$x = \frac{1}{\sqrt{7}-\sqrt{5}}$、$y = \frac{1}{\sqrt{7}+\sqrt{5}}$ のとき、次の式の値を求めます。 (1) $x+y$, $xy$ (2) $x^2+y^2$

1. 問題の内容

$x = \frac{1}{\sqrt{7}-\sqrt{5}}$、$y = \frac{1}{\sqrt{7}+\sqrt{5}}$ のとき、次の式の値を求めます。

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

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

2. 解き方の手順

まず、$x$と$y$の分母を有理化します。

$x = \frac{1}{\sqrt{7}-\sqrt{5}} = \frac{\sqrt{7}+\sqrt{5}}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})} = \frac{\sqrt{7}+\sqrt{5}}{7-5} = \frac{\sqrt{7}+\sqrt{5}}{2}$

$y = \frac{1}{\sqrt{7}+\sqrt{5}} = \frac{\sqrt{7}-\sqrt{5}}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})} = \frac{\sqrt{7}-\sqrt{5}}{7-5} = \frac{\sqrt{7}-\sqrt{5}}{2}$

(1)

$x+y = \frac{\sqrt{7}+\sqrt{5}}{2} + \frac{\sqrt{7}-\sqrt{5}}{2} = \frac{\sqrt{7}+\sqrt{5}+\sqrt{7}-\sqrt{5}}{2} = \frac{2\sqrt{7}}{2} = \sqrt{7}$

$xy = \frac{\sqrt{7}+\sqrt{5}}{2} \cdot \frac{\sqrt{7}-\sqrt{5}}{2} = \frac{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}{4} = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$

(2)

$x^2+y^2 = (x+y)^2 - 2xy = (\sqrt{7})^2 - 2(\frac{1}{2}) = 7 - 1 = 6$

3. 最終的な答え

(1) $x+y = \sqrt{7}$, $xy = \frac{1}{2}$

(2) $x^2+y^2 = 6$