次の3つの式を計算します。 (1) $\sqrt{-5}\sqrt{-6}$ (2) $(2+\sqrt{-5})^2$ (3) $\frac{\sqrt{12}}{\sqrt{-3}}$

1. 問題の内容

次の3つの式を計算します。

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

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

(3) $\frac{\sqrt{12}}{\sqrt{-3}}$

2. 解き方の手順

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

$\sqrt{-5} = \sqrt{5}i$ および $\sqrt{-6} = \sqrt{6}i$ と表せるので、

$\sqrt{-5}\sqrt{-6} = \sqrt{5}i \cdot \sqrt{6}i = \sqrt{30}i^2 = -\sqrt{30}$

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

$(2+\sqrt{-5})^2 = (2+\sqrt{5}i)^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{5}i + (\sqrt{5}i)^2 = 4 + 4\sqrt{5}i + 5i^2 = 4 + 4\sqrt{5}i - 5 = -1 + 4\sqrt{5}i$

(3) $\frac{\sqrt{12}}{\sqrt{-3}}$

$\frac{\sqrt{12}}{\sqrt{-3}} = \frac{\sqrt{12}}{\sqrt{3}i} = \frac{\sqrt{4}\sqrt{3}}{\sqrt{3}i} = \frac{2\sqrt{3}}{\sqrt{3}i} = \frac{2}{i} = \frac{2i}{i^2} = \frac{2i}{-1} = -2i$

3. 最終的な答え

(1) $-\sqrt{30}$

(2) $-1 + 4\sqrt{5}i$

(3) $-2i$