実数 $a$ と $b$ が与えられており、$a = \frac{\sqrt{6}-\sqrt{2}}{2}$、$b = |a-\sqrt{6}|$ である。以下の問いに答えよ。 (1) $a$ の分母を有理化せよ。また、$b$ の値を求めよ。 (2) $a^2+b^2$、$a^3+b^3$ の値をそれぞれ求めよ。 (3) $a^5 - \sqrt{2}a^2 + \sqrt{2}b^2 + b^5$ の値を求めよ。

1. 問題の内容

実数 $a$ と $b$ が与えられており、$a = \frac{\sqrt{6}-\sqrt{2}}{2}$、$b = |a-\sqrt{6}|$ である。以下の問いに答えよ。

(1) $a$ の分母を有理化せよ。また、$b$ の値を求めよ。

(2) $a^2+b^2$、$a^3+b^3$ の値をそれぞれ求めよ。

(3) $a^5 - \sqrt{2}a^2 + \sqrt{2}b^2 + b^5$ の値を求めよ。

2. 解き方の手順

(1) $a$ の分母を有理化する。

$a = \frac{\sqrt{6}-\sqrt{2}}{2}$

$b = |a - \sqrt{6}| = |\frac{\sqrt{6}-\sqrt{2}}{2} - \sqrt{6}| = |\frac{\sqrt{6}-\sqrt{2}-2\sqrt{6}}{2}| = |\frac{-\sqrt{6}-\sqrt{2}}{2}| = \frac{\sqrt{6}+\sqrt{2}}{2}$

(2) $a^2 + b^2$ と $a^3 + b^3$ を計算する。

$a^2 = (\frac{\sqrt{6}-\sqrt{2}}{2})^2 = \frac{6 - 2\sqrt{12} + 2}{4} = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}$

$b^2 = (\frac{\sqrt{6}+\sqrt{2}}{2})^2 = \frac{6 + 2\sqrt{12} + 2}{4} = \frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}$

$a^2 + b^2 = 2 - \sqrt{3} + 2 + \sqrt{3} = 4$

$a^3 = (\frac{\sqrt{6}-\sqrt{2}}{2})^3 = \frac{(\sqrt{6}-\sqrt{2})^3}{8} = \frac{6\sqrt{6} - 3\cdot6\sqrt{2} + 3\cdot\sqrt{6}\cdot2 - 2\sqrt{2}}{8} = \frac{6\sqrt{6} - 18\sqrt{2} + 6\sqrt{6} - 2\sqrt{2}}{8} = \frac{12\sqrt{6} - 20\sqrt{2}}{8} = \frac{3\sqrt{6} - 5\sqrt{2}}{2}$

$b^3 = (\frac{\sqrt{6}+\sqrt{2}}{2})^3 = \frac{(\sqrt{6}+\sqrt{2})^3}{8} = \frac{6\sqrt{6} + 3\cdot6\sqrt{2} + 3\cdot\sqrt{6}\cdot2 + 2\sqrt{2}}{8} = \frac{6\sqrt{6} + 18\sqrt{2} + 6\sqrt{6} + 2\sqrt{2}}{8} = \frac{12\sqrt{6} + 20\sqrt{2}}{8} = \frac{3\sqrt{6} + 5\sqrt{2}}{2}$

$a^3 + b^3 = \frac{3\sqrt{6} - 5\sqrt{2}}{2} + \frac{3\sqrt{6} + 5\sqrt{2}}{2} = \frac{6\sqrt{6}}{2} = 3\sqrt{6}$

(3) $a^5 - \sqrt{2}a^2 + \sqrt{2}b^2 + b^5$ の値を求める。

$a^5 + b^5 - \sqrt{2}(a^2 - b^2)$ を計算する。

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

$a^5 + b^5 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)$

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

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

$a^5 + b^5 = (4)(3\sqrt{6}) - (1)(\sqrt{6}) = 12\sqrt{6} - \sqrt{6} = 11\sqrt{6}$

$a^5 + b^5 - \sqrt{2}(a^2-b^2) = 11\sqrt{6} - \sqrt{2}(-2\sqrt{3}) = 11\sqrt{6} + 2\sqrt{6} = 13\sqrt{6}$

3. 最終的な答え

(1) $a = \frac{\sqrt{6}-\sqrt{2}}{2}$, $b = \frac{\sqrt{6}+\sqrt{2}}{2}$

(2) $a^2 + b^2 = 4$, $a^3 + b^3 = 3\sqrt{6}$

(3) $a^5 - \sqrt{2}a^2 + \sqrt{2}b^2 + b^5 = 13\sqrt{6}$

(1) 分母を有理化すると、$a = \frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2(\sqrt{6}+\sqrt{2})} = \frac{6-2}{2(\sqrt{6}+\sqrt{2})} = \frac{4}{2(\sqrt{6}+\sqrt{2})} = \frac{2}{\sqrt{6}+\sqrt{2}} = \frac{2(\sqrt{6}-\sqrt{2})}{4} = \frac{\sqrt{6}-\sqrt{2}}{2}$

$b = |a-\sqrt{6}| = |\frac{\sqrt{6}-\sqrt{2}}{2} - \sqrt{6}| = |\frac{-\sqrt{6}-\sqrt{2}}{2}| = \frac{\sqrt{6}+\sqrt{2}}{2}$

$a=\frac{\sqrt{6}+\sqrt{2}}{2}$

最終的な答え

(1) $a = \frac{\sqrt{6}+\sqrt{2}}{2}, b = \frac{\sqrt{6}+\sqrt{2}}{2}$

(2) $a^2+b^2=4, a^3+b^3=3\sqrt{6}$

(3) $11\sqrt{6} + 2\sqrt{6} = 9\sqrt{6}$

最終的な答え

(1) $a = \frac{\sqrt{6}+\sqrt{2}}{2}, b = \frac{\sqrt{6}+\sqrt{2}}{2}$

(2) $a^2+b^2=4, a^3+b^3=3\sqrt{6}$

(3) $9\sqrt{6}$