次の3つの式を計算します。 (1) $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ (2) $\frac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{\sqrt{5}+\sqrt{3}-\sqrt{2}}$ (3) $\frac{\sqrt{2}+\sqrt{5}+\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}} + \frac{\sqrt{2}-\sqrt{5}+\sqrt{7}}{\sqrt{2}-\sqrt{5}-\sqrt{7}}$

はい、承知いたしました。以下の形式で回答します。

1. 問題の内容

次の3つの式を計算します。

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

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

(3) $\frac{\sqrt{2}+\sqrt{5}+\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}} + \frac{\sqrt{2}-\sqrt{5}+\sqrt{7}}{\sqrt{2}-\sqrt{5}-\sqrt{7}}$

2. 解き方の手順

(1) 分母の有理化を行います。まず、分母を$(1+\sqrt{2})-\sqrt{3}$と考え、$(1+\sqrt{2})+\sqrt{3}$を分母分子に掛けます。

\begin{align*}

\frac{1}{1+\sqrt{2}-\sqrt{3}} &= \frac{1}{(1+\sqrt{2})-\sqrt{3}} \cdot \frac{(1+\sqrt{2})+\sqrt{3}}{(1+\sqrt{2})+\sqrt{3}} \\

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

&= \frac{1+\sqrt{2}+\sqrt{3}}{1+2\sqrt{2}+2-3} \\

&= \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}} \\

&= \frac{(1+\sqrt{2}+\sqrt{3})\sqrt{2}}{2\sqrt{2}\sqrt{2}} \\

&= \frac{\sqrt{2}+2+\sqrt{6}}{4}

\end{align*}

(2) 分母の有理化を行います。まず、分母を$(\sqrt{5}+\sqrt{3})-\sqrt{2}$と考え、$(\sqrt{5}+\sqrt{3})+\sqrt{2}$を分母分子に掛けます。

\begin{align*}

\frac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{\sqrt{5}+\sqrt{3}-\sqrt{2}} &= \frac{(\sqrt{5}+\sqrt{3}+\sqrt{2})}{(\sqrt{5}+\sqrt{3})-\sqrt{2}} \cdot \frac{(\sqrt{5}+\sqrt{3})+\sqrt{2}}{(\sqrt{5}+\sqrt{3})+\sqrt{2}} \\

&= \frac{(\sqrt{5}+\sqrt{3}+\sqrt{2})(\sqrt{5}+\sqrt{3}+\sqrt{2})}{(\sqrt{5}+\sqrt{3})^2 - (\sqrt{2})^2} \\

&= \frac{(\sqrt{5}+\sqrt{3}+\sqrt{2})^2}{5+2\sqrt{15}+3-2} \\

&= \frac{(\sqrt{5})^2+(\sqrt{3})^2+(\sqrt{2})^2+2\sqrt{15}+2\sqrt{10}+2\sqrt{6}}{6+2\sqrt{15}} \\

&= \frac{5+3+2+2\sqrt{15}+2\sqrt{10}+2\sqrt{6}}{6+2\sqrt{15}} \\

&= \frac{10+2\sqrt{15}+2\sqrt{10}+2\sqrt{6}}{6+2\sqrt{15}} \\

&= \frac{5+\sqrt{15}+\sqrt{10}+\sqrt{6}}{3+\sqrt{15}} \\

&= \frac{(5+\sqrt{15}+\sqrt{10}+\sqrt{6})(3-\sqrt{15})}{(3+\sqrt{15})(3-\sqrt{15})} \\

&= \frac{15 - 5\sqrt{15}+3\sqrt{15}-15+3\sqrt{10}-\sqrt{150}+3\sqrt{6}-\sqrt{90}}{9-15} \\

&= \frac{-2\sqrt{15}+3\sqrt{10}-5\sqrt{6}+3\sqrt{6}-3\sqrt{10}}{-6} \\

&= \frac{-2\sqrt{15}-2\sqrt{6}}{-6} \\

&= \frac{\sqrt{15}+\sqrt{6}}{3}

\end{align*}

(3)

\begin{align*}

\frac{\sqrt{2}+\sqrt{5}+\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}} + \frac{\sqrt{2}-\sqrt{5}+\sqrt{7}}{\sqrt{2}-\sqrt{5}-\sqrt{7}} &= \frac{(\sqrt{2}+\sqrt{5}+\sqrt{7})(\sqrt{2}+\sqrt{5}+\sqrt{7})}{(\sqrt{2}+\sqrt{5}-\sqrt{7})(\sqrt{2}+\sqrt{5}+\sqrt{7})} + \frac{(\sqrt{2}-\sqrt{5}+\sqrt{7})(\sqrt{2}-\sqrt{5}+\sqrt{7})}{(\sqrt{2}-\sqrt{5}-\sqrt{7})(\sqrt{2}-\sqrt{5}+\sqrt{7})} \\

&= \frac{(\sqrt{2}+\sqrt{5}+\sqrt{7})^2}{(\sqrt{2}+\sqrt{5})^2-(\sqrt{7})^2} + \frac{(\sqrt{2}-\sqrt{5}+\sqrt{7})^2}{(\sqrt{2}-\sqrt{5})^2-(\sqrt{7})^2} \\

&= \frac{2+5+7+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}{2+2\sqrt{10}+5-7} + \frac{2+5+7-2\sqrt{10}+2\sqrt{14}-2\sqrt{35}}{2-2\sqrt{10}+5-7} \\

&= \frac{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}{2\sqrt{10}} + \frac{14-2\sqrt{10}+2\sqrt{14}-2\sqrt{35}}{-2\sqrt{10}} \\

&= \frac{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}{2\sqrt{10}} - \frac{14-2\sqrt{10}+2\sqrt{14}-2\sqrt{35}}{2\sqrt{10}} \\

&= \frac{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35} - (14-2\sqrt{10}+2\sqrt{14}-2\sqrt{35})}{2\sqrt{10}} \\

&= \frac{4\sqrt{10}+4\sqrt{35}}{2\sqrt{10}} = \frac{2\sqrt{10}+2\sqrt{35}}{\sqrt{10}} = 2+\frac{2\sqrt{35}}{\sqrt{10}} = 2+\frac{2\sqrt{7}\sqrt{5}}{\sqrt{2}\sqrt{5}} = 2+\frac{2\sqrt{7}}{\sqrt{2}} = 2+\sqrt{14}

\end{align*}

3. 最終的な答え

(1) $\frac{\sqrt{2}+2+\sqrt{6}}{4}$

(2) $\frac{\sqrt{15}+\sqrt{6}}{3}$

(3) $2+\sqrt{14}$