Ta có: \(F=\sqrt{2+\sqrt3}+\sqrt{14-5\sqrt3}+\sqrt2\)
\(=\frac{1}{\sqrt2}\left(\sqrt{4+2\sqrt3}+\sqrt{28-10\sqrt3}+2\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3+1\right)^2}+\sqrt{\left(5-\sqrt3\right)^2}+2\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt3+1+5-\sqrt3+2\right)=\frac{1}{\sqrt2}\cdot8=4\sqrt2\)