Đáp án:
$a^{2n+1}<b^{2n+1}$
Giải thích các bước giải:
$a^{2n+1}<b^{2n+1}$
$↔a^{2n+1}-b^{2n+1}<0$
$↔b^{2n+1}-a^{2n+1}>0$
$↔b^{2(n+\dfrac{1}{2})}-a^{2(n+\dfrac{1}{2})}>0$
$↔(b^{n+\dfrac{1}{2}}-a^{n+\dfrac{1}{2}})(b^{n+\dfrac{1}{2}}+a^{n+\dfrac{1}{2}})>0$ luôn đúng
vì $\begin{cases}b>a↔b^{n+\dfrac{1}{2}}>a^{n+\dfrac{1}{2}}\\b^{n+\dfrac{1}{2}}+a^{n+\dfrac{1}{2}}>0∀n \in N\\\end{cases}$
$→a^{2n+1}<b^{2n+1}$
