数学をもう一度・・・

\( a>0, b>0 \) とするとき, \( \dfrac{b}{\sqrt{a}} \)および\( \dfrac{c}{\sqrt{a}+\sqrt{b}} \)の分母の根号を外すことを分母の有理化という. 分母の有理化をおこなうためには，多項式の乗法公式Ⅰ（展開公式Ⅰ）を応用する．

\( \dfrac{b}{\sqrt{a}}=\dfrac{b}{\sqrt{a}}\cdot\dfrac{\sqrt{a}}{\sqrt{a}}=\dfrac{b\sqrt{a}}{\big(~\sqrt{a}~\big)^2}=\dfrac{b\sqrt{a}}{a} \)

\( \dfrac{c}{\sqrt{\mathstrut a}+{\sqrt{b}}}=\dfrac{c}{\sqrt{\mathstrut a}+{\sqrt{b}}}\cdot\dfrac{\sqrt{\mathstrut a}-{\sqrt{b}}}{\sqrt{\mathstrut a}-{\sqrt{b}}} =\dfrac{c\big(\sqrt{\mathstrut a}-{\sqrt{b}~\big)}}{\big(\sqrt{\mathstrut a}\big)^2-\big({\sqrt{b}\big)^2}}=\dfrac{c \sqrt{\mathstrut a}-c \sqrt{b}}{a-b} \)

以下に実際の計算例を示す．
例

1. \( \dfrac{2}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{b\sqrt{3}}{\big(~\sqrt{3}~\big)^2}=\dfrac{2\sqrt{3}}{3} \)
2. \( \dfrac{5}{\sqrt{2}+\sqrt{3}}\dfrac{5}{\sqrt{2}+\sqrt{3}} \)
   \(
   \begin{eqnarray*}
   & = & \dfrac{5}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} \\
   & = & \dfrac{5\big(\sqrt{2}-\sqrt{3}~\big)}{\big(\sqrt{2}\big)^2-\big(\sqrt{3}\big)^2} \\
   & = & \dfrac{5 \sqrt{2}-5\sqrt{3}}{2-3}\\
   & = & \dfrac{5 \sqrt{2}-5\sqrt{3}}{-1}\\
   & = & 5\sqrt{3}-5\sqrt{2}
   \end{eqnarray*} \)
3. \( \dfrac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \)
   \(
   \begin{eqnarray*}
   & = & \dfrac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \cdot \dfrac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}} \\
   & = & \dfrac{\big(\sqrt{2}+\sqrt{3}\big)-\sqrt{5}}{\big(\sqrt{2}+\sqrt{3}\big)^2-\big(\sqrt{5}\big)^2} \\
   & = & \dfrac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\big(\sqrt{2}\big)^2+2\sqrt{2}\sqrt{3}+\big(\sqrt{3}\big)^2-\big(\sqrt{5}\big)^2} \\
   & = & \dfrac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+2\sqrt{6}+3-5} \\
   & = & \dfrac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}} \\
   & = & \dfrac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\cdot\dfrac{\sqrt{6}}{\sqrt{6}} \\
   & = & \dfrac{\big(\sqrt{2}+\sqrt{3}-\sqrt{5}\big)\sqrt{6}}{2\big(\sqrt{6}\big)^2} \\
   & = &\dfrac{\sqrt{2}\cdot\sqrt{6}+\sqrt{3}\cdot\sqrt{6}-\sqrt{5}\cdot\sqrt{6}}{2 \cdot 6} \\
   & = & \dfrac{\sqrt{2 \cdot6}+\sqrt{3 \cdot 6}-\sqrt{5 \cdot 6}}{12} \\
   & = & \dfrac{\sqrt{2^2 \cdot3}+\sqrt{3^2 \cdot 2}-\sqrt{30}}{12} \\
   & = & \dfrac{\sqrt{2^2} \cdot\sqrt{3}+\sqrt{3^2} \cdot \sqrt{2}-\sqrt{30}}{12} \\
   & = & \dfrac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}
   \end{eqnarray*} \)