$\sqrt{2019} - \sqrt{2018}$
$ = \dfrac{(\sqrt{2019} - \sqrt{2018})(\sqrt{2019} + \sqrt{2018})}{\sqrt{2019} + \sqrt{2018}}$
$=\dfrac{2019 - 2018}{\sqrt{2019} + \sqrt{2018}}$
$=\dfrac{1}{\sqrt{2019} + \sqrt{2018}}$
Tương tự ta có: $\sqrt{2018} - \sqrt{2017} = \dfrac{1}{\sqrt{2018} + \sqrt{2017}}$
Vì $2019 > 2017 > 0$
⇒ $\sqrt{2019} > \sqrt{2017}$
⇔ $\sqrt{2019} + \sqrt{2018} > \sqrt{2017} + \sqrt{2018}$
⇔ $\dfrac{1}{\sqrt{2019} + \sqrt{2018}} < \dfrac{1}{\sqrt{2018} + \sqrt{2017}}$
hay $\sqrt{2019} - \sqrt{2018} < \sqrt{2018} - \sqrt{2017}$
