与えられた数式を計算する問題です。 (9) $(1+\sqrt{3})\sqrt{3}$ (10) $\sqrt{5}(\sqrt{10}+2) + 3\sqrt{45}$ (11) $(3\sqrt{5}-2)(4\sqrt{5}-3)$ (12) $(2\sqrt{3}+\sqrt{2})(2\sqrt{3}-\sqrt{2})$ (13) $(3\sqrt{2}+2\sqrt{5})^2$ (14) $\sqrt{6}(\sqrt{6}-\sqrt{2})^2$ (15) $(\sqrt{10}+6)(\sqrt{10}-2)$

はい、承知いたしました。

1. 問題の内容

与えられた数式を計算する問題です。

(9) $(1+\sqrt{3})\sqrt{3}$

(10) $\sqrt{5}(\sqrt{10}+2) + 3\sqrt{45}$

(11) $(3\sqrt{5}-2)(4\sqrt{5}-3)$

(12) $(2\sqrt{3}+\sqrt{2})(2\sqrt{3}-\sqrt{2})$

(13) $(3\sqrt{2}+2\sqrt{5})^2$

(14) $\sqrt{6}(\sqrt{6}-\sqrt{2})^2$

(15) $(\sqrt{10}+6)(\sqrt{10}-2)$

2. 解き方の手順

(9) $(1+\sqrt{3})\sqrt{3} = \sqrt{3} + (\sqrt{3})^2 = \sqrt{3} + 3 = 3 + \sqrt{3}$

(10) $\sqrt{5}(\sqrt{10}+2) + 3\sqrt{45} = \sqrt{5}\sqrt{10} + 2\sqrt{5} + 3\sqrt{9\cdot 5} = \sqrt{50} + 2\sqrt{5} + 3\cdot 3\sqrt{5} = \sqrt{25\cdot 2} + 2\sqrt{5} + 9\sqrt{5} = 5\sqrt{2} + 11\sqrt{5}$

(11) $(3\sqrt{5}-2)(4\sqrt{5}-3) = (3\sqrt{5})(4\sqrt{5}) - (3\sqrt{5})(3) - 2(4\sqrt{5}) + (-2)(-3) = 12 \cdot 5 - 9\sqrt{5} - 8\sqrt{5} + 6 = 60 - 17\sqrt{5} + 6 = 66 - 17\sqrt{5}$

(12) $(2\sqrt{3}+\sqrt{2})(2\sqrt{3}-\sqrt{2}) = (2\sqrt{3})^2 - (\sqrt{2})^2 = 4 \cdot 3 - 2 = 12 - 2 = 10$

(13) $(3\sqrt{2}+2\sqrt{5})^2 = (3\sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{5}) + (2\sqrt{5})^2 = 9 \cdot 2 + 12\sqrt{10} + 4 \cdot 5 = 18 + 12\sqrt{10} + 20 = 38 + 12\sqrt{10}$

(14) $\sqrt{6}(\sqrt{6}-\sqrt{2})^2 = \sqrt{6}((\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2) = \sqrt{6}(6 - 2\sqrt{12} + 2) = \sqrt{6}(8 - 2\sqrt{4 \cdot 3}) = \sqrt{6}(8 - 2\cdot 2\sqrt{3}) = \sqrt{6}(8 - 4\sqrt{3}) = 8\sqrt{6} - 4\sqrt{18} = 8\sqrt{6} - 4\sqrt{9\cdot 2} = 8\sqrt{6} - 4\cdot 3\sqrt{2} = 8\sqrt{6} - 12\sqrt{2}$

(15) $(\sqrt{10}+6)(\sqrt{10}-2) = (\sqrt{10})^2 + 6\sqrt{10} - 2\sqrt{10} - 12 = 10 + 4\sqrt{10} - 12 = -2 + 4\sqrt{10}$

3. 最終的な答え

(9) $3 + \sqrt{3}$

(10) $5\sqrt{2} + 11\sqrt{5}$

(11) $66 - 17\sqrt{5}$

(12) $10$

(13) $38 + 12\sqrt{10}$

(14) $8\sqrt{6} - 12\sqrt{2}$

(15) $-2 + 4\sqrt{10}$