Đặt \(A=\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
=> \(A^2=\frac{\left(3-\sqrt{5}\right).\left(3+\sqrt{5}\right)^2}{12+4\sqrt{5}}\)
\(=\frac{\left(9-5\right).\left(3+\sqrt{5}\right)}{4.\left(3+\sqrt{5}\right)}\)
= 1
=> A = 1