\(A=\frac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}+4040\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+4040\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{3}+1+\sqrt{3}-1+4040\right)\)
\(=\frac{1}{\sqrt{2}}\left(2\sqrt{3}+4040\right)=\sqrt{6}+2020\sqrt{2}\)