$a = \frac{4}{3\sqrt{2} - \sqrt{10}}$ とする。 (1) $a$ の分母を有理化し、簡単にせよ。 (2) $a + \frac{2}{a}$ の値を求めよ。また、$a^2 + \frac{4}{a^2}$ の値を求めよ。 (3) $\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1}$ の値を求めよ。

1. 問題の内容

$a = \frac{4}{3\sqrt{2} - \sqrt{10}}$ とする。

(1) $a$ の分母を有理化し、簡単にせよ。

(2) $a + \frac{2}{a}$ の値を求めよ。また、$a^2 + \frac{4}{a^2}$ の値を求めよ。

(3) $\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1}$ の値を求めよ。

2. 解き方の手順

(1) $a$ の分母を有理化する。

$a = \frac{4}{3\sqrt{2} - \sqrt{10}} = \frac{4(3\sqrt{2} + \sqrt{10})}{(3\sqrt{2} - \sqrt{10})(3\sqrt{2} + \sqrt{10})}$

$= \frac{4(3\sqrt{2} + \sqrt{10})}{(3\sqrt{2})^2 - (\sqrt{10})^2} = \frac{4(3\sqrt{2} + \sqrt{10})}{18 - 10} = \frac{4(3\sqrt{2} + \sqrt{10})}{8} = \frac{3\sqrt{2} + \sqrt{10}}{2}$

(2) $a + \frac{2}{a}$ を求める。

$\frac{2}{a} = \frac{2}{\frac{3\sqrt{2} + \sqrt{10}}{2}} = \frac{4}{3\sqrt{2} + \sqrt{10}} = \frac{4(3\sqrt{2} - \sqrt{10})}{(3\sqrt{2} + \sqrt{10})(3\sqrt{2} - \sqrt{10})} = \frac{4(3\sqrt{2} - \sqrt{10})}{18 - 10} = \frac{4(3\sqrt{2} - \sqrt{10})}{8} = \frac{3\sqrt{2} - \sqrt{10}}{2}$

$a + \frac{2}{a} = \frac{3\sqrt{2} + \sqrt{10}}{2} + \frac{3\sqrt{2} - \sqrt{10}}{2} = \frac{3\sqrt{2} + \sqrt{10} + 3\sqrt{2} - \sqrt{10}}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$

$a^2 + \frac{4}{a^2}$ を求める。

$(a + \frac{2}{a})^2 = a^2 + 2(a)(\frac{2}{a}) + \frac{4}{a^2} = a^2 + 4 + \frac{4}{a^2}$

$a^2 + \frac{4}{a^2} = (a + \frac{2}{a})^2 - 4 = (3\sqrt{2})^2 - 4 = 9(2) - 4 = 18 - 4 = 14$

(3) $\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1}$ を求める。

$a^4 - 16 = (a^2)^2 - 4^2 = (a^2 - 4)(a^2 + 4)$

$a^2 - \frac{8}{a^2} - 1 = a^2 + \frac{4}{a^2} - \frac{12}{a^2} - 1 = 14 - \frac{12}{a^2} - 1 = 13 - \frac{12}{a^2}$

$\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1} = \frac{(a^2 - 4)(a^2 + 4)}{a^2 - \frac{8}{a^2} - 1} = \frac{(a^2 - 4)(a^2 + 4)}{\frac{a^4 - a^2 - 8}{a^2}}$

$a^2 = (\frac{3\sqrt{2} + \sqrt{10}}{2})^2 = \frac{(3\sqrt{2})^2 + 2(3\sqrt{2})(\sqrt{10}) + (\sqrt{10})^2}{4} = \frac{18 + 6\sqrt{20} + 10}{4} = \frac{28 + 12\sqrt{5}}{4} = 7 + 3\sqrt{5}$

$a^4 = (7 + 3\sqrt{5})^2 = 49 + 42\sqrt{5} + 9(5) = 49 + 42\sqrt{5} + 45 = 94 + 42\sqrt{5}$

