与えられた複数の計算問題（根号を含む）を解き、空欄を埋める問題です。問題は、根号の計算、分母の有理化、式の展開など、様々な種類の計算を含んでいます。

1. 問題の内容

与えられた複数の計算問題（根号を含む）を解き、空欄を埋める問題です。問題は、根号の計算、分母の有理化、式の展開など、様々な種類の計算を含んでいます。

2. 解き方の手順

問1

(1) $\sqrt{6} \times \sqrt{7} = \sqrt{6 \times 7} = \sqrt{42}$

(2) $\sqrt{15} \div \sqrt{3} = \sqrt{\frac{15}{3}} = \sqrt{5}$

(3) $\sqrt{7} \times (-\sqrt{5}) = -\sqrt{7 \times 5} = -\sqrt{35}$

(4) $(-\sqrt{72}) \div \sqrt{12} = -\sqrt{\frac{72}{12}} = -\sqrt{6}$

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

(6) $(-\sqrt{85}) \div (-\sqrt{5}) = \sqrt{\frac{85}{5}} = \sqrt{17}$

問2

(1) $3\sqrt{5} = \sqrt{3^2 \times 5} = \sqrt{9 \times 5} = \sqrt{45}$

(2) $\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$

(3) $\sqrt{\frac{2}{25}} = \frac{\sqrt{2}}{\sqrt{25}} = \frac{\sqrt{2}}{5}$

(4) $\sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{\sqrt{36}} = \frac{\sqrt{5}}{6}$

(5) $\sqrt{30} \times \sqrt{6} = \sqrt{30 \times 6} = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$

(6) $\sqrt{12} \times \sqrt{5} = \sqrt{12 \times 5} = \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$

問3

(1) $\frac{3}{\sqrt{7}} = \frac{3 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{3\sqrt{7}}{7}$

(2) $\frac{\sqrt{5}}{\sqrt{10}} = \frac{\sqrt{5} \times \sqrt{10}}{\sqrt{10} \times \sqrt{10}} = \frac{\sqrt{50}}{10} = \frac{\sqrt{25 \times 2}}{10} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2}$

(3) $\frac{7}{\sqrt{14}} = \frac{7 \times \sqrt{14}}{\sqrt{14} \times \sqrt{14}} = \frac{7\sqrt{14}}{14} = \frac{\sqrt{14}}{2}$

(4) $\frac{\sqrt{3}}{2\sqrt{5}} = \frac{\sqrt{3} \times \sqrt{5}}{2\sqrt{5} \times \sqrt{5}} = \frac{\sqrt{15}}{2 \times 5} = \frac{\sqrt{15}}{10}$

問4

(1) $4\sqrt{5} + 3\sqrt{5} = (4+3)\sqrt{5} = 7\sqrt{5}$

(2) $\sqrt{24} + \sqrt{54} = \sqrt{4 \times 6} + \sqrt{9 \times 6} = 2\sqrt{6} + 3\sqrt{6} = (2+3)\sqrt{6} = 5\sqrt{6}$

問5

(1) $3\sqrt{6}(\sqrt{6} - 2) = 3\sqrt{6} \times \sqrt{6} - 3\sqrt{6} \times 2 = 3 \times 6 - 6\sqrt{6} = 18 - 6\sqrt{6}$

(2) $(\sqrt{2} + \sqrt{7})^2 = (\sqrt{2})^2 + 2\sqrt{2}\sqrt{7} + (\sqrt{7})^2 = 2 + 2\sqrt{14} + 7 = 9 + 2\sqrt{14}$

(3) $(5\sqrt{2} + 1)(2\sqrt{2} + 3) = 5\sqrt{2} \times 2\sqrt{2} + 5\sqrt{2} \times 3 + 1 \times 2\sqrt{2} + 1 \times 3 = 20 + 15\sqrt{2} + 2\sqrt{2} + 3 = 23 + 17\sqrt{2}$

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

(5) $\sqrt{32} + \frac{4}{\sqrt{2}} = \sqrt{16 \times 2} + \frac{4\sqrt{2}}{2} = 4\sqrt{2} + 2\sqrt{2} = 6\sqrt{2}$

(6) $\frac{5 - \sqrt{5}}{\sqrt{5}} = \frac{(5 - \sqrt{5})\sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{5\sqrt{5} - 5}{5} = \sqrt{5} - 1$

3. 最終的な答え

問1:

ア: $\sqrt{42}$

イ: $\sqrt{5}$

ウ: $-\sqrt{35}$

エ: $-\sqrt{6}$

オ: $\sqrt{14}$

カ: $\sqrt{17}$

問2:

キ: 45

ク: 2

ケ: 6

コ: 5

サ: 6

シ: 6

ス: 5

セ: 2

ソ: 15

問3:

タ: $\frac{3\sqrt{7}}{7}$

チ: $\frac{\sqrt{2}}{2}$

ツ: $\frac{\sqrt{14}}{2}$

テ: $\frac{\sqrt{15}}{10}$

問4:

ト: $7\sqrt{5}$

ナ: $5\sqrt{6}$

問5:

ニ: $18 - 6\sqrt{6}$

ヌ: $9 + 2\sqrt{14}$

ネ: $23 + 17\sqrt{2}$

ノ: 3

ハ: $6\sqrt{2}$

ヒ: $\sqrt{5} - 1$