以下の8つの計算問題を解いてください。 (1) $2\sqrt{3} + \frac{1}{\sqrt{3}}$ (2) $\frac{3}{\sqrt{5}} - \frac{\sqrt{5}}{10}$ (3) $\sqrt{54} + \frac{5\sqrt{2}}{\sqrt{3}}$ (4) $8 - \frac{1}{\sqrt{3}}$ (5) $\sqrt{72} - \frac{2}{\sqrt{8}}$ (6) $\sqrt{48} - \frac{1}{\sqrt{3}}$ (7) $\frac{1}{\sqrt{3}} + \sqrt{3} - \frac{3\sqrt{6}}{\sqrt{2}}$ (8) $\sqrt{32} - \frac{\sqrt{8}}{4} - \frac{2}{\sqrt{4}}$

はい、承知いたしました。画像に写っている計算問題を解いていきます。

1. 問題の内容

以下の8つの計算問題を解いてください。

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

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

(3) $\sqrt{54} + \frac{5\sqrt{2}}{\sqrt{3}}$

(4) $8 - \frac{1}{\sqrt{3}}$

(5) $\sqrt{72} - \frac{2}{\sqrt{8}}$

(6) $\sqrt{48} - \frac{1}{\sqrt{3}}$

(7) $\frac{1}{\sqrt{3}} + \sqrt{3} - \frac{3\sqrt{6}}{\sqrt{2}}$

(8) $\sqrt{32} - \frac{\sqrt{8}}{4} - \frac{2}{\sqrt{4}}$

2. 解き方の手順

各問題ごとに解き方を説明します。

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

分母を有理化します。

$2\sqrt{3} + \frac{1}{\sqrt{3}} = 2\sqrt{3} + \frac{\sqrt{3}}{3} = \frac{6\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{7\sqrt{3}}{3}$

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

分母を有理化します。

$\frac{3}{\sqrt{5}} - \frac{\sqrt{5}}{10} = \frac{3\sqrt{5}}{5} - \frac{\sqrt{5}}{10} = \frac{6\sqrt{5}}{10} - \frac{\sqrt{5}}{10} = \frac{5\sqrt{5}}{10} = \frac{\sqrt{5}}{2}$

(3) $\sqrt{54} + \frac{5\sqrt{2}}{\sqrt{3}}$

$\sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6}$

$\frac{5\sqrt{2}}{\sqrt{3}} = \frac{5\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{5\sqrt{6}}{3}$

$3\sqrt{6} + \frac{5\sqrt{6}}{3} = \frac{9\sqrt{6}}{3} + \frac{5\sqrt{6}}{3} = \frac{14\sqrt{6}}{3}$

(4) $8 - \frac{1}{\sqrt{3}}$

分母を有理化します。

$8 - \frac{1}{\sqrt{3}} = 8 - \frac{\sqrt{3}}{3} = \frac{24}{3} - \frac{\sqrt{3}}{3} = \frac{24 - \sqrt{3}}{3}$

(5) $\sqrt{72} - \frac{2}{\sqrt{8}}$

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

$\frac{2}{\sqrt{8}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$6\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{12\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = \frac{11\sqrt{2}}{2}$

(6) $\sqrt{48} - \frac{1}{\sqrt{3}}$

$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$

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

$4\sqrt{3} - \frac{\sqrt{3}}{3} = \frac{12\sqrt{3}}{3} - \frac{\sqrt{3}}{3} = \frac{11\sqrt{3}}{3}$

(7) $\frac{1}{\sqrt{3}} + \sqrt{3} - \frac{3\sqrt{6}}{\sqrt{2}}$

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

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

$\frac{\sqrt{3}}{3} + \sqrt{3} - 3\sqrt{3} = \frac{\sqrt{3}}{3} - 2\sqrt{3} = \frac{\sqrt{3}}{3} - \frac{6\sqrt{3}}{3} = -\frac{5\sqrt{3}}{3}$

(8) $\sqrt{32} - \frac{\sqrt{8}}{4} - \frac{2}{\sqrt{4}}$

$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

$\sqrt{8} = 2\sqrt{2}$

$\sqrt{4} = 2$

$4\sqrt{2} - \frac{2\sqrt{2}}{4} - \frac{2}{2} = 4\sqrt{2} - \frac{\sqrt{2}}{2} - 1 = \frac{8\sqrt{2}}{2} - \frac{\sqrt{2}}{2} - 1 = \frac{7\sqrt{2}}{2} - 1$

3. 最終的な答え

(1) $\frac{7\sqrt{3}}{3}$

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

(3) $\frac{14\sqrt{6}}{3}$

(4) $\frac{24 - \sqrt{3}}{3}$

(5) $\frac{11\sqrt{2}}{2}$

(6) $\frac{11\sqrt{3}}{3}$

(7) $-\frac{5\sqrt{3}}{3}$

(8) $\frac{7\sqrt{2}}{2} - 1$