a: Ta có: \(A=\sqrt{4-\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
\(=2>\sqrt{3}\)