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

1. 問題の内容

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

(1) $x+y$

(2) $xy$

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

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

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

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化します。

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

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

(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) $x^3+y^3=(x+y)^3 - 3xy(x+y)=4^3 - 3(1)(4)=64-12=52$

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