(1) $a = \frac{2}{2\sqrt{2} - \sqrt{6}}$, $b = \frac{2}{2\sqrt{2} + \sqrt{6}}$のとき、$a^2 + b^2$と$a^3 + b^3$の値を求める。 (2) $\sqrt{2}$の整数部分を$a$, 小数部分を$b$とするとき、$a^2 - ab + b^2$と$\frac{1}{b} + \frac{2}{b+2}$の値を求める。

1. 問題の内容

(1) $a = \frac{2}{2\sqrt{2} - \sqrt{6}}$, $b = \frac{2}{2\sqrt{2} + \sqrt{6}}$のとき、$a^2 + b^2$と$a^3 + b^3$の値を求める。

(2) $\sqrt{2}$の整数部分を$a$, 小数部分を$b$とするとき、$a^2 - ab + b^2$と$\frac{1}{b} + \frac{2}{b+2}$の値を求める。

2. 解き方の手順

(1) まず、$a$と$b$をそれぞれ有理化する。

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

$b = \frac{2}{2\sqrt{2} + \sqrt{6}} = \frac{2(2\sqrt{2} - \sqrt{6})}{(2\sqrt{2})^2 - (\sqrt{6})^2} = \frac{2(2\sqrt{2} - \sqrt{6})}{8 - 6} = \frac{2(2\sqrt{2} - \sqrt{6})}{2} = 2\sqrt{2} - \sqrt{6}$

次に、$a+b$と$ab$を計算する。

$a+b = (2\sqrt{2} + \sqrt{6}) + (2\sqrt{2} - \sqrt{6}) = 4\sqrt{2}$

$ab = (2\sqrt{2} + \sqrt{6})(2\sqrt{2} - \sqrt{6}) = (2\sqrt{2})^2 - (\sqrt{6})^2 = 8 - 6 = 2$

$a^2 + b^2 = (a+b)^2 - 2ab = (4\sqrt{2})^2 - 2(2) = 16(2) - 4 = 32 - 4 = 28$

$a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2 - 3ab) = (4\sqrt{2})((4\sqrt{2})^2 - 3(2)) = 4\sqrt{2}(32 - 6) = 4\sqrt{2}(26) = 104\sqrt{2}$

(2) $\sqrt{2} \approx 1.414$より、$\sqrt{2}$の整数部分は$a = 1$, 小数部分は$b = \sqrt{2} - 1$となる。

$a^2 - ab + b^2 = (1)^2 - (1)(\sqrt{2} - 1) + (\sqrt{2} - 1)^2 = 1 - \sqrt{2} + 1 + (2 - 2\sqrt{2} + 1) = 1 - \sqrt{2} + 1 + 3 - 2\sqrt{2} = 5 - 3\sqrt{2}$

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

3. 最終的な答え

(1) $a^2 + b^2 = 28$, $a^3 + b^3 = 104\sqrt{2}$

(2) $a^2 - ab + b^2 = 5 - 3\sqrt{2}$, $\frac{1}{b} + \frac{2}{b+2} = 3\sqrt{2} - 1$