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

1. 問題の内容

$x = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}$、 $y = \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}}$ のとき、次の式の値を求めます。

(1) $x+y$, $xy$

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

2. 解き方の手順

まず、$x$と$y$の分母を有理化します。

$x = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{(\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3})}{(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})} = \frac{2 + 2\sqrt{6} + 3}{2 - 3} = \frac{5 + 2\sqrt{6}}{-1} = -5 - 2\sqrt{6}$

$y = \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}} = \frac{(\sqrt{2} - \sqrt{3})(\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} = \frac{2 - 2\sqrt{6} + 3}{2 - 3} = \frac{5 - 2\sqrt{6}}{-1} = -5 + 2\sqrt{6}$

(1) $x+y = (-5 - 2\sqrt{6}) + (-5 + 2\sqrt{6}) = -10$

$xy = (-5 - 2\sqrt{6})(-5 + 2\sqrt{6}) = (-5)^2 - (2\sqrt{6})^2 = 25 - 4 \cdot 6 = 25 - 24 = 1$

(2) $x^2 - xy + y^2 = (x+y)^2 - 3xy$

$x^2 - xy + y^2 = (-10)^2 - 3(1) = 100 - 3 = 97$

3. 最終的な答え

(1) $x+y = -10$, $xy = 1$

(2) $x^2 - xy + y^2 = 97$