$x = \frac{1}{1 + \sqrt{2} + \sqrt{3}}$、 $y = \frac{1}{1 + \sqrt{2} - \sqrt{3}}$ のとき、$\frac{1}{x+y}$ の値を求めよ。

1. 問題の内容

$x = \frac{1}{1 + \sqrt{2} + \sqrt{3}}$、 $y = \frac{1}{1 + \sqrt{2} - \sqrt{3}}$ のとき、$\frac{1}{x+y}$ の値を求めよ。

2. 解き方の手順

まず、$x+y$ を計算します。

$x + y = \frac{1}{1 + \sqrt{2} + \sqrt{3}} + \frac{1}{1 + \sqrt{2} - \sqrt{3}}$

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

$x + y = \frac{2 + 2\sqrt{2}}{(1 + \sqrt{2})^2 - (\sqrt{3})^2}$

$x + y = \frac{2 + 2\sqrt{2}}{1 + 2\sqrt{2} + 2 - 3}$

$x + y = \frac{2 + 2\sqrt{2}}{2\sqrt{2}}$

$x + y = \frac{2(1 + \sqrt{2})}{2\sqrt{2}}$

$x + y = \frac{1 + \sqrt{2}}{\sqrt{2}}$

$x + y = \frac{(1 + \sqrt{2})\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$

$x + y = \frac{\sqrt{2} + 2}{2}$

次に、$\frac{1}{x+y}$ を計算します。

$\frac{1}{x+y} = \frac{1}{\frac{\sqrt{2} + 2}{2}}$

$\frac{1}{x+y} = \frac{2}{\sqrt{2} + 2}$

$\frac{1}{x+y} = \frac{2(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})}$

$\frac{1}{x+y} = \frac{2(2 - \sqrt{2})}{4 - 2}$

$\frac{1}{x+y} = \frac{2(2 - \sqrt{2})}{2}$

$\frac{1}{x+y} = 2 - \sqrt{2}$

3. 最終的な答え

$2 - \sqrt{2}$