## 1. 問題の内容

##

1. 問題の内容

与えられた5つの計算問題を解く。

(1) $\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{12}} - \frac{1}{\sqrt{27}}$

(2) $\frac{2\sqrt{5}-5\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

(3) $\frac{\sqrt{3}+2\sqrt{2}}{2\sqrt{3}-\sqrt{2}}$

(4) $\frac{\sqrt{5}-3}{\sqrt{5}+1} - \frac{\sqrt{5}+1}{\sqrt{5}-3}$

(5) $\frac{1}{1-\sqrt{2}} - \frac{1}{\sqrt{2}-\sqrt{3}} + \frac{1}{\sqrt{3}-2}$

##

2. 解き方の手順

**(1)**

$\sqrt{12} = 2\sqrt{3}$、$\sqrt{27} = 3\sqrt{3}$ なので、

\begin{align*}

\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{12}} - \frac{1}{\sqrt{27}} &= \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{3}} - \frac{1}{3\sqrt{3}} \\

&= \frac{6 - 3 - 2}{6\sqrt{3}} \\

&= \frac{1}{6\sqrt{3}} \\

&= \frac{\sqrt{3}}{18}

\end{align*}

**(2)**

分母を有理化する。

\begin{align*}

\frac{2\sqrt{5}-5\sqrt{2}}{\sqrt{5}-\sqrt{2}} &= \frac{(2\sqrt{5}-5\sqrt{2})(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})} \\

&= \frac{2(\sqrt{5})^2 + 2\sqrt{10} - 5\sqrt{10} - 5(\sqrt{2})^2}{5-2} \\

&= \frac{2(5) - 3\sqrt{10} - 5(2)}{3} \\

&= \frac{10 - 3\sqrt{10} - 10}{3} \\

&= \frac{-3\sqrt{10}}{3} \\

&= -\sqrt{10}

\end{align*}

**(3)**

分母を有理化する。

\begin{align*}

\frac{\sqrt{3}+2\sqrt{2}}{2\sqrt{3}-\sqrt{2}} &= \frac{(\sqrt{3}+2\sqrt{2})(2\sqrt{3}+\sqrt{2})}{(2\sqrt{3}-\sqrt{2})(2\sqrt{3}+\sqrt{2})} \\

&= \frac{2(\sqrt{3})^2 + \sqrt{6} + 4\sqrt{6} + 2(\sqrt{2})^2}{4(\sqrt{3})^2 - (\sqrt{2})^2} \\

&= \frac{2(3) + 5\sqrt{6} + 2(2)}{4(3) - 2} \\

&= \frac{6 + 5\sqrt{6} + 4}{12-2} \\

&= \frac{10 + 5\sqrt{6}}{10} \\

&= \frac{2+\sqrt{6}}{2}

\end{align*}

**(4)**

通分して計算する。

\begin{align*}

\frac{\sqrt{5}-3}{\sqrt{5}+1} - \frac{\sqrt{5}+1}{\sqrt{5}-3} &= \frac{(\sqrt{5}-3)(\sqrt{5}-3) - (\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}+1)(\sqrt{5}-3)} \\

&= \frac{(\sqrt{5}-3)^2 - (\sqrt{5}+1)^2}{5 - 3\sqrt{5} + \sqrt{5} - 3} \\

&= \frac{5 - 6\sqrt{5} + 9 - (5 + 2\sqrt{5} + 1)}{2 - 2\sqrt{5}} \\

&= \frac{14 - 6\sqrt{5} - 6 - 2\sqrt{5}}{2 - 2\sqrt{5}} \\

&= \frac{8 - 8\sqrt{5}}{2 - 2\sqrt{5}} \\

&= \frac{8(1 - \sqrt{5})}{2(1 - \sqrt{5})} \\

&= 4

\end{align*}

**(5)**

それぞれ有理化してから計算する。

\begin{align*}

\frac{1}{1-\sqrt{2}} &= \frac{1+\sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2})} = \frac{1+\sqrt{2}}{1-2} = -1-\sqrt{2} \\

\frac{1}{\sqrt{2}-\sqrt{3}} &= \frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})} = \frac{\sqrt{2}+\sqrt{3}}{2-3} = -\sqrt{2}-\sqrt{3} \\

\frac{1}{\sqrt{3}-2} &= \frac{\sqrt{3}+2}{(\sqrt{3}-2)(\sqrt{3}+2)} = \frac{\sqrt{3}+2}{3-4} = -\sqrt{3}-2

\end{align*}

よって、

\begin{align*}

\frac{1}{1-\sqrt{2}} - \frac{1}{\sqrt{2}-\sqrt{3}} + \frac{1}{\sqrt{3}-2} &= (-1-\sqrt{2}) - (-\sqrt{2}-\sqrt{3}) + (-\sqrt{3}-2) \\

&= -1-\sqrt{2} + \sqrt{2} + \sqrt{3} - \sqrt{3} - 2 \\

&= -3

\end{align*}

##

3. 最終的な答え

(1) $\frac{\sqrt{3}}{18}$

(2) $-\sqrt{10}$

(3) $\frac{2+\sqrt{6}}{2}$

(4) $4$

(5) $-3$