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

1. 問題の内容

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

(1) $x+y$, $xy$

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

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

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化します。

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

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

(1)

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

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

(2)

$x^2+y^2 = (x+y)^2 - 2xy = (2\sqrt{2})^2 - 2(1) = 8-2 = 6$

(3)

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

3. 最終的な答え

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

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

(3) $x^2y+xy^2 = 2\sqrt{2}$