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

1. 問題の内容

$x = \frac{2}{\sqrt{3}+1}$、 $y = \frac{2}{\sqrt{3}-1}$ のとき、以下の値を求めます。

(1) $x+y$

(2) $xy$

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

(4) $x^2y + xy^2$

2. 解き方の手順

まず、$x$と$y$の分母を有理化します。

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

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

(1) $x+y = (\sqrt{3}-1) + (\sqrt{3}+1) = 2\sqrt{3}$

(2) $xy = (\sqrt{3}-1)(\sqrt{3}+1) = (\sqrt{3})^2 - 1^2 = 3-1 = 2$

(3) $x^2 + y^2 = (x+y)^2 - 2xy = (2\sqrt{3})^2 - 2(2) = 4(3) - 4 = 12 - 4 = 8$

(4) $x^2y + xy^2 = xy(x+y) = 2(2\sqrt{3}) = 4\sqrt{3}$

3. 最終的な答え

(1) $x+y = 2\sqrt{3}$

(2) $xy = 2$

(3) $x^2+y^2 = 8$

(4) $x^2y+xy^2 = 4\sqrt{3}$