(1) $\frac{1}{1+\sqrt{2}+\sqrt{3}}+\frac{1}{1+\sqrt{2}-\sqrt{3}}-\frac{1}{1-\sqrt{2}+\sqrt{3}}-\frac{1}{1-\sqrt{2}-\sqrt{3}}$ を簡単にせよ。 (2) $a\star b = \sqrt{\frac{a}{a-b}+\frac{a+b}{a}}$ で定義する。 $(\sqrt{6}+1)\star 2$ を分母に根号を含まない数で表せ。

1. 問題の内容

(1) $\frac{1}{1+\sqrt{2}+\sqrt{3}}+\frac{1}{1+\sqrt{2}-\sqrt{3}}-\frac{1}{1-\sqrt{2}+\sqrt{3}}-\frac{1}{1-\sqrt{2}-\sqrt{3}}$ を簡単にせよ。

(2) $a\star b = \sqrt{\frac{a}{a-b}+\frac{a+b}{a}}$ で定義する。 $(\sqrt{6}+1)\star 2$ を分母に根号を含まない数で表せ。

2. 解き方の手順

(1)

まず、最初の2つの分数を計算します。

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

次に、後ろの2つの分数を計算します。

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

よって、求める式は

$\frac{\sqrt{2}+2}{2} - \frac{2-\sqrt{2}}{2} = \frac{\sqrt{2}+2-2+\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

(2)

$a\star b = \sqrt{\frac{a}{a-b}+\frac{a+b}{a}}$ に $a = \sqrt{6}+1$, $b=2$ を代入すると、

$(\sqrt{6}+1)\star 2 = \sqrt{\frac{\sqrt{6}+1}{\sqrt{6}+1-2}+\frac{\sqrt{6}+1+2}{\sqrt{6}+1}} = \sqrt{\frac{\sqrt{6}+1}{\sqrt{6}-1}+\frac{\sqrt{6}+3}{\sqrt{6}+1}}$

$\frac{\sqrt{6}+1}{\sqrt{6}-1} = \frac{(\sqrt{6}+1)(\sqrt{6}+1)}{(\sqrt{6}-1)(\sqrt{6}+1)} = \frac{6+2\sqrt{6}+1}{6-1} = \frac{7+2\sqrt{6}}{5}$

$\frac{\sqrt{6}+3}{\sqrt{6}+1} = \frac{(\sqrt{6}+3)(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)} = \frac{6- \sqrt{6}+3\sqrt{6}-3}{6-1} = \frac{3+2\sqrt{6}}{5}$

$(\sqrt{6}+1)\star 2 = \sqrt{\frac{7+2\sqrt{6}}{5} + \frac{3+2\sqrt{6}}{5}} = \sqrt{\frac{10+4\sqrt{6}}{5}} = \sqrt{2+\frac{4}{5}\sqrt{6}} = \sqrt{\frac{10+4\sqrt{6}}{5}}$

ここで、$2+\frac{4}{5}\sqrt{6} = \frac{10+4\sqrt{6}}{5} = \frac{10+2\sqrt{24}}{5} = \frac{(\sqrt{6}+\sqrt{4})^2}{5} = \frac{(\sqrt{6}+2)^2}{5}$

$(\sqrt{6}+1)\star 2 = \sqrt{\frac{(\sqrt{6}+2)^2}{5}} = \frac{\sqrt{6}+2}{\sqrt{5}} = \frac{(\sqrt{6}+2)\sqrt{5}}{5} = \frac{\sqrt{30}+2\sqrt{5}}{5}$

3. 最終的な答え

(1) $\sqrt{2}$

(2) $\frac{\sqrt{30}+2\sqrt{5}}{5}$