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

1. 問題の内容

$x = \frac{1}{1+\sqrt{2}}$、$y = \frac{1}{1-\sqrt{2}}$ のとき、以下の式の値を求める問題です。

(1) $x+y$

(2) $xy$

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

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

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化します。

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

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

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

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

(3) $x^2 + y^2 = (x+y)^2 - 2xy = (-2)^2 - 2(-1) = 4 + 2 = 6$

(4) $x^3+y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = (-2)((-2)^2 - 3(-1)) = (-2)(4+3) = (-2)(7) = -14$

3. 最終的な答え

(1) $x+y = -2$

(2) $xy = -1$

(3) $x^2 + y^2 = 6$

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