以下の8つの計算問題を解きます。 (1) $\sqrt{3} \times (-\sqrt{5})$ (2) $\sqrt{18} \times \sqrt{48}$ (3) $\sqrt{21} \times \sqrt{35}$ (4) $\frac{\sqrt{6}}{2\sqrt{3}}$ (5) $(-3\sqrt{3})^2 \div \sqrt{18} \times \sqrt{12}$ (6) $3\sqrt{7} - 2\sqrt{2} + 6\sqrt{2} + \sqrt{7}$ (7) $\sqrt{32} - \sqrt{48} + \sqrt{50}$ (8) $\frac{6}{\sqrt{12}} - 4\sqrt{3} + \frac{9}{\sqrt{3}}$

1. 問題の内容

以下の8つの計算問題を解きます。

(1) $\sqrt{3} \times (-\sqrt{5})$

(2) $\sqrt{18} \times \sqrt{48}$

(3) $\sqrt{21} \times \sqrt{35}$

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

(5) $(-3\sqrt{3})^2 \div \sqrt{18} \times \sqrt{12}$

(6) $3\sqrt{7} - 2\sqrt{2} + 6\sqrt{2} + \sqrt{7}$

(7) $\sqrt{32} - \sqrt{48} + \sqrt{50}$

(8) $\frac{6}{\sqrt{12}} - 4\sqrt{3} + \frac{9}{\sqrt{3}}$

2. 解き方の手順

(1) $\sqrt{3} \times (-\sqrt{5}) = -\sqrt{3 \times 5} = -\sqrt{15}$

(2) $\sqrt{18} \times \sqrt{48} = \sqrt{18 \times 48} = \sqrt{2 \times 3^2 \times 2^4 \times 3} = \sqrt{2^5 \times 3^3} = \sqrt{2^4 \times 2 \times 3^2 \times 3} = 2^2 \times 3 \times \sqrt{2 \times 3} = 4 \times 3 \times \sqrt{6} = 12\sqrt{6}$

(3) $\sqrt{21} \times \sqrt{35} = \sqrt{21 \times 35} = \sqrt{3 \times 7 \times 5 \times 7} = \sqrt{3 \times 5 \times 7^2} = 7\sqrt{3 \times 5} = 7\sqrt{15}$

(4) $\frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2 \times 3}}{2\sqrt{3}} = \frac{\sqrt{2} \times \sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{2}}{2}$

(5) $(-3\sqrt{3})^2 \div \sqrt{18} \times \sqrt{12} = (-3)^2 \times (\sqrt{3})^2 \div \sqrt{2 \times 3^2} \times \sqrt{2^2 \times 3} = 9 \times 3 \div (3\sqrt{2}) \times (2\sqrt{3}) = 27 \div (3\sqrt{2}) \times (2\sqrt{3}) = \frac{27}{3\sqrt{2}} \times 2\sqrt{3} = \frac{9}{\sqrt{2}} \times 2\sqrt{3} = 9\sqrt{2} \times \frac{1}{\sqrt{2}} \times 2\sqrt{3} = 9\sqrt{2}\times \frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}\times2\sqrt{3} = \frac{18\sqrt{6}}{2}=9\sqrt{6}$

(6) $3\sqrt{7} - 2\sqrt{2} + 6\sqrt{2} + \sqrt{7} = (3+1)\sqrt{7} + (-2+6)\sqrt{2} = 4\sqrt{7} + 4\sqrt{2}$

(7) $\sqrt{32} - \sqrt{48} + \sqrt{50} = \sqrt{2^5} - \sqrt{2^4 \times 3} + \sqrt{2 \times 5^2} = \sqrt{2^4 \times 2} - \sqrt{2^4 \times 3} + \sqrt{5^2 \times 2} = 2^2\sqrt{2} - 2^2\sqrt{3} + 5\sqrt{2} = 4\sqrt{2} - 4\sqrt{3} + 5\sqrt{2} = (4+5)\sqrt{2} - 4\sqrt{3} = 9\sqrt{2} - 4\sqrt{3}$

(8) $\frac{6}{\sqrt{12}} - 4\sqrt{3} + \frac{9}{\sqrt{3}} = \frac{6}{2\sqrt{3}} - 4\sqrt{3} + \frac{9}{\sqrt{3}} = \frac{3}{\sqrt{3}} - 4\sqrt{3} + \frac{9}{\sqrt{3}} = \frac{3\sqrt{3}}{3} - 4\sqrt{3} + \frac{9\sqrt{3}}{3} = \sqrt{3} - 4\sqrt{3} + 3\sqrt{3} = (1-4+3)\sqrt{3} = 0\sqrt{3} = 0$

3. 最終的な答え

(1) $-\sqrt{15}$

(2) $12\sqrt{6}$

(3) $7\sqrt{15}$

(4) $\frac{\sqrt{2}}{2}$

(5) $9\sqrt{6}$

(6) $4\sqrt{7} + 4\sqrt{2}$

(7) $9\sqrt{2} - 4\sqrt{3}$

(8) $0$