Đặt A = \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}\)
=> \(A^2=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}}{2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}}\)\(=\frac{4+2\sqrt{4-3}}{4-2\sqrt{4-3}}=\frac{4+2}{4-2}=\frac{6}{2}=3\)
=>A = \(\sqrt{3}\)
Chứng minh tương tự B = \(\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}=\frac{1}{\sqrt{3}}\)
=> A + B = \(\sqrt{3}+\frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}\)