与えられた3つの式をそれぞれ簡単にします。 (1) $(\sqrt{3} + \sqrt{5} + \sqrt{7})(\sqrt{3} + \sqrt{5} - \sqrt{7})(\sqrt{3} - \sqrt{5} + \sqrt{7})(-\sqrt{3} + \sqrt{5} + \sqrt{7})$ (2) $\frac{1 + \sqrt{2} - \sqrt{3}}{1 + \sqrt{2} + \sqrt{3}} + \frac{1 - \sqrt{2} - \sqrt{3}}{1 - \sqrt{2} + \sqrt{3}}$ (3) $\sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}}$

1. 問題の内容

与えられた3つの式をそれぞれ簡単にします。

(1) $(\sqrt{3} + \sqrt{5} + \sqrt{7})(\sqrt{3} + \sqrt{5} - \sqrt{7})(\sqrt{3} - \sqrt{5} + \sqrt{7})(-\sqrt{3} + \sqrt{5} + \sqrt{7})$

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

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

2. 解き方の手順

(1) まず、$A = \sqrt{3} + \sqrt{5}$ とおくと、

$(\sqrt{3} + \sqrt{5} + \sqrt{7})(\sqrt{3} + \sqrt{5} - \sqrt{7}) = (A + \sqrt{7})(A - \sqrt{7}) = A^2 - 7 = (\sqrt{3} + \sqrt{5})^2 - 7 = 3 + 2\sqrt{15} + 5 - 7 = 1 + 2\sqrt{15}$

次に、$B = \sqrt{5} + \sqrt{7}$ とおくと、

$(\sqrt{3} - \sqrt{5} + \sqrt{7})(-\sqrt{3} + \sqrt{5} + \sqrt{7}) = (B - \sqrt{3})(B + \sqrt{3}) = B^2 - 3 = (\sqrt{5} + \sqrt{7})^2 - 3 = 5 + 2\sqrt{35} + 7 - 3 = 9 + 2\sqrt{35}$

したがって、与えられた式は

$(1 + 2\sqrt{15})(9 + 2\sqrt{35}) = 9 + 2\sqrt{35} + 18\sqrt{15} + 4\sqrt{525} = 9 + 2\sqrt{35} + 18\sqrt{15} + 4\sqrt{25 \cdot 21} = 9 + 2\sqrt{35} + 18\sqrt{15} + 4 \cdot 5 \sqrt{21} = 9 + 2\sqrt{35} + 18\sqrt{15} + 20\sqrt{21}$

ここで、別の考え方として、

$a = \sqrt{3}, b = \sqrt{5}, c = \sqrt{7}$ とおくと、

$(a+b+c)(a+b-c)(a-b+c)(-a+b+c) = ((a+b)^2 - c^2)((b+c)^2 - a^2) = (a^2+2ab+b^2 - c^2)(b^2+2bc+c^2 - a^2) = (3+2\sqrt{15}+5-7)(5+2\sqrt{35}+7-3) = (1+2\sqrt{15})(9+2\sqrt{35}) = 9+2\sqrt{35}+18\sqrt{15}+4\sqrt{525} = 9+2\sqrt{35}+18\sqrt{15}+20\sqrt{21}$

より、答えは $9 + 2\sqrt{35} + 18\sqrt{15} + 20\sqrt{21}$

(2)

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

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

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

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

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

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

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

$= \frac{4\sqrt{2} - 4\sqrt{6}}{2\sqrt{2}} = 2 - 2\sqrt{3}$

(3)

$x = \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}}$ とおくと、

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

$x > 0$ なので、$x = \sqrt{6}$

3. 最終的な答え

(1) $9 + 2\sqrt{35} + 18\sqrt{15} + 20\sqrt{21}$

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

(3) $\sqrt{6}$