数学をもう一度・・・

\( a = x + y,~ b = xy \) を満たす 正の整数 \( x,~y\) が存在すれば，\(\sqrt{a+2\sqrt{b}}\) は次のように変形できる．
\[ \sqrt{a+2\sqrt{b}} = \sqrt{x} + \sqrt{y} .\]
その理由を考えてみよう．
\[
\begin{eqnarray*}
\sqrt{a+2\sqrt{b}} & = & \sqrt{x+y + 2\sqrt{xy}}\\
& = & \sqrt{(\sqrt{x})^2+(\sqrt{y})^2 + 2\sqrt{xy}} \\
& = & \sqrt{(\sqrt{x})^2+(\sqrt{y})^2 + 2\sqrt{x}\sqrt{y} } \\
& = & \sqrt{(\sqrt{x})^2 + 2\sqrt{x}\sqrt{y} + (\sqrt{y})^2} \\
& = & \sqrt{(\sqrt{x} + \sqrt{y})^2} \\
& = & \bigg|\sqrt{x} + \sqrt{y}) \bigg| \\
& = & \sqrt{x} + \sqrt{y}
\end{eqnarray*}
\]

(例)
\[
\begin{eqnarray*}
\sqrt{5+2\sqrt{6}} & = & \sqrt{5+2\sqrt{2 \cdot 3}} \\
& = & \sqrt{2 + 3 + 2 \sqrt{2 \cdot 3}} \\
& = & \sqrt{\big(\sqrt{2}\big)^2 + \big(\sqrt{3}\big)^2 + 2 \sqrt{2} \cdot \sqrt{3}} \\
& = & \sqrt{\big(\sqrt{2} + \sqrt{3}\big)^2 } \\
& = & \big|~ \sqrt{2} + \sqrt{3}~ \big|
\end{eqnarray*}
\]
\( \sqrt{2} + \sqrt{3} > 0 \)より　　 \(\sqrt{5+2\sqrt{6}}=\sqrt{2} + \sqrt{3} \)