$x = \frac{2}{\sqrt{3} + \sqrt{2}}$、 $y = \frac{2}{\sqrt{3} - \sqrt{2}}$ のとき、$x^2 + y^2$ の値を求めよ。

1. 問題の内容

$x = \frac{2}{\sqrt{3} + \sqrt{2}}$、 $y = \frac{2}{\sqrt{3} - \sqrt{2}}$ のとき、$x^2 + y^2$ の値を求めよ。

2. 解き方の手順

まず、$x$と$y$の分母を有理化します。

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

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

次に、$x^2$と$y^2$を計算します。

$x^2 = [2(\sqrt{3} - \sqrt{2})]^2 = 4(\sqrt{3} - \sqrt{2})^2 = 4(3 - 2\sqrt{6} + 2) = 4(5 - 2\sqrt{6}) = 20 - 8\sqrt{6}$

$y^2 = [2(\sqrt{3} + \sqrt{2})]^2 = 4(\sqrt{3} + \sqrt{2})^2 = 4(3 + 2\sqrt{6} + 2) = 4(5 + 2\sqrt{6}) = 20 + 8\sqrt{6}$

最後に、$x^2 + y^2$を計算します。

$x^2 + y^2 = (20 - 8\sqrt{6}) + (20 + 8\sqrt{6}) = 40$

3. 最終的な答え

40