与えられた6つの式を計算し、それぞれを簡略化すること。 (1) $\sqrt{13} \times \sqrt{65} \times \sqrt{15}$ (2) $\sqrt{12}+\sqrt{48}-\sqrt{27}$ (3) $(\sqrt{7}+\sqrt{6})^2$ (4) $(2\sqrt{3}-\sqrt{5})^2$ (5) $(\sqrt{2}-3)(3\sqrt{2}+1)$ (6) $8(4-\sqrt{7})-(4-\sqrt{7})^2$

## 問題の回答

1. 問題の内容

与えられた6つの式を計算し、それぞれを簡略化すること。

(1) $\sqrt{13} \times \sqrt{65} \times \sqrt{15}$

(2) $\sqrt{12}+\sqrt{48}-\sqrt{27}$

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

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

(5) $(\sqrt{2}-3)(3\sqrt{2}+1)$

(6) $8(4-\sqrt{7})-(4-\sqrt{7})^2$

2. 解き方の手順

(1)

$\sqrt{13} \times \sqrt{65} \times \sqrt{15} = \sqrt{13} \times \sqrt{13 \times 5} \times \sqrt{3 \times 5} = \sqrt{13^2 \times 5^2 \times 3} = 13 \times 5 \times \sqrt{3} = 65\sqrt{3}$

(2)

$\sqrt{12}+\sqrt{48}-\sqrt{27} = \sqrt{4 \times 3}+\sqrt{16 \times 3}-\sqrt{9 \times 3} = 2\sqrt{3}+4\sqrt{3}-3\sqrt{3} = (2+4-3)\sqrt{3} = 3\sqrt{3}$

(3)

$(\sqrt{7}+\sqrt{6})^2 = (\sqrt{7})^2 + 2\sqrt{7}\sqrt{6} + (\sqrt{6})^2 = 7 + 2\sqrt{42} + 6 = 13 + 2\sqrt{42}$

(4)

$(2\sqrt{3}-\sqrt{5})^2 = (2\sqrt{3})^2 - 2(2\sqrt{3})(\sqrt{5}) + (\sqrt{5})^2 = 4 \times 3 - 4\sqrt{15} + 5 = 12 - 4\sqrt{15} + 5 = 17 - 4\sqrt{15}$

(5)

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

(6)

$8(4-\sqrt{7})-(4-\sqrt{7})^2 = 32 - 8\sqrt{7} - (16 - 8\sqrt{7} + 7) = 32 - 8\sqrt{7} - (23 - 8\sqrt{7}) = 32 - 8\sqrt{7} - 23 + 8\sqrt{7} = 32 - 23 = 9$

3. 最終的な答え

(1) $65\sqrt{3}$

(2) $3\sqrt{3}$

(3) $13 + 2\sqrt{42}$

(4) $17 - 4\sqrt{15}$

(5) $3 - 8\sqrt{2}$

(6) $9$