与えられた根号を含む式の乗除計算（問題4）と加減計算（問題5）を行う。

1. 問題の内容

与えられた根号を含む式の乗除計算（問題4）と加減計算（問題5）を行う。

2. 解き方の手順

**問題4**

(1) $\sqrt{14} \times \sqrt{5} = \sqrt{14 \times 5} = \sqrt{70}$

(2) $\sqrt{27} \times \sqrt{8} = \sqrt{27 \times 8} = \sqrt{216} = \sqrt{36 \times 6} = 6\sqrt{6}$

(3) $\sqrt{6} \times \sqrt{42} = \sqrt{6 \times 42} = \sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7}$

(4) $\sqrt{32} \times \sqrt{2} = \sqrt{32 \times 2} = \sqrt{64} = 8$

(5) $\sqrt{75} \div \sqrt{3} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5$

(6) $6\sqrt{5} \div \sqrt{6} = \frac{6\sqrt{5}}{\sqrt{6}} = \frac{6\sqrt{5}\sqrt{6}}{6} = \sqrt{30}$

**問題5**

(1) $8\sqrt{2} + 4\sqrt{2} = (8+4)\sqrt{2} = 12\sqrt{2}$

(2) $8\sqrt{3} - 5\sqrt{3} = (8-5)\sqrt{3} = 3\sqrt{3}$

(3) $4\sqrt{5} + 3\sqrt{7} - 2\sqrt{5} + \sqrt{7} = (4-2)\sqrt{5} + (3+1)\sqrt{7} = 2\sqrt{5} + 4\sqrt{7}$

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

(5) $\sqrt{32} + \sqrt{8} = \sqrt{16 \times 2} + \sqrt{4 \times 2} = 4\sqrt{2} + 2\sqrt{2} = (4+2)\sqrt{2} = 6\sqrt{2}$

(6) $\sqrt{45} - \sqrt{20} = \sqrt{9 \times 5} - \sqrt{4 \times 5} = 3\sqrt{5} - 2\sqrt{5} = (3-2)\sqrt{5} = \sqrt{5}$

(7) $\sqrt{5}(\sqrt{10}-1) = \sqrt{50} - \sqrt{5} = \sqrt{25 \times 2} - \sqrt{5} = 5\sqrt{2} - \sqrt{5}$

3. 最終的な答え

**問題4**

(1) $\sqrt{70}$

(2) $6\sqrt{6}$

(3) $6\sqrt{7}$

(4) $8$

(5) $5$

(6) $\sqrt{30}$

**問題5**

(1) $12\sqrt{2}$

(2) $3\sqrt{3}$

(3) $2\sqrt{5} + 4\sqrt{7}$

(4) $3\sqrt{3} - \sqrt{6}$

(5) $6\sqrt{2}$

(6) $\sqrt{5}$

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