Ta có: \(\sqrt{14-4\sqrt{6}}-\frac{\sqrt{5}+1}{\sqrt{2}}+\sqrt{11+\sqrt{96}}+\sqrt{3-\sqrt{5}}\)
\(=\sqrt{12-2\cdot2\sqrt{3}\cdot\sqrt{2}+2}-\frac{\sqrt{5}+1}{\sqrt{2}}+\frac{\sqrt{22+2\sqrt{96}}}{\sqrt{2}}+\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}+\frac{-\sqrt{5}-1+\sqrt{16+2\cdot4\cdot\sqrt{6}+6}+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}{\sqrt{2}}\)
\(=\left|2\sqrt{3}-\sqrt{2}\right|+\frac{-\sqrt{5}-1+\sqrt{\left(4+\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)
\(=2\sqrt{3}-\sqrt{2}+\frac{-\sqrt{5}-1+\left|4+\sqrt{6}\right|+\left|\sqrt{5}-1\right|}{\sqrt{2}}\)(Vì \(2\sqrt{3}>\sqrt{2}\))
\(=2\sqrt{3}-\sqrt{2}+\frac{-\sqrt{5}-1+4+\sqrt{6}+\sqrt{5}-1}{\sqrt{2}}\)(Vì \(\left\{{}\begin{matrix}4>\sqrt{6}>0\\\sqrt{5}>1\end{matrix}\right.\))
\(=2\sqrt{3}-\sqrt{2}+\frac{2+\sqrt{6}}{\sqrt{2}}\)
\(=2\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(=3\sqrt{3}\)