$x = \frac{1}{1+\sqrt{2}+\sqrt{3}}$、 $y = \frac{1}{1+\sqrt{2}-\sqrt{3}}$ 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、$x+y$ を計算する。

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

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

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

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

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

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

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

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

次に、$\frac{1}{x+y}$ を計算する。

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

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

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

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

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

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

3. 最終的な答え

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