(2) 次の式を計算し、分母を有理化せよ。 $\frac{6\sqrt{2}+2}{\sqrt{3}} \times \frac{6\sqrt{6}-2\sqrt{3}}{\sqrt{2}}$ (3) $\frac{1}{\sqrt{6}+1} - \frac{\sqrt{6}}{6-\sqrt{6}}$ (4) $\frac{1}{\sqrt{2} + \sqrt{3}+\sqrt{5}} + \frac{1}{\sqrt{2} + \sqrt{3}-\sqrt{5}} + \frac{1}{\sqrt{2} - \sqrt{3}+\sqrt{5}} + \frac{1}{\sqrt{2} - \sqrt{3}-\sqrt{5}}$

1. 問題の内容

(2) 次の式を計算し、分母を有理化せよ。

$\frac{6\sqrt{2}+2}{\sqrt{3}} \times \frac{6\sqrt{6}-2\sqrt{3}}{\sqrt{2}}$

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

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

2. 解き方の手順

(2)

まず、それぞれの分数を有理化します。

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

$\frac{6\sqrt{6}-2\sqrt{3}}{\sqrt{2}} = \frac{(6\sqrt{6}-2\sqrt{3})\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{6\sqrt{12}-2\sqrt{6}}{2} = \frac{6(2\sqrt{3})-2\sqrt{6}}{2} = \frac{12\sqrt{3}-2\sqrt{6}}{2} = 6\sqrt{3}-\sqrt{6}$

次に、掛け算を行います。

$(2\sqrt{6}+\frac{2\sqrt{3}}{3})(6\sqrt{3}-\sqrt{6}) = 2\sqrt{6}(6\sqrt{3}-\sqrt{6}) + \frac{2\sqrt{3}}{3}(6\sqrt{3}-\sqrt{6}) = 12\sqrt{18}-2(6)+ \frac{12(3)}{3}-\frac{2\sqrt{18}}{3} = 12(3\sqrt{2})-12+12-\frac{2(3\sqrt{2})}{3} = 36\sqrt{2}-2\sqrt{2}=34\sqrt{2}$

(3)

$\frac{1}{\sqrt{6}+1} - \frac{\sqrt{6}}{6-\sqrt{6}} = \frac{\sqrt{6}-1}{(\sqrt{6}+1)(\sqrt{6}-1)} - \frac{\sqrt{6}(6+\sqrt{6})}{(6-\sqrt{6})(6+\sqrt{6})} = \frac{\sqrt{6}-1}{6-1} - \frac{6\sqrt{6}+6}{36-6} = \frac{\sqrt{6}-1}{5} - \frac{6\sqrt{6}+6}{30} = \frac{6(\sqrt{6}-1)}{30} - \frac{6\sqrt{6}+6}{30} = \frac{6\sqrt{6}-6 - 6\sqrt{6}-6}{30} = \frac{-12}{30} = -\frac{2}{5}$

(4)

$A=\sqrt{2}+\sqrt{3}$

$B=\sqrt{5}$

$\frac{1}{A+B}+\frac{1}{A-B}+\frac{1}{-A+B}+\frac{1}{-A-B} = (\frac{1}{A+B}+\frac{1}{A-B}) + (\frac{1}{-A+B}+\frac{1}{-A-B})$

$= \frac{A-B+A+B}{A^2-B^2} + \frac{-A-B-A+B}{A^2-B^2} = \frac{2A}{A^2-B^2}+\frac{-2A}{A^2-B^2} = 0$

よって、答えは

0.

3. 最終的な答え

(2): 34√2

(3): -2/5

(4): 0