与えられた式を計算し、簡単にします。 (1) $\frac{\sqrt{5}}{\sqrt{3}-1} - \frac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ (2) $\frac{\sqrt{7}+1}{\sqrt{7}-\sqrt{3}} - \frac{\sqrt{7}-1}{\sqrt{7}+\sqrt{3}}$ (3) $\frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+2}$ (4) $\frac{\sqrt{2}+1}{\sqrt{2}-1} + \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} + \frac{4(1+\sqrt{2})}{1-\sqrt{3}}$

はい、承知いたしました。それでは、問題(1)から(4)まで順に解いていきます。

1. 問題の内容

与えられた式を計算し、簡単にします。

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

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

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

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

2. 解き方の手順

(1) 各項をそれぞれ有理化します。

$\frac{\sqrt{5}}{\sqrt{3}-1} = \frac{\sqrt{5}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{\sqrt{15}+\sqrt{5}}{3-1} = \frac{\sqrt{15}+\sqrt{5}}{2}$

$\frac{\sqrt{3}}{\sqrt{5}+\sqrt{3}} = \frac{\sqrt{3}(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{\sqrt{15}-3}{5-3} = \frac{\sqrt{15}-3}{2}$

したがって、

$\frac{\sqrt{5}}{\sqrt{3}-1} - \frac{\sqrt{3}}{\sqrt{5}+\sqrt{3}} = \frac{\sqrt{15}+\sqrt{5}}{2} - \frac{\sqrt{15}-3}{2} = \frac{\sqrt{15}+\sqrt{5}-\sqrt{15}+3}{2} = \frac{\sqrt{5}+3}{2}$

(2) 各項をそれぞれ有理化します。

$\frac{\sqrt{7}+1}{\sqrt{7}-\sqrt{3}} = \frac{(\sqrt{7}+1)(\sqrt{7}+\sqrt{3})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})} = \frac{7+\sqrt{21}+\sqrt{7}+\sqrt{3}}{7-3} = \frac{7+\sqrt{21}+\sqrt{7}+\sqrt{3}}{4}$

$\frac{\sqrt{7}-1}{\sqrt{7}+\sqrt{3}} = \frac{(\sqrt{7}-1)(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})} = \frac{7-\sqrt{21}-\sqrt{7}+\sqrt{3}}{7-3} = \frac{7-\sqrt{21}-\sqrt{7}+\sqrt{3}}{4}$

したがって、

$\frac{\sqrt{7}+1}{\sqrt{7}-\sqrt{3}} - \frac{\sqrt{7}-1}{\sqrt{7}+\sqrt{3}} = \frac{7+\sqrt{21}+\sqrt{7}+\sqrt{3}}{4} - \frac{7-\sqrt{21}-\sqrt{7}+\sqrt{3}}{4} = \frac{7+\sqrt{21}+\sqrt{7}+\sqrt{3}-7+\sqrt{21}+\sqrt{7}-\sqrt{3}}{4} = \frac{2\sqrt{21}+2\sqrt{7}}{4} = \frac{\sqrt{21}+\sqrt{7}}{2}$

(3) 各項をそれぞれ有理化します。

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

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

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

したがって、

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

(4) 各項をそれぞれ有理化します。

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

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

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

したがって、

$\frac{\sqrt{2}+1}{\sqrt{2}-1} + \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} + \frac{4(1+\sqrt{2})}{1-\sqrt{3}} = (3+2\sqrt{2}) + (5+2\sqrt{6}) + (-2-2\sqrt{3}-2\sqrt{2}-2\sqrt{6}) = 3+2\sqrt{2}+5+2\sqrt{6}-2-2\sqrt{3}-2\sqrt{2}-2\sqrt{6} = 6-2\sqrt{3}$

3. 最終的な答え

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

(2) $\frac{\sqrt{21}+\sqrt{7}}{2}$

(3) 1

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