Ta có :
\(A=\dfrac{1+\dfrac{\sqrt{3}}{2}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\dfrac{\sqrt{3}}{2}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)
\(=\dfrac{\dfrac{2+\sqrt{3}}{2}}{1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}}+\dfrac{\dfrac{2-\sqrt{3}}{2}}{1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}}\)
\(=\dfrac{\dfrac{2+\sqrt{3}}{2}}{1+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{4}}}+\dfrac{\dfrac{2-\sqrt{3}}{2}}{1-\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{4}}}\)
\(=\dfrac{\dfrac{2+\sqrt{3}}{2}}{1+\dfrac{\sqrt{3}+1}{2}}+\dfrac{\dfrac{2-\sqrt{3}}{2}}{1-\dfrac{\sqrt{3}-1}{2}}\)
\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)
\(=\dfrac{6}{6}\)
\(=1\)
