\(VT=\sqrt{14}-\sqrt{13}=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)
\(VP=2\sqrt{3}-\sqrt{11}=\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
Ta thấy: \(\sqrt{14}+\sqrt{13}>\sqrt{12}+\sqrt{11}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{14}+\sqrt{13}}< \dfrac{1}{\sqrt{12}+\sqrt{11}}\)
Hay \(VT< VP\)
Vậy \(\sqrt{14}-\sqrt{13}< 2\sqrt{3}-\sqrt{11}\)