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

1. 問題の内容

$x = \frac{2}{\sqrt{5} + \sqrt{3}}$、 $y = \frac{2}{\sqrt{5} - \sqrt{3}}$のとき、以下の式の値を求めよ。

(1) $x + y$

(2) $xy$

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

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

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

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化する。

$x = \frac{2}{\sqrt{5} + \sqrt{3}} = \frac{2(\sqrt{5} - \sqrt{3})}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})} = \frac{2(\sqrt{5} - \sqrt{3})}{5 - 3} = \frac{2(\sqrt{5} - \sqrt{3})}{2} = \sqrt{5} - \sqrt{3}$

$y = \frac{2}{\sqrt{5} - \sqrt{3}} = \frac{2(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} = \frac{2(\sqrt{5} + \sqrt{3})}{5 - 3} = \frac{2(\sqrt{5} + \sqrt{3})}{2} = \sqrt{5} + \sqrt{3}$

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

(2) $xy = (\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = 5 - 3 = 2$

(3) $x^2 + y^2 = (x + y)^2 - 2xy = (2\sqrt{5})^2 - 2(2) = 4(5) - 4 = 20 - 4 = 16$

(4) $x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x + y)((x + y)^2 - 3xy) = (2\sqrt{5})((2\sqrt{5})^2 - 3(2)) = (2\sqrt{5})(20 - 6) = (2\sqrt{5})(14) = 28\sqrt{5}$

(5) $x^3y - x^2y^2 + xy^3 = xy(x^2 - xy + y^2) = xy((x+y)^2 - 3xy) = 2((2\sqrt{5})^2 - 3(2)) = 2(20 - 6) = 2(14) = 28$

3. 最終的な答え

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

(2) $xy = 2$

(3) $x^2 + y^2 = 16$

(4) $x^3 + y^3 = 28\sqrt{5}$

(5) $x^3y - x^2y^2 + xy^3 = 28$