与えられた6つの式をそれぞれ計算し、簡単化します。 (1) $\sqrt{-2} - \sqrt{-8}$ (2) $-\sqrt{-3} - \sqrt{-4}$ (3) $(2 + \sqrt{-3})^2$ (4) $(-1 + \sqrt{-2})^2$ (5) $\frac{\sqrt{-25}}{5}$ (6) $\frac{\sqrt{6}}{\sqrt{-2}}$

はい、承知いたしました。問題を解いていきましょう。

1. 問題の内容

与えられた6つの式をそれぞれ計算し、簡単化します。

(1) $\sqrt{-2} - \sqrt{-8}$

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

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

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

(5) $\frac{\sqrt{-25}}{5}$

(6) $\frac{\sqrt{6}}{\sqrt{-2}}$

2. 解き方の手順

(1) $\sqrt{-2} - \sqrt{-8}$

$\sqrt{-2} = \sqrt{2}i$

$\sqrt{-8} = \sqrt{8}i = 2\sqrt{2}i$

$\sqrt{-2} - \sqrt{-8} = \sqrt{2}i - 2\sqrt{2}i = -\sqrt{2}i$

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

$\sqrt{-3} = \sqrt{3}i$

$\sqrt{-4} = 2i$

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

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

$\sqrt{-3} = \sqrt{3}i$

$(2 + \sqrt{-3})^2 = (2 + \sqrt{3}i)^2 = 4 + 4\sqrt{3}i + (\sqrt{3}i)^2 = 4 + 4\sqrt{3}i - 3 = 1 + 4\sqrt{3}i$

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

$\sqrt{-2} = \sqrt{2}i$

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

(5) $\frac{\sqrt{-25}}{5}$

$\sqrt{-25} = 5i$

$\frac{\sqrt{-25}}{5} = \frac{5i}{5} = i$

(6) $\frac{\sqrt{6}}{\sqrt{-2}}$

$\sqrt{-2} = \sqrt{2}i$

$\frac{\sqrt{6}}{\sqrt{-2}} = \frac{\sqrt{6}}{\sqrt{2}i} = \frac{\sqrt{3}}{i} = \frac{\sqrt{3}i}{i^2} = -\sqrt{3}i$

3. 最終的な答え

(1) $-\sqrt{2}i$

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

(3) $1 + 4\sqrt{3}i$

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

(5) $i$

(6) $-\sqrt{3}i$