与えられた6つの根号を含む計算問題を解く。 (1) $\sqrt{15} \times \sqrt{45}$ (2) $\sqrt{18} \times (-\sqrt{8})$ (3) $\sqrt{24} \times \sqrt{21}$ (4) $-6\sqrt{6} \div \sqrt{18}$ (5) $\sqrt{180} \div \sqrt{3}$ (6) $7\sqrt{2} \div \sqrt{14}$

1. 問題の内容

与えられた6つの根号を含む計算問題を解く。

(1) $\sqrt{15} \times \sqrt{45}$

(2) $\sqrt{18} \times (-\sqrt{8})$

(3) $\sqrt{24} \times \sqrt{21}$

(4) $-6\sqrt{6} \div \sqrt{18}$

(5) $\sqrt{180} \div \sqrt{3}$

(6) $7\sqrt{2} \div \sqrt{14}$

2. 解き方の手順

(1)

$\sqrt{15} \times \sqrt{45} = \sqrt{15 \times 45} = \sqrt{15 \times 15 \times 3} = \sqrt{15^2 \times 3} = 15\sqrt{3}$

(2)

$\sqrt{18} \times (-\sqrt{8}) = -\sqrt{18 \times 8} = -\sqrt{2 \times 9 \times 8} = -\sqrt{2 \times 9 \times 2 \times 4} = -\sqrt{4 \times 9 \times 4} = -\sqrt{2^2 \times 3^2 \times 2^2} = -(2 \times 3 \times 2) = -12$

(3)

$\sqrt{24} \times \sqrt{21} = \sqrt{24 \times 21} = \sqrt{8 \times 3 \times 7 \times 3} = \sqrt{2^3 \times 3^2 \times 7} = \sqrt{2^2 \times 2 \times 3^2 \times 7} = 2 \times 3 \sqrt{2 \times 7} = 6\sqrt{14}$

(4)

$-6\sqrt{6} \div \sqrt{18} = -6\sqrt{\frac{6}{18}} = -6\sqrt{\frac{1}{3}} = -6 \times \frac{1}{\sqrt{3}} = -6 \times \frac{\sqrt{3}}{3} = -2\sqrt{3}$

(5)

$\sqrt{180} \div \sqrt{3} = \sqrt{\frac{180}{3}} = \sqrt{60} = \sqrt{4 \times 15} = \sqrt{2^2 \times 15} = 2\sqrt{15}$

(6)

$7\sqrt{2} \div \sqrt{14} = 7\sqrt{\frac{2}{14}} = 7\sqrt{\frac{1}{7}} = 7 \times \frac{1}{\sqrt{7}} = 7 \times \frac{\sqrt{7}}{7} = \sqrt{7}$

3. 最終的な答え

(1) $15\sqrt{3}$

(2) $-12$

(3) $6\sqrt{14}$

(4) $-2\sqrt{3}$

(5) $2\sqrt{15}$

(6) $\sqrt{7}$