問題
$x=\dfrac{\sqrt{5}+2}{\sqrt{5}-2},\ y=\dfrac{\sqrt{5}-2}{\sqrt{5}+2}$ のとき, 次の式の値を求めよ.
解説
- (1) $x+y$
- $y=\dfrac{1}{x}$ より,
$x+y=\dfrac{(\sqrt{5}+2)^{2}+(\sqrt{5}-2)^{2}}{5-4}
=\dfrac{2\cdot5+2\cdot4}{1}=$$18$
- $y=\dfrac{1}{x}$ より,
- (2) $xy$
- $=\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\cdot\dfrac{\sqrt{5}-2}{\sqrt{5}+2}
=$$1$
- $=\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\cdot\dfrac{\sqrt{5}-2}{\sqrt{5}+2}
- (3) $x^{2}y+xy^{2}$
- $=xy(x+y)=$$18$
- (4) $x^{2}+y^{2}$
- $=(x+y)^{2}-2xy=18^{2}-2=$$322$
- (5) $x^{3}+y^{3}$
- $=(x+y)^{3}-3xy(x+y)=18^{3}-3\cdot1\cdot18=$$5778$
