ta có : \(A=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(4^2-\left(\sqrt{15}\right)^2\right)\)
\(=2\left(16-15\right)=2\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(16-15\right)=2\)
