$x = \frac{2}{\sqrt{6}-2}$、$y = \frac{2}{\sqrt{6}+2}$ のとき、以下の値を求めよ。 (1) $x+y$ (2) $xy$ (3) $x^2 + y^2$ (4) $x^3 + y^3$

1. 問題の内容

$x = \frac{2}{\sqrt{6}-2}$、$y = \frac{2}{\sqrt{6}+2}$ のとき、以下の値を求めよ。

(1) $x+y$

(2) $xy$

(3) $x^2 + y^2$

(4) $x^3 + y^3$

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化する。

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

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

(1) $x+y = (\sqrt{6}+2) + (\sqrt{6}-2) = 2\sqrt{6}$

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

(3) $x^2 + y^2 = (x+y)^2 - 2xy = (2\sqrt{6})^2 - 2(2) = 4 \cdot 6 - 4 = 24 - 4 = 20$

(4) $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = 2\sqrt{6}((2\sqrt{6})^2 - 3(2)) = 2\sqrt{6}(24-6) = 2\sqrt{6}(18) = 36\sqrt{6}$

3. 最終的な答え

(1) $x+y = 2\sqrt{6}$

(2) $xy = 2$

(3) $x^2 + y^2 = 20$

(4) $x^3 + y^3 = 36\sqrt{6}$