$x = \frac{\sqrt{7} + \sqrt{5}}{2}$、 $y = \frac{\sqrt{7} - \sqrt{5}}{2}$ のとき、以下の式の値を求めよ。 (1) $x+y$ (2) $xy$ (3) $x^2+y^2$ (4) $x^3y+xy^3$

1. 問題の内容

$x = \frac{\sqrt{7} + \sqrt{5}}{2}$、 $y = \frac{\sqrt{7} - \sqrt{5}}{2}$ のとき、以下の式の値を求めよ。

(1) $x+y$

(2) $xy$

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

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

2. 解き方の手順

(1) $x+y$

$x+y = \frac{\sqrt{7} + \sqrt{5}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2}$

$x+y = \frac{\sqrt{7} + \sqrt{5} + \sqrt{7} - \sqrt{5}}{2}$

$x+y = \frac{2\sqrt{7}}{2}$

$x+y = \sqrt{7}$

(2) $xy$

$xy = \frac{\sqrt{7} + \sqrt{5}}{2} \times \frac{\sqrt{7} - \sqrt{5}}{2}$

$xy = \frac{(\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5})}{4}$

$xy = \frac{(\sqrt{7})^2 - (\sqrt{5})^2}{4}$

$xy = \frac{7 - 5}{4}$

$xy = \frac{2}{4}$

$xy = \frac{1}{2}$

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

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

$x^2+y^2 = (\sqrt{7})^2 - 2 \times \frac{1}{2}$

$x^2+y^2 = 7 - 1$

$x^2+y^2 = 6$

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

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

$x^3y+xy^3 = \frac{1}{2} \times 6$

$x^3y+xy^3 = 3$

3. 最終的な答え

(1) $x+y = \sqrt{7}$

(2) $xy = \frac{1}{2}$

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

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