【数学I】例題1.2.3：分母の有理化（One More）★★

% 例題I1.2.3：分母の有理化 （One More）★★
次の式の分母を有理化して簡単にせよ. (1)$\frac{2}{\sqrt{2}}$(2)$\frac{4}{\sqrt{7}+\sqrt{3}}$(3)$\frac{\sqrt{5}}{\sqrt{3}-1}-\frac{1}{\sqrt{7}-\sqrt{5}}$(4)$\frac{1}{1+\sqrt{2}+\sqrt{3}}$

% 解答（例題I1.2.3）
(1)$\dfrac{2}{\sqrt{2}}=\dfrac{2\times\sqrt{2}}{\sqrt{2}\times{\sqrt{2}}}=\dfrac{2\sqrt{2}}{2}=\sqrt{2}$別解：$\dfrac{2}{\sqrt{2}}=\dfrac{\sqrt{2}\times\sqrt{2}}{\sqrt{2}}=\sqrt{2}$(2)$\dfrac{4}{\sqrt{7}+\sqrt{3}}=\dfrac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}=\dfrac{4(\sqrt{7}-\sqrt{3})}{7-3}=\sqrt{7}-\sqrt{3}$(3)$\begin{aligned} \frac{\sqrt{5}}{\sqrt{3}-1}-\frac{1}{\sqrt{7}-\sqrt{5}}&=\frac{\sqrt{5}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}-\frac{\sqrt{7}+\sqrt{5}}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}\\ &=\frac{\sqrt{15}+\sqrt{5}}{3-1}-\frac{\sqrt{7}+\sqrt{5}}{7-5}=\frac{\sqrt{15}-\sqrt{7}}{2} \end{aligned}$(4)$\begin{aligned} \frac{1}{1+\sqrt{2}+\sqrt{3}}&=\frac{(1+\sqrt{2})-\sqrt{3}}{\{(1+\sqrt{2})+\sqrt{3}\}\{(1+\sqrt{2})-\sqrt{3}\}}=\frac{1+\sqrt{2}-\sqrt{3}}{(1+\sqrt{2})^2-(\sqrt{3})^2}\\ &=\frac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}} =\frac{(1+\sqrt{2}-\sqrt{3})\times\sqrt{2}}{2\sqrt{2}\times\sqrt{2}}\\ &=\frac{\sqrt{2}+2-\sqrt{6}}{4} \end{aligned}$

% 問題I1.2.3
次の式の分母を有理化して簡単にせよ. (1)$\frac{3}{\sqrt{3}}$(2)$\frac{5}{\sqrt{6}+\sqrt{2}}$(3)$\frac{\sqrt{10}}{\sqrt{5}-2}-\frac{2}{\sqrt{11}-\sqrt{10}}$(4)$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

% 解答I1.2.3
(1)$\dfrac{3}{\sqrt{3}}=\dfrac{3\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\dfrac{3\sqrt{3}}{3}=\sqrt{3}$別解：$\dfrac{3}{\sqrt{3}}=\dfrac{\sqrt{3}\times\sqrt{3}}{\sqrt{3}}=\sqrt{3}$(2)$\dfrac{5}{\sqrt{6}+\sqrt{2}}=\dfrac{5(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\dfrac{5(\sqrt{6}-\sqrt{2})}{6-2}=\dfrac{5\sqrt{6}-5\sqrt{2}}{4}$(3)$\begin{aligned} \frac{\sqrt{10}}{\sqrt{5}-2}-\frac{2}{\sqrt{11}-\sqrt{10}}&=\frac{\sqrt{10}(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}-\frac{2(\sqrt{11}+\sqrt{10})}{(\sqrt{11}-\sqrt{10})(\sqrt{11}+\sqrt{10})}\\ &=\frac{\sqrt{50}+2\sqrt{10}}{5-4}-\frac{2(\sqrt{11}+\sqrt{10})}{11-10}\\ &=5\sqrt{2}+2\sqrt{10}-2(\sqrt{11}+\sqrt{10})\\ &=5\sqrt{2}-2\sqrt{11} \end{aligned}$(4)$\begin{aligned} \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&=\frac{(\sqrt{2}+\sqrt{3})-\sqrt{5}}{\{(\sqrt{2}+\sqrt{3})+\sqrt{5}\}\{(\sqrt{2}+\sqrt{3})-\sqrt{5}\}}\\ &=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}\\ &=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\\ &=\frac{(\sqrt{2}+\sqrt{3}-\sqrt{5})\times\sqrt{6}}{2\sqrt{6}\times\sqrt{6}}\\ &=\frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12} \end{aligned}$