画像に示された8つの根号を含む計算問題を解きます。

1. 問題の内容

画像に示された8つの根号を含む計算問題を解きます。

2. 解き方の手順

(1) $\sqrt{3} + \sqrt{6} \times \sqrt{2}$

$\sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3}$

$\sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$

(2) $\sqrt{3} \times \sqrt{15} - \sqrt{20}$

$\sqrt{3} \times \sqrt{15} = \sqrt{45} = 3\sqrt{5}$

$\sqrt{20} = 2\sqrt{5}$

$3\sqrt{5} - 2\sqrt{5} = \sqrt{5}$

(3) $\sqrt{6} \times \sqrt{12} + \sqrt{8}$

$\sqrt{6} \times \sqrt{12} = \sqrt{72} = 6\sqrt{2}$

$\sqrt{8} = 2\sqrt{2}$

$6\sqrt{2} + 2\sqrt{2} = 8\sqrt{2}$

(4) $3\sqrt{20} - \sqrt{2} \times \sqrt{10}$

$3\sqrt{20} = 3 \times 2\sqrt{5} = 6\sqrt{5}$

$\sqrt{2} \times \sqrt{10} = \sqrt{20} = 2\sqrt{5}$

$6\sqrt{5} - 2\sqrt{5} = 4\sqrt{5}$

(5) $2\sqrt{3} + \sqrt{54} \div \sqrt{2}$

$\sqrt{54} = 3\sqrt{6}$

$3\sqrt{6} \div \sqrt{2} = 3\sqrt{3}$

$2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$

(6) $\sqrt{28} - \sqrt{56} \div \sqrt{8}$

$\sqrt{28} = 2\sqrt{7}$

$\sqrt{56} \div \sqrt{8} = \sqrt{7}$

$2\sqrt{7} - \sqrt{7} = \sqrt{7}$

(7) $\sqrt{40} \div \sqrt{5} + \sqrt{32}$

$\sqrt{40} \div \sqrt{5} = \sqrt{8} = 2\sqrt{2}$

$\sqrt{32} = 4\sqrt{2}$

$2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2}$

(8) $\sqrt{42} \div \sqrt{7} - \sqrt{96}$

$\sqrt{42} \div \sqrt{7} = \sqrt{6}$

$\sqrt{96} = 4\sqrt{6}$

$\sqrt{6} - 4\sqrt{6} = -3\sqrt{6}$

3. 最終的な答え

(1) $3\sqrt{3}$

(2) $\sqrt{5}$

(3) $8\sqrt{2}$

(4) $4\sqrt{5}$

(5) $5\sqrt{3}$

(6) $\sqrt{7}$

(7) $6\sqrt{2}$

(8) $-3\sqrt{6}$