次の計算をしなさい。 (1) $\sqrt{45} + 2\sqrt{5}$ (2) $2\sqrt{3} + \sqrt{27}$ (3) $5\sqrt{2} - \sqrt{18}$ (4) $\sqrt{63} - 4\sqrt{7}$

はい、承知いたしました。以下の形式で問題(1), (2), (3), (4)を解いていきます。

**問題51**

1. 問題の内容

次の計算をしなさい。

(1) $\sqrt{45} + 2\sqrt{5}$

(2) $2\sqrt{3} + \sqrt{27}$

(3) $5\sqrt{2} - \sqrt{18}$

(4) $\sqrt{63} - 4\sqrt{7}$

2. 解き方の手順

(1) $\sqrt{45} + 2\sqrt{5}$

$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$

よって、$\sqrt{45} + 2\sqrt{5} = 3\sqrt{5} + 2\sqrt{5} = (3+2)\sqrt{5} = 5\sqrt{5}$

(2) $2\sqrt{3} + \sqrt{27}$

$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

よって、$2\sqrt{3} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = (2+3)\sqrt{3} = 5\sqrt{3}$

(3) $5\sqrt{2} - \sqrt{18}$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

よって、$5\sqrt{2} - \sqrt{18} = 5\sqrt{2} - 3\sqrt{2} = (5-3)\sqrt{2} = 2\sqrt{2}$

(4) $\sqrt{63} - 4\sqrt{7}$

$\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$

よって、$\sqrt{63} - 4\sqrt{7} = 3\sqrt{7} - 4\sqrt{7} = (3-4)\sqrt{7} = -\sqrt{7}$

3. 最終的な答え

(1) $5\sqrt{5}$

(2) $5\sqrt{3}$

(3) $2\sqrt{2}$

(4) $-\sqrt{7}$

**問題52**

1. 問題の内容

次の計算をしなさい。

(1) $\frac{5}{4\sqrt{3}} + \frac{\sqrt{3}}{12}$

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

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

(4) $\frac{1}{3\sqrt{3}} - \frac{2\sqrt{3}}{9}$

2. 解き方の手順

(1) $\frac{5}{4\sqrt{3}} + \frac{\sqrt{3}}{12}$

まず、$\frac{5}{4\sqrt{3}}$ を有理化します。

$\frac{5}{4\sqrt{3}} = \frac{5\sqrt{3}}{4\sqrt{3}\sqrt{3}} = \frac{5\sqrt{3}}{4 \times 3} = \frac{5\sqrt{3}}{12}$

よって、$\frac{5}{4\sqrt{3}} + \frac{\sqrt{3}}{12} = \frac{5\sqrt{3}}{12} + \frac{\sqrt{3}}{12} = \frac{5\sqrt{3} + \sqrt{3}}{12} = \frac{6\sqrt{3}}{12} = \frac{\sqrt{3}}{2}$

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

$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3}\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{\sqrt{6}}{2}$

よって、$\frac{\sqrt{3}}{\sqrt{2}} + \frac{3\sqrt{6}}{2} = \frac{\sqrt{6}}{2} + \frac{3\sqrt{6}}{2} = \frac{\sqrt{6} + 3\sqrt{6}}{2} = \frac{4\sqrt{6}}{2} = 2\sqrt{6}$

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

$\frac{2}{5\sqrt{2}} = \frac{2\sqrt{2}}{5\sqrt{2}\sqrt{2}} = \frac{2\sqrt{2}}{5 \times 2} = \frac{2\sqrt{2}}{10} = \frac{\sqrt{2}}{5}$

よって、$\frac{2}{5\sqrt{2}} - \frac{\sqrt{2}}{10} = \frac{\sqrt{2}}{5} - \frac{\sqrt{2}}{10} = \frac{2\sqrt{2}}{10} - \frac{\sqrt{2}}{10} = \frac{2\sqrt{2} - \sqrt{2}}{10} = \frac{\sqrt{2}}{10}$

(4) $\frac{1}{3\sqrt{3}} - \frac{2\sqrt{3}}{9}$

$\frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{3\sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{3 \times 3} = \frac{\sqrt{3}}{9}$

よって、$\frac{1}{3\sqrt{3}} - \frac{2\sqrt{3}}{9} = \frac{\sqrt{3}}{9} - \frac{2\sqrt{3}}{9} = \frac{\sqrt{3} - 2\sqrt{3}}{9} = \frac{-\sqrt{3}}{9} = -\frac{\sqrt{3}}{9}$

3. 最終的な答え

(1) $\frac{\sqrt{3}}{2}$

(2) $2\sqrt{6}$

(3) $\frac{\sqrt{2}}{10}$

(4) $-\frac{\sqrt{3}}{9}$