$x = \frac{1}{\sqrt{7} + \sqrt{5}}$, $y = \frac{1}{\sqrt{7} - \sqrt{5}}$ のとき、以下の式の値を求めよ。 (1) $x+y$ (2) $xy$ (3) $x^2+y^2$ (4) $x^2y+xy^2$

1. 問題の内容

$x = \frac{1}{\sqrt{7} + \sqrt{5}}$, $y = \frac{1}{\sqrt{7} - \sqrt{5}}$ のとき、以下の式の値を求めよ。

(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$y$の分母を有理化する。

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

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

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

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

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

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

3. 最終的な答え

(1) $x+y = \sqrt{7}$

(2) $xy = \frac{1}{2}$

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

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