$x = \frac{2}{\sqrt{3}+1}$ と $y = \frac{2}{\sqrt{3}-1}$ 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) $x^3y + xy^3$

1. 問題の内容

$x = \frac{2}{\sqrt{3}+1}$ と $y = \frac{2}{\sqrt{3}-1}$ 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) $x^3y + xy^3$

2. 解き方の手順

まず、$x$と$y$をそれぞれ簡単にします。

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

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

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

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

(3) $x^2 + y^2 = (x+y)^2 - 2xy = (2\sqrt{3})^2 - 2(2) = 4(3) - 4 = 12 - 4 = 8$

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

または

$x^3y + xy^3 = xy(x^2+y^2) = 2 \cdot 8 = 16$

または

$x^3y + xy^3 = xy(x^2+y^2) = xy((x+y)^2-2xy) = 2((2\sqrt{3})^2 - 2(2)) = 2(12-4) = 2(8) = 16$

3. 最終的な答え

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

(2) $xy = 2$

(3) $x^2 + y^2 = 8$

(4) $x^3y + xy^3 = 16$