与えられた8つの平方根に関する式を簡単にします。

1. 問題の内容

与えられた8つの平方根に関する式を簡単にします。

2. 解き方の手順

(1) $\sqrt{5} \cdot \sqrt{6}$

$\sqrt{5} \cdot \sqrt{6} = \sqrt{5 \cdot 6} = \sqrt{30}$

(2) $\frac{\sqrt{21}}{\sqrt{3}}$

$\frac{\sqrt{21}}{\sqrt{3}} = \sqrt{\frac{21}{3}} = \sqrt{7}$

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

$(-\sqrt{2})^3 = (-\sqrt{2}) \cdot (-\sqrt{2}) \cdot (-\sqrt{2}) = -2\sqrt{2}$

(4) $4\sqrt{3} + \sqrt{75} - \sqrt{12}$

$4\sqrt{3} + \sqrt{75} - \sqrt{12} = 4\sqrt{3} + \sqrt{25 \cdot 3} - \sqrt{4 \cdot 3} = 4\sqrt{3} + 5\sqrt{3} - 2\sqrt{3} = (4+5-2)\sqrt{3} = 7\sqrt{3}$

(5) $\sqrt{5}(\sqrt{2} - \sqrt{20})$

$\sqrt{5}(\sqrt{2} - \sqrt{20}) = \sqrt{5}\sqrt{2} - \sqrt{5}\sqrt{20} = \sqrt{10} - \sqrt{100} = \sqrt{10} - 10$

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

$(2\sqrt{3} + \sqrt{2})(\sqrt{3} + 3\sqrt{2}) = 2\sqrt{3}\sqrt{3} + 6\sqrt{3}\sqrt{2} + \sqrt{2}\sqrt{3} + 3\sqrt{2}\sqrt{2} = 2(3) + 6\sqrt{6} + \sqrt{6} + 3(2) = 6 + 7\sqrt{6} + 6 = 12 + 7\sqrt{6}$

(7) $(5\sqrt{6} - \sqrt{5})^2$

$(5\sqrt{6} - \sqrt{5})^2 = (5\sqrt{6})^2 - 2(5\sqrt{6})(\sqrt{5}) + (\sqrt{5})^2 = 25(6) - 10\sqrt{30} + 5 = 150 - 10\sqrt{30} + 5 = 155 - 10\sqrt{30}$

(8) $(\sqrt{10} + 2\sqrt{3})(\sqrt{10} - 2\sqrt{3})$

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

3. 最終的な答え

(1) $\sqrt{30}$

(2) $\sqrt{7}$

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

(4) $7\sqrt{3}$

(5) $\sqrt{10} - 10$

(6) $12 + 7\sqrt{6}$

(7) $155 - 10\sqrt{30}$

(8) $-2$