Đặt \(D=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(\Rightarrow D^2=\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)^2\)
\(=2-\sqrt{3}+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
\(=4+2\sqrt{4-3}\)
\(=4+2=6\)
=> \(D=\sqrt{6}\)
Vậy \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}=\sqrt{6}\)