\(A=\dfrac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}+\dfrac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}\\ =\dfrac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}+\dfrac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}\\ =\dfrac{2-\sqrt{3}}{1+\sqrt{\sqrt{3}^2+2.\sqrt{3}.1+1^2}}+\dfrac{2+\sqrt{3}}{1-\sqrt{\sqrt{3}^2-2.\sqrt{3}.1+1^2}}\)
\(=\dfrac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{2-\sqrt{3}}{1+\left|\sqrt{3}+1\right|}+\dfrac{2+\sqrt{3}}{1-\left|\sqrt{3}-1\right|}\\ =\dfrac{2-\sqrt{3}}{1+\sqrt{3}+1}+\dfrac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}\\ =\dfrac{2-\sqrt{3}}{2+\sqrt{3}}+\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\\ =\dfrac{\left(2-\sqrt{3}\right)^2+\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ =\dfrac{4-4\sqrt{3}+3+4+4\sqrt{3}+3}{2^2-\sqrt{3}^2}\\ =\dfrac{14}{4-3}=14\)