$a = \frac{4}{3\sqrt{2}-\sqrt{10}}$ 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) $a^4 - \frac{16}{a^4} - \frac{8}{a^2} - 1$ 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) $a^4 - \frac{16}{a^4} - \frac{8}{a^2} - 1$ の値を求めよ。

2. 解き方の手順

(1) $a$ の分母を有理化する。

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

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

(2) $a + \frac{2}{a}$ の値を求める。

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

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

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

次に、$a^2 + \frac{4}{a^2}$ の値を求める。

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

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

$a^2 + \frac{4}{a^2} = 14$

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

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

$a^4 + \frac{16}{a^4} = (a^2 + \frac{4}{a^2})^2 - 8 = (14)^2 - 8 = 196 - 8 = 188$

$a^4 - \frac{16}{a^4} = a^4 + \frac{16}{a^4} - 2 \cdot \frac{16}{a^4} = 188 - \frac{32}{a^4}$

$a^4 - \frac{16}{a^4} = 188 - \frac{32}{a^4}$

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

$a^2 + \frac{4}{a^2} = 14$

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

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

$a^2 - \frac{4}{a^2} = 3\sqrt{2} \cdot \sqrt{10} = 3\sqrt{20} = 3 \cdot 2 \sqrt{5} = 6\sqrt{5}$

$a^4 - \frac{16}{a^4} - \frac{8}{a^2} - 1 = (a^2+\frac{4}{a^2})(a^2-\frac{4}{a^2}) - \frac{8}{a^2}-1=14(6\sqrt{5})-\frac{8}{a^2}-1$

$a^4 - \frac{16}{a^4} - \frac{8}{a^2} - 1 = a^4 - (\frac{4}{a^2})^2 - 2 (\frac{2}{a})^2 - 1 = (a^2)^2 - (2/a)^4 -2(2/a)^2 - 1 = (a^2 - \frac{4}{a^2})^2+4a^2\frac{4}{a^2}- \frac{8}{a^2}-1 = 36*5 + 16- 8/a^2 -1 = 180 + 16 -8/a^2-1 = 195-8/a^2$

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

$195-\frac{8}{7+3\sqrt{5}} = 195 - \frac{8(7-3\sqrt{5})}{49-45} = 195 - \frac{8(7-3\sqrt{5})}{4} = 195 - 2(7-3\sqrt{5}) = 195 - 14+6\sqrt{5} = 181+6\sqrt{5}$

$a^4 - \frac{16}{a^4} - \frac{8}{a^2} - 1 = 0$

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) $a^4 - \frac{16}{a^4} - \frac{8}{a^2} - 1 = 0$