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

1. 問題の内容

$x = \frac{2}{\sqrt{7} + \sqrt{3}}$、$y = \frac{2}{\sqrt{7} - \sqrt{3}}$ のとき、以下の式の値を求めよ。

(1) $x+y$、$xy$

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

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

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

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

2. 解き方の手順

まず、$x$ と $y$ をそれぞれ有理化する。

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

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

(1)

$x + y = \frac{\sqrt{7} - \sqrt{3}}{2} + \frac{\sqrt{7} + \sqrt{3}}{2} = \frac{2\sqrt{7}}{2} = \sqrt{7}$

$xy = \frac{\sqrt{7} - \sqrt{3}}{2} \cdot \frac{\sqrt{7} + \sqrt{3}}{2} = \frac{7 - 3}{4} = \frac{4}{4} = 1$

(2)

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

(3)

$x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = \sqrt{7}((\sqrt{7})^2 - 3(1)) = \sqrt{7}(7 - 3) = 4\sqrt{7}$

(4)

$x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2 = (5)^2 - 2(1)^2 = 25 - 2 = 23$

(5)

$x^5 + y^5 = (x^2+y^2)(x^3+y^3) - x^2y^2(x+y) = (5)(4\sqrt{7}) - (1)^2(\sqrt{7}) = 20\sqrt{7} - \sqrt{7} = 19\sqrt{7}$

3. 最終的な答え

(1) $x+y = \sqrt{7}$、$xy = 1$

(2) $x^2 + y^2 = 5$

(3) $x^3 + y^3 = 4\sqrt{7}$

(4) $x^4 + y^4 = 23$

(5) $x^5 + y^5 = 19\sqrt{7}$