2} - \frac{2\sqrt{2}}{2} - \frac{2\sqrt{6}\sqrt{3}}{3} = 2\sqrt{2} - \sqrt{2} - \frac{2\sqrt{18}}{3} = \sqrt{2} - \frac{2 \cdot 3\sqrt{2}}{3} = \sqrt{2} - 2\sqrt{2} = -\sqrt{2}$

5. (5)

$\sqrt{7} - \frac{\sqrt{2}}{\sqrt{14}} = \sqrt{7} - \frac{\sqrt{2}\sqrt{14}}{14} = \sqrt{7} - \frac{\sqrt{28}}{14} = \sqrt{7} - \frac{2\sqrt{7}}{14} = \sqrt{7} - \frac{\sqrt{7}}{7} = \frac{7\sqrt{7} - \sqrt{7}}{7} = \frac{6\sqrt{7}}{7}$

6. (1)

$(\sqrt{27} + \sqrt{90}) \div \sqrt{3} = \frac{\sqrt{27}}{\sqrt{3}} + \frac{\sqrt{90}}{\sqrt{3}} = \sqrt{\frac{27}{3}} + \sqrt{\frac{90}{3}} = \sqrt{9} + \sqrt{30} = 3 + \sqrt{30}$

7. (2)

$\sqrt{5}(\sqrt{15} - 2) = \sqrt{5}\sqrt{15} - 2\sqrt{5} = \sqrt{75} - 2\sqrt{5} = 5\sqrt{3} - 2\sqrt{5}$

8. (3)

$-\sqrt{3}(\sqrt{27} + 2\sqrt{6}) = -\sqrt{3}\sqrt{27} - 2\sqrt{3}\sqrt{6} = -\sqrt{81} - 2\sqrt{18} = -9 - 2 \cdot 3\sqrt{2} = -9 - 6\sqrt{2}$

9. (4)

$(\sqrt{3} - 2\sqrt{12}) \div 2\sqrt{6} = \frac{\sqrt{3} - 2\sqrt{12}}{2\sqrt{6}} = \frac{\sqrt{3} - 4\sqrt{3}}{2\sqrt{6}} = \frac{-3\sqrt{3}}{2\sqrt{6}} = \frac{-3\sqrt{3}\sqrt{6}}{2 \cdot 6} = \frac{-3\sqrt{18}}{12} = \frac{-3 \cdot 3\sqrt{2}}{12} = \frac{-9\sqrt{2}}{12} = -\frac{3\sqrt{2}}{4}$

3. 最終的な答え

1. (1) $\frac{\sqrt{3}}{3}$

2. (2) $\frac{7\sqrt{10}}{2}$

3. (3) $0$

4. (4) $-\sqrt{2}$

5. (5) $\frac{6\sqrt{7}}{7}$

6. (1) $3 + \sqrt{30}$

7. (2) $5\sqrt{3} - 2\sqrt{5}$

8. (3) $-9 - 6\sqrt{2}$

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