次の6つの式を計算します。 (1) $\sqrt{6} \times \sqrt{15}$ (2) $\frac{\sqrt{48}}{\sqrt{2}}$ (3) $\sqrt{3} + \sqrt{27}$ (4) $(3 - \sqrt{2})(2 + 3\sqrt{2})$ (5) $(\sqrt{3} - 3\sqrt{2})^2$ (6) $(\sqrt{7} + \sqrt{10})(\sqrt{7} - \sqrt{10})$

1. 問題の内容

次の6つの式を計算します。

(1) $\sqrt{6} \times \sqrt{15}$

(2) $\frac{\sqrt{48}}{\sqrt{2}}$

(3) $\sqrt{3} + \sqrt{27}$

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

(5) $(\sqrt{3} - 3\sqrt{2})^2$

(6) $(\sqrt{7} + \sqrt{10})(\sqrt{7} - \sqrt{10})$

2. 解き方の手順

(1) $\sqrt{6} \times \sqrt{15} = \sqrt{6 \times 15} = \sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}$

(2) $\frac{\sqrt{48}}{\sqrt{2}} = \sqrt{\frac{48}{2}} = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$

(3) $\sqrt{3} + \sqrt{27} = \sqrt{3} + \sqrt{9 \times 3} = \sqrt{3} + 3\sqrt{3} = 4\sqrt{3}$

(4) $(3 - \sqrt{2})(2 + 3\sqrt{2}) = 3 \times 2 + 3 \times 3\sqrt{2} - \sqrt{2} \times 2 - \sqrt{2} \times 3\sqrt{2} = 6 + 9\sqrt{2} - 2\sqrt{2} - 3 \times 2 = 6 + 7\sqrt{2} - 6 = 7\sqrt{2}$

(5) $(\sqrt{3} - 3\sqrt{2})^2 = (\sqrt{3})^2 - 2 \times \sqrt{3} \times 3\sqrt{2} + (3\sqrt{2})^2 = 3 - 6\sqrt{6} + 9 \times 2 = 3 - 6\sqrt{6} + 18 = 21 - 6\sqrt{6}$

(6) $(\sqrt{7} + \sqrt{10})(\sqrt{7} - \sqrt{10}) = (\sqrt{7})^2 - (\sqrt{10})^2 = 7 - 10 = -3$

3. 最終的な答え

(1) $3\sqrt{10}$

(2) $2\sqrt{6}$

(3) $4\sqrt{3}$

(4) $7\sqrt{2}$

(5) $21 - 6\sqrt{6}$

(6) $-3$