問題
$\dfrac{\sqrt{2}}{\sqrt{2}-1}$ の整数部分を $a$, 小数部分を $b$ とする．次の式の値を求めよ．
解説
- (1) $a$
- $\dfrac{\sqrt{2}}{\sqrt{2}-1}
=\dfrac{\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}
=\dfrac{2+\sqrt{2}}{2-1}
=2+\sqrt{2}$.
$1<\sqrt{2}<2$ より，$2<2+\sqrt{2}<3$.
したがって整数部分は $a=$$3$．
- $\dfrac{\sqrt{2}}{\sqrt{2}-1}
- (2) $b$
- 小数部分は $b=(2+\sqrt{2})-3=$$\sqrt{2}-1$.
- (3) $a+b+b^{2}$
- $=3+(\sqrt{2}-1)+(\sqrt{2}-1)^{2}$
$=3+(\sqrt{2}-1)+(3-2\sqrt{2})$
$=$$5-\sqrt{2}$
- $=3+(\sqrt{2}-1)+(\sqrt{2}-1)^{2}$
