$x = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$、$y = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$のとき、次の値を求めます。 (1) $x+y$ (2) $xy$ (3) $x^2 + y^2$

1. 問題の内容

$x = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$、$y = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$のとき、次の値を求めます。

(1) $x+y$

(2) $xy$

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

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化します。

$x = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}} = \frac{(\sqrt{7} - \sqrt{3})(\sqrt{7} - \sqrt{3})}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})} = \frac{7 - 2\sqrt{21} + 3}{7 - 3} = \frac{10 - 2\sqrt{21}}{4} = \frac{5 - \sqrt{21}}{2}$

$y = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} = \frac{(\sqrt{7} + \sqrt{3})(\sqrt{7} + \sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})} = \frac{7 + 2\sqrt{21} + 3}{7 - 3} = \frac{10 + 2\sqrt{21}}{4} = \frac{5 + \sqrt{21}}{2}$

(1) $x+y = \frac{5 - \sqrt{21}}{2} + \frac{5 + \sqrt{21}}{2} = \frac{5 - \sqrt{21} + 5 + \sqrt{21}}{2} = \frac{10}{2} = 5$

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

または、$xy = \frac{5 - \sqrt{21}}{2} \cdot \frac{5 + \sqrt{21}}{2} = \frac{25 - 21}{4} = \frac{4}{4} = 1$

(3) $x^2 + y^2 = (x+y)^2 - 2xy = 5^2 - 2(1) = 25 - 2 = 23$

3. 最終的な答え

(1) $x+y = 5$

(2) $xy = 1$

(3) $x^2 + y^2 = 23$