Tính \(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\). Nhân cả tử và mẫu với \(\sqrt{2}\) ta có: \(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{20}+2}\) \(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\) \(=\dfrac{\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)\(=\dfrac{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{2.4}=1\).