$a^4 - a^2 - 8 = (94 + 42\sqrt{5}) - (7 + 3\sqrt{5}) - 8 = 94 - 7 - 8 + 42\sqrt{5} - 3\sqrt{5} = 79 + 39\sqrt{5}$

$\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1} = \frac{a^2(a^2 - 4)(a^2 + 4)}{a^4 - a^2 - 8} = \frac{(7 + 3\sqrt{5})((7 + 3\sqrt{5}) - 4)((7 + 3\sqrt{5}) + 4)}{79 + 39\sqrt{5}} = \frac{(7 + 3\sqrt{5})(3 + 3\sqrt{5})(11 + 3\sqrt{5})}{79 + 39\sqrt{5}} = \frac{(7 + 3\sqrt{5})(33 + 9\sqrt{5} + 33\sqrt{5} + 45)}{79 + 39\sqrt{5}} = \frac{(7 + 3\sqrt{5})(78 + 42\sqrt{5})}{79 + 39\sqrt{5}} = \frac{546 + 294\sqrt{5} + 234\sqrt{5} + 126(5)}{79 + 39\sqrt{5}} = \frac{546 + 780 + 528\sqrt{5}}{79 + 39\sqrt{5}} = \frac{1326 + 528\sqrt{5}}{79 + 39\sqrt{5}} = \frac{6(221 + 88\sqrt{5})}{79 + 39\sqrt{5}} = \frac{6(221 + 88\sqrt{5})(79 - 39\sqrt{5})}{(79 + 39\sqrt{5})(79 - 39\sqrt{5})} = \frac{6(221(79) - 221(39)\sqrt{5} + 88(79)\sqrt{5} - 88(39)(5))}{79^2 - (39^2)(5)} = \frac{6(17459 - 8619\sqrt{5} + 6952\sqrt{5} - 17160)}{6241 - 1521(5)} = \frac{6(17459 - 17160 - (8619 - 6952)\sqrt{5})}{6241 - 7605} = \frac{6(299 - 1667\sqrt{5})}{-1364}$

別のやり方:

$\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1} = \frac{(a^2-4)(a^2+4)}{a^2 - \frac{8}{a^2} - 1} = \frac{(a - \frac{2}{a})(a + \frac{2}{a})a^2(a^2 + \frac{4}{a^2} - \frac{12}{a^2})}{a^2(a^2 + \frac{4}{a^2} - \frac{12}{a^2} - 1)} = \frac{(a - \frac{2}{a})(3\sqrt{2})(14 - \frac{12}{7 + 3\sqrt{5}})}{13} = \frac{[(3/2)\sqrt{2} + \sqrt{10} -(3/2)\sqrt{2} + \sqrt{10}]3\sqrt{2}(7 + 3\sqrt{5})(14(7 + 3\sqrt{5}) - 12}{79 + 39\sqrt{5}}$

$a - \frac{2}{a} = \frac{3\sqrt{2} + \sqrt{10}}{2} - \frac{3\sqrt{2} - \sqrt{10}}{2} = \frac{3\sqrt{2} + \sqrt{10} - 3\sqrt{2} + \sqrt{10}}{2} = \frac{2\sqrt{10}}{2} = \sqrt{10}$

$\frac{(a^2 +4)(a^2-4)}{\frac{a^4-a^2-8}{a^2}} = \frac{(7+3\sqrt{5} +4)(7+3\sqrt{5}-4)}{\frac{(7+3\sqrt{5})^2-7-3\sqrt{5}-8}{(7+3\sqrt{5})}}$

(3)

$\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1} = a^2 + \frac{4}{a^2} = \frac{(a^2 - 4)(a^2 + 4)}{\frac{a^4-a^2-8}{a^2}}$

3. 最終的な答え

(1) $a = \frac{3\sqrt{2} + \sqrt{10}}{2}$

(2) $a + \frac{2}{a} = 3\sqrt{2}$, $a^2 + \frac{4}{a^2} = 14$

(3) $\frac{a^4 - 16}{a^2 - \frac{8}{a^2} - 1}$