$x = \frac{3}{\sqrt{5}+\sqrt{2}}$、 $y = \frac{3}{\sqrt{5}-\sqrt{2}}$ のとき、以下の式の値を求めます。 (i) $x+y$ (ii) $xy$ (iii) $x^2+y^2$ (iv) $x^3+y^3$

1. 問題の内容

$x = \frac{3}{\sqrt{5}+\sqrt{2}}$、 $y = \frac{3}{\sqrt{5}-\sqrt{2}}$ のとき、以下の式の値を求めます。

(i) $x+y$

(ii) $xy$

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

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

2. 解き方の手順

まず、$x$と$y$を有理化します。

$x = \frac{3}{\sqrt{5}+\sqrt{2}} = \frac{3(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})} = \frac{3(\sqrt{5}-\sqrt{2})}{5-2} = \sqrt{5}-\sqrt{2}$

$y = \frac{3}{\sqrt{5}-\sqrt{2}} = \frac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})} = \frac{3(\sqrt{5}+\sqrt{2})}{5-2} = \sqrt{5}+\sqrt{2}$

(i) $x+y = (\sqrt{5}-\sqrt{2})+(\sqrt{5}+\sqrt{2}) = 2\sqrt{5}$

(ii) $xy = (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2}) = 5-2 = 3$

(iii) $x^2+y^2 = (x+y)^2 - 2xy = (2\sqrt{5})^2 - 2(3) = 4(5)-6 = 20-6 = 14$

(iv) $x^3+y^3 = (x+y)(x^2-xy+y^2) = (x+y)((x+y)^2-3xy) = (2\sqrt{5})((2\sqrt{5})^2-3(3)) = (2\sqrt{5})(20-9) = 2\sqrt{5}(11) = 22\sqrt{5}$

3. 最終的な答え

(i) $x+y = 2\sqrt{5}$

(ii) $xy = 3$

(iii) $x^2+y^2 = 14$

(iv) $x^3+y^3 = 22\sqrt{5}$