$x = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, $y = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ のとき、$x+y$, $xy$, $x^2+y^2$, $x^3+y^3$, $x^4+y^4$, $x^5+y^5$ の値を求める。

1. 問題の内容

$x = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, $y = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ のとき、$x+y$, $xy$, $x^2+y^2$, $x^3+y^3$, $x^4+y^4$, $x^5+y^5$ の値を求める。

2. 解き方の手順

まず、$x$ と $y$ をそれぞれ有理化する。

$x = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \frac{3 - 2\sqrt{6} + 2}{3 - 2} = 5 - 2\sqrt{6}$

$y = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} = \frac{3 + 2\sqrt{6} + 2}{3 - 2} = 5 + 2\sqrt{6}$

したがって、$x+y = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10$

$xy = (5 - 2\sqrt{6})(5 + 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 4 \cdot 6 = 25 - 24 = 1$

$x^2 + y^2 = (x+y)^2 - 2xy = 10^2 - 2(1) = 100 - 2 = 98$

$x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = 10(10^2 - 3(1)) = 10(100 - 3) = 10 \cdot 97 = 970$

$x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2 = (98)^2 - 2(1)^2 = 9604 - 2 = 9602$

$x^5 + y^5 = (x^2+y^2)(x^3+y^3) - x^2y^2(x+y) = (98)(970) - (1)^2(10) = 95060 - 10 = 95050$

3. 最終的な答え

$x+y = 10$

$xy = 1$

$x^2 + y^2 = 98$

$x^3 + y^3 = 970$

$x^4 + y^4 = 9602$

$x^5 + y^5 = 95050$