\(x=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=2\)
Với \(x=2\):
\(\frac{\sqrt{\frac{1}{x}+4+4x}}{\sqrt{x}\left(2x^2-x-1\right)}=\frac{\sqrt{\frac{1}{2}+4+8}}{\sqrt{2}\left(8-2-1\right)}=\frac{1}{2}\)