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

## 数学の問題

1. 問題の内容

$x = \frac{\sqrt{5}+2}{\sqrt{5}-2}$、 $y = \frac{\sqrt{5}-2}{\sqrt{5}+2}$ のとき、以下の式の値を求めます。

(1) $x+y$

(2) $xy$

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

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

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

2. 解き方の手順

まず、$x$ と $y$ の分母を有理化します。

$x = \frac{\sqrt{5}+2}{\sqrt{5}-2} = \frac{(\sqrt{5}+2)(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)} = \frac{(\sqrt{5}+2)^2}{5-4} = (\sqrt{5}+2)^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}$

$y = \frac{\sqrt{5}-2}{\sqrt{5}+2} = \frac{(\sqrt{5}-2)(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)} = \frac{(\sqrt{5}-2)^2}{5-4} = (\sqrt{5}-2)^2 = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5}$

(1) $x+y = (9 + 4\sqrt{5}) + (9 - 4\sqrt{5}) = 18$

(2) $xy = (9 + 4\sqrt{5})(9 - 4\sqrt{5}) = 81 - (16 \times 5) = 81 - 80 = 1$

(3) $x^2y + xy^2 = xy(x+y) = 1 \times 18 = 18$

(4) $x^2 + y^2 = (x+y)^2 - 2xy = (18)^2 - 2(1) = 324 - 2 = 322$

(5) $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = 18(18^2 - 3(1)) = 18(324 - 3) = 18(321) = 5778$

3. 最終的な答え

(1) $x+y = 18$

(2) $xy = 1$

(3) $x^2y + xy^2 = 18$

(4) $x^2 + y^2 = 322$

(5) $x^3 + y^3 = 5778$