$x = \frac{1}{2 - \sqrt{3}}$, $y = \frac{1}{2 + \sqrt{3}}$ のとき、次の式の値を求めよ。 (1) $x+y$ (2) $xy$ (3) $x^2+y^2$ (4) $x^3y + xy^3$ (5) $x^3+y^3$ (6) $x^5y^2 + x^2y^5$

1. 問題の内容

$x = \frac{1}{2 - \sqrt{3}}$, $y = \frac{1}{2 + \sqrt{3}}$ のとき、次の式の値を求めよ。

(1) $x+y$

(2) $xy$

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

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

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

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

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化する。

$x = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}$

$y = \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$

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

(2) $xy = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$

(3) $x^2 + y^2 = (x+y)^2 - 2xy = 4^2 - 2(1) = 16 - 2 = 14$

(4) $x^3y + xy^3 = xy(x^2 + y^2) = (1)(14) = 14$

(5) $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = 4(4^2 - 3(1)) = 4(16 - 3) = 4(13) = 52$

(6) $x^5y^2 + x^2y^5 = x^2y^2(x^3 + y^3) = (xy)^2(x^3 + y^3) = 1^2 (52) = 52$

3. 最終的な答え

(1) $x+y = 4$

(2) $xy = 1$

(3) $x^2+y^2 = 14$

(4) $x^3y + xy^3 = 14$

(5) $x^3+y^3 = 52$

(6) $x^5y^2 + x^2y^5 = 52$