問題
$x=\sqrt{2}-1$ のとき, 次の式の値を求めよ.
解説
- (1) $x+\dfrac{1}{x}$
- $\dfrac{1}{x}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1$ より,
$x+\dfrac{1}{x}=$$2\sqrt{2}$
- $\dfrac{1}{x}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1$ より,
- (2) $x^{2}+\dfrac{1}{x^{2}}$
- $=\bigg(x+\dfrac{1}{x}\bigg)^{2}-2=(2\sqrt{2})^{2}-2=$$6$
- (3) $x^{3}+\dfrac{1}{x^{3}}$
- $=\bigg(x+\dfrac{1}{x}\bigg)^{3}-3\!\left(x+\dfrac{1}{x}\right)
=(2\sqrt{2})^{3}-3(2\sqrt{2})=$$10\sqrt{2}$
- $=\bigg(x+\dfrac{1}{x}\bigg)^{3}-3\!\left(x+\dfrac{1}{x}\right)
- (4) $x^{4}+\dfrac{1}{x^{4}}$
- $\left(x^{2}+\dfrac{1}{x^{2}}\right)^{2}-2=6^{2}-2=$$34$
- (5) $x^{5}+\dfrac{1}{x^{5}}$
- $\bigg(x+\dfrac{1}{x}\bigg)^5-5\bigg(x+\dfrac{1}{x}\bigg)^3+5\bigg(x+\dfrac{1}{x}\bigg)$
$=(2\sqrt{2})^5-5(2\sqrt{2})^3+5(2\sqrt{2})=$$58\sqrt{2}$
- $\bigg(x+\dfrac{1}{x}\bigg)^5-5\bigg(x+\dfrac{1}{x}\bigg)^3+5\bigg(x+\dfrac{1}{x}\bigg)$
