ルートを含む分数の有理化② $\dfrac{1}{1+\sqrt{2}+\sqrt{3}}$
コツ
平方根（ルート）を「平方の差」に変形します。つまり、$$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$$を利用します。２回の有理化が必要です。
解答例
$$\begin{align}
& \color{red}{\frac{1}{1+\sqrt{2}+\sqrt{3}}} \\
& = \frac{1}{(1+\sqrt{2})+\sqrt{3}} \times \color{blue}{\frac{ (1+\sqrt{2})-\sqrt{3} }{ (1+\sqrt{2})-\sqrt{3} } } \\
& = \frac{1+\sqrt{2}-\sqrt{3}}{(1+\sqrt{2})^2-(\sqrt{3})^2} \\
& = \frac{1+\sqrt{2}-\sqrt{3}}{(1+2\sqrt{2}+2)-3} \\
& = \frac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}} \\
& =\frac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}} \color{blue}{ \times \frac{\sqrt{2}}{\sqrt{2}} } \\
& = \frac{\sqrt{2} + 2 – \sqrt{6}}{4} \\
& = \color{red}{\frac{2 + \sqrt{2} – \sqrt{6}}{4}}
\end{align}$$