Đặt \(A=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\Leftrightarrow\sqrt{2}.A=\sqrt{2}\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)\Leftrightarrow\sqrt{2}\sqrt{A}=\sqrt{2}.\sqrt{3+\sqrt{5}}-\sqrt{2}.\sqrt{3-\sqrt{5}}\Leftrightarrow\sqrt{2}.A=\sqrt{2\left(3+\sqrt{5}\right)}-\sqrt{2\left(3-\sqrt{5}\right)}\Leftrightarrow\sqrt{2}.A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\Leftrightarrow\sqrt{2}.A=\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}\Leftrightarrow\sqrt{2}.A=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\Leftrightarrow\sqrt{2}.A=\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|\Leftrightarrow\sqrt{2}.A=\sqrt{5}+1-\sqrt{5}+1\Leftrightarrow\sqrt{2}.A=2\Leftrightarrow A=\sqrt{2}\)Vậy \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}\)