与えられた8つの平方根の計算問題を解きます。 (1) $2\sqrt{3} + \frac{1}{\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3}} - \frac{1\sqrt{3}}{3} + \frac{6\sqrt{2}}{3}$ (2) $\frac{3}{\sqrt{5}} - \frac{\sqrt{5}}{10} + \frac{\sqrt{5}}{10}$ (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) $\sqrt{32} - \frac{\sqrt{8}}{4} - \frac{2}{\sqrt{4}}$

1. 問題の内容

与えられた8つの平方根の計算問題を解きます。

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

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

(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

(1)

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

(2)

$\frac{3}{\sqrt{5}} - \frac{\sqrt{5}}{10} + \frac{\sqrt{5}}{10} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$

(3)

$\sqrt{54} + \frac{5\sqrt{2}}{\sqrt{3}} = \sqrt{9 \cdot 6} + \frac{5\sqrt{2}\sqrt{3}}{3} = 3\sqrt{6} + \frac{5\sqrt{6}}{3} = \frac{9\sqrt{6}}{3} + \frac{5\sqrt{6}}{3} = \frac{14\sqrt{6}}{3}$

(4)

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

(5)

$\sqrt{72} - \frac{2}{\sqrt{8}} = \sqrt{36 \cdot 2} - \frac{2}{2\sqrt{2}} = 6\sqrt{2} - \frac{1}{\sqrt{2}} = 6\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{12\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = \frac{11\sqrt{2}}{2}$

(6)

$\sqrt{48} - \frac{1}{\sqrt{3}} = \sqrt{16 \cdot 3} - \frac{\sqrt{3}}{3} = 4\sqrt{3} - \frac{\sqrt{3}}{3} = \frac{12\sqrt{3}}{3} - \frac{\sqrt{3}}{3} = \frac{11\sqrt{3}}{3}$

(7)

$\frac{1}{\sqrt{3}} + \sqrt{3} - \frac{3\sqrt{6}}{\sqrt{2}} = \frac{\sqrt{3}}{3} + \sqrt{3} - 3\sqrt{3} = \frac{\sqrt{3}}{3} + \frac{3\sqrt{3}}{3} - \frac{9\sqrt{3}}{3} = \frac{-5\sqrt{3}}{3}$

(8)

$\sqrt{32} - \frac{\sqrt{8}}{4} - \frac{2}{\sqrt{4}} = \sqrt{16 \cdot 2} - \frac{2\sqrt{2}}{4} - \frac{2}{2} = 4\sqrt{2} - \frac{\sqrt{2}}{2} - 1 = \frac{8\sqrt{2}}{2} - \frac{\sqrt{2}}{2} - 1 = \frac{7\sqrt{2}}{2} - 1$

3. 最終的な答え

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

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

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