(1), (2)の式を計算し、(3)~(6)の式の分母を有理化する問題です。 (1) $2\sqrt{27} - 3\sqrt{12} + \sqrt{54}$ (2) $(\sqrt{3} + \sqrt{6})^2$ (3) $\frac{\sqrt{3}-1}{\sqrt{8}}$ (4) $\frac{2\sqrt{3} + \sqrt{2}}{\sqrt{2} + \sqrt{3}}$ (5) $\frac{3}{2 - \sqrt{7}}$ (6) $\frac{3 + \sqrt{3}}{\sqrt{6}(1 + \sqrt{3})}$

はい、承知いたしました。問題を解いていきます。

1. 問題の内容

(1), (2)の式を計算し、(3)~(6)の式の分母を有理化する問題です。

(1) $2\sqrt{27} - 3\sqrt{12} + \sqrt{54}$

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

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

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

(5) $\frac{3}{2 - \sqrt{7}}$

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

2. 解き方の手順

(1)

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

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

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

よって、

$2\sqrt{27} - 3\sqrt{12} + \sqrt{54} = 2(3\sqrt{3}) - 3(2\sqrt{3}) + 3\sqrt{6} = 6\sqrt{3} - 6\sqrt{3} + 3\sqrt{6} = 3\sqrt{6}$

(2)

$(\sqrt{3} + \sqrt{6})^2 = (\sqrt{3})^2 + 2\sqrt{3}\sqrt{6} + (\sqrt{6})^2 = 3 + 2\sqrt{18} + 6 = 9 + 2\sqrt{9 \times 2} = 9 + 2(3\sqrt{2}) = 9 + 6\sqrt{2}$

(3)

$\frac{\sqrt{3}-1}{\sqrt{8}} = \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{(\sqrt{3}-1)\sqrt{2}}{2\sqrt{2}\sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$

(4)

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

(5)

$\frac{3}{2 - \sqrt{7}} = \frac{3(2 + \sqrt{7})}{(2 - \sqrt{7})(2 + \sqrt{7})} = \frac{6 + 3\sqrt{7}}{4 - 7} = \frac{6 + 3\sqrt{7}}{-3} = -2 - \sqrt{7}$

(6)

$\frac{3 + \sqrt{3}}{\sqrt{6}(1 + \sqrt{3})} = \frac{(3 + \sqrt{3})(1 - \sqrt{3})}{\sqrt{6}(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{3 - 3\sqrt{3} + \sqrt{3} - (\sqrt{3})^2}{\sqrt{6}(1 - 3)} = \frac{3 - 2\sqrt{3} - 3}{-2\sqrt{6}} = \frac{-2\sqrt{3}}{-2\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}\sqrt{6}}{\sqrt{6}\sqrt{6}} = \frac{\sqrt{18}}{6} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}$

3. 最終的な答え

(1) $3\sqrt{6}$

(2) $9 + 6\sqrt{2}$

(3) $\frac{\sqrt{6} - \sqrt{2}}{4}$

(4) $4 - \sqrt{6}$

(5) $-2 - \sqrt{7}$

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