$x = \frac{1}{\sqrt{5}-2}$ と $y = \frac{1}{\sqrt{5}+2}$ が与えられたとき、次の式の値を求めます。 (1) $x+y$ (2) $xy$ (3) $x^2 - y^2$ (4) $\frac{x}{y} + \frac{y}{x}$

1. 問題の内容

$x = \frac{1}{\sqrt{5}-2}$ と $y = \frac{1}{\sqrt{5}+2}$ が与えられたとき、次の式の値を求めます。

(1) $x+y$

(2) $xy$

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

(4) $\frac{x}{y} + \frac{y}{x}$

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化します。

$x = \frac{1}{\sqrt{5}-2} = \frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)} = \frac{\sqrt{5}+2}{5-4} = \sqrt{5}+2$

$y = \frac{1}{\sqrt{5}+2} = \frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)} = \frac{\sqrt{5}-2}{5-4} = \sqrt{5}-2$

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

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

(3) $x^2 - y^2 = (x+y)(x-y) = (2\sqrt{5})((\sqrt{5}+2)-(\sqrt{5}-2)) = (2\sqrt{5})(4) = 8\sqrt{5}$

(4) $\frac{x}{y} + \frac{y}{x} = \frac{x^2+y^2}{xy} = \frac{(x+y)^2 - 2xy}{xy} = \frac{(2\sqrt{5})^2 - 2(1)}{1} = \frac{20-2}{1} = 18$

3. 最終的な答え

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

(2) $xy = 1$

(3) $x^2 - y^2 = 8\sqrt{5}$

(4) $\frac{x}{y} + \frac{y}{x} = 18$