\(F^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{16-7}\)
\(\Leftrightarrow F^2=8-2.3=2\)
Dễ thấy \(\sqrt{4+\sqrt{7}}>\sqrt{4-\sqrt{7}}\)
Nên \(F>0\)
\(\Rightarrow F=\sqrt{2}\)
\(F=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)}{\sqrt{2}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\left|\sqrt{7}+1\right|-\left|\sqrt{7}-1\right|}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
