$x = \frac{2}{\sqrt{5}-1}$, $y = \frac{2}{\sqrt{5}+1}$ のとき、以下の式の値を求める問題です。 (1) $x+y$ (2) $x^2+y^2$ (3) $\frac{y}{x}+\frac{x}{y}$ (4) $x^3+y^3$ (5) $x^4+y^4$ (6) $x^5+y^5$

1. 問題の内容

$x = \frac{2}{\sqrt{5}-1}$, $y = \frac{2}{\sqrt{5}+1}$ のとき、以下の式の値を求める問題です。

(1) $x+y$

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

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

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

(5) $x^4+y^4$

(6) $x^5+y^5$

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化します。

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

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

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

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

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

$x^2+y^2 = (\sqrt{5})^2 - 2(1) = 5 - 2 = 3$

(3) $\frac{y}{x}+\frac{x}{y} = \frac{y^2+x^2}{xy} = \frac{x^2+y^2}{xy} = \frac{3}{1} = 3$

(4) $x^3+y^3 = (x+y)(x^2-xy+y^2) = (x+y)((x+y)^2-3xy) = \sqrt{5}((\sqrt{5})^2-3(1)) = \sqrt{5}(5-3) = 2\sqrt{5}$

(5) $x^4+y^4 = (x^2+y^2)^2 - 2x^2y^2 = (3)^2 - 2(1)^2 = 9 - 2 = 7$

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

3. 最終的な答え

(1) $\sqrt{5}$

(2) $3$

(3) $3$

(4) $2\sqrt{5}$

(5) $7$

(6) $5\sqrt{5}$