【数学I】例題1.2.3：分母の有理化（One More）★★

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% 例題I1.2.3：分母の有理化（One More）★★
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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) $\frac{1}{1+\sqrt{2}+\sqrt{3}}$

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% 例題I1.2.3の解答（One More）★★
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(1) $\frac{2}{\sqrt{2}}=\frac{2\times\sqrt{2}}{\sqrt{2}\times{\sqrt{2}}}=\frac{2\sqrt{2}}{2}=\sqrt{2}$

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% 例題I1.2.3の別解
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$\frac{2}{\sqrt{2}}=\frac{\sqrt{2}\times\sqrt{2}}{\sqrt{2}}=\sqrt{2}$

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% 例題I1.2.3の解答（One More）★★
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(2) $\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}=\frac{4(\sqrt{7}-\sqrt{3})}{7-3}=\sqrt{7}-\sqrt{3}$

(3)

\begin{align*}
\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{align*}

(4)

\begin{align*}
\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{align*}

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% 問題I1.2.3：分母の有理化（One More）★★
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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) $\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

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% 問題I1.2.3の解答（One More）★★
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(1) $\frac{3}{\sqrt{3}}=\frac{3\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}$

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% 問題I1.2.3の別解
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$\frac{3}{\sqrt{3}}=\frac{\sqrt{3}\times\sqrt{3}}{\sqrt{3}}=\sqrt{3}$

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% 問題I1.2.3の解答（One More）★★
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(2) $\frac{5}{\sqrt{6}+\sqrt{2}}=\frac{5(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\frac{5(\sqrt{6}-\sqrt{2})}{6-2}=\frac{5\sqrt{6}-5\sqrt{2}}{4}$

(3)

\begin{align*}
\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{align*}

(4)

\begin{align*}
\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{align*}

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% 例題I1.2.3の側注：分母の有理化（One More）★★
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% 側注1
分母，分子に$\sqrt{2}$を掛ける．

% 側注2
\begin{align*}
&(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})\\
=&(\sqrt{7})^2-(\sqrt{3})^2=4
\end{align*}

% 側注3
第1項と第2項の分母，分子にそれぞれ，$\sqrt{3}+1,\sqrt{7}+\sqrt{5}$を掛ける．

% 側注4
$1^2+(\sqrt{2})^2=(\sqrt{3})^2$より，$(1+\sqrt{2})+\sqrt{3}$と項を分けて，$(1+\sqrt{2})-\sqrt{3}$を掛ける．

% 側注5
更に分母を有理化する．

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% 問題I1.2.3の側注：分母の有理化（One More）★★
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% 側注1
分母，分子に$\sqrt{3}$を掛ける．

% 側注2
\begin{align*}
&(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})\\
=&(\sqrt{6})^2-(\sqrt{2})^2=4
\end{align*}

% 側注3
第1項と第2項の分母，分子にそれぞれ，$\sqrt{5}+2,\sqrt{11}+\sqrt{10}$を掛ける．

% 側注4
$(\sqrt{2})^2+(\sqrt{3})^2=(\sqrt{5})^2$より，$(\sqrt{2}+\sqrt{3})+\sqrt{5}$と項を分けて，$(\sqrt{2}+\sqrt{3})-\sqrt{5}$を掛ける．

% 側注5
更に分母を有理化する．

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% 例題I1.2.3の考え方・補足など：分母の有理化（One More）★★
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% 考え方
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(1) 分母と分子に同じ数を掛け，$(\sqrt{a})^2=a$を用いる．

(2)，(3)$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=(\sqrt{a})^2-(\sqrt{b})^2=a-b$を用いる．

(4) 分母を2つの項と1つの項に分けて，分母と分子に同じ数を掛けることを考える．しかし，ここでは1回の操作で有理化することができない．そこで，2回の操作で計算することを考える．なお，1回目の同じ数を掛ける操作において，$1+(\sqrt{2}+\sqrt{3})$と分けて分母と分子に$1-(\sqrt{2}+\sqrt{3})$を掛けるより，$(1+\sqrt{2})+\sqrt{3}$と分けて分母と分子に$(1+\sqrt{2})-\sqrt{3}$を掛けた方が計算が楽になる（$1^2+(\sqrt{2})^2=(\sqrt{3})^2$に着目して，項を分けるとよい）．$1+(\sqrt{2}+\sqrt{3})$に分けた場合

\[
\frac{1-(\sqrt{2}+\sqrt{3})}{1-(\sqrt{2}+\sqrt{3})^2}=\frac{1-\sqrt{2}-\sqrt{3}}{-4-2\sqrt{6}}
\]

2回目の操作において，$-4+2\sqrt{6}$を掛けなければならない．

$(1+\sqrt{2})+\sqrt{3}$に分けた場合

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

2回目の操作において，単に$\sqrt{2}$を掛ければよい．

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% One Point
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分母の有理化は，$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$を用いる．