与えられた6つの式を計算する問題です。 (1) $(1 + \sqrt{5} + \sqrt{6})(1 + \sqrt{5} - \sqrt{6})$ (2) $\frac{1 + \sqrt{3}}{2 - \sqrt{3}} - \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ (3) $(\sqrt{-3} + \sqrt{2})(\sqrt{-18} - \sqrt{12})$ (4) $\frac{3 + \sqrt{-2}}{3 - \sqrt{-2}} + \frac{3 - \sqrt{-2}}{3 + \sqrt{-2}}$ (5) $(1 - i)^3$ (6) $\frac{\sqrt{3} + i}{\sqrt{3} - i} - \frac{\sqrt{3} - i}{\sqrt{3} + i}$

はい、承知いたしました。与えられた数学の問題を解いていきます。

1. 問題の内容

与えられた6つの式を計算する問題です。

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

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

(3) $(\sqrt{-3} + \sqrt{2})(\sqrt{-18} - \sqrt{12})$

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

(5) $(1 - i)^3$

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

2. 解き方の手順

(1)

$A = 1 + \sqrt{5}$とおくと、

$(A + \sqrt{6})(A - \sqrt{6}) = A^2 - (\sqrt{6})^2 = (1 + \sqrt{5})^2 - 6 = 1 + 2\sqrt{5} + 5 - 6 = 2\sqrt{5}$

(2)

$\frac{1 + \sqrt{3}}{2 - \sqrt{3}} - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(1 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} - \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{2 + \sqrt{3} + 2\sqrt{3} + 3}{4 - 3} - \frac{3 - 2\sqrt{3} + 1}{3 - 1} = (5 + 3\sqrt{3}) - \frac{4 - 2\sqrt{3}}{2} = 5 + 3\sqrt{3} - (2 - \sqrt{3}) = 3 + 4\sqrt{3}$

(3)

$\sqrt{-3} = \sqrt{3}i$, $\sqrt{-18} = \sqrt{18}i = 3\sqrt{2}i$, $\sqrt{12} = 2\sqrt{3}$

$(\sqrt{-3} + \sqrt{2})(\sqrt{-18} - \sqrt{12}) = (\sqrt{3}i + \sqrt{2})(3\sqrt{2}i - 2\sqrt{3}) = 3\sqrt{6}i^2 - 2\sqrt{9}i + 3\sqrt{4}i - 2\sqrt{6} = -3\sqrt{6} - 6i + 6i - 2\sqrt{6} = -5\sqrt{6}$

(4)

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

$\frac{3 + \sqrt{-2}}{3 - \sqrt{-2}} + \frac{3 - \sqrt{-2}}{3 + \sqrt{-2}} = \frac{3 + \sqrt{2}i}{3 - \sqrt{2}i} + \frac{3 - \sqrt{2}i}{3 + \sqrt{2}i} = \frac{(3 + \sqrt{2}i)^2 + (3 - \sqrt{2}i)^2}{(3 - \sqrt{2}i)(3 + \sqrt{2}i)} = \frac{(9 + 6\sqrt{2}i - 2) + (9 - 6\sqrt{2}i - 2)}{9 + 2} = \frac{7 + 6\sqrt{2}i + 7 - 6\sqrt{2}i}{11} = \frac{14}{11}$

(5)

$(1 - i)^3 = (1 - i)^2(1 - i) = (1 - 2i + i^2)(1 - i) = (1 - 2i - 1)(1 - i) = -2i(1 - i) = -2i + 2i^2 = -2i - 2 = -2 - 2i$

(6)

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

3. 最終的な答え

(1) $2\sqrt{5}$

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

(3) $-5\sqrt{6}$

(4) $\frac{14}{11}$

(5) $-2 - 2i$

(6) $\sqrt{3}i$