Ta có:
\(\dfrac{1}{\sqrt{a}-\sqrt[]{a+1}}=\dfrac{\sqrt{a}+\sqrt{a+1}}{a-a+1}=\sqrt{a}+\sqrt{a+1}\)
\(\Rightarrow\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\sqrt{2013}+\sqrt{2014}-\sqrt{2014}-\sqrt{2015}=\sqrt{2013}-\sqrt{2015}\)
Ta có công thức tổng quát
\(\dfrac{1}{\sqrt{n}-\sqrt{n+1}}=\dfrac{\sqrt{n}+\sqrt{n+1}}{n-n-1}=-\left(\sqrt{n}+\sqrt{n+1}\right)=-\sqrt{n}-\sqrt{n-1}\)
Vậy \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=-\sqrt{2013}-\sqrt{2014}-\left(-\sqrt{2014}-\sqrt{2015}\right)=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
