x = \(6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
==> \(\sqrt{x}\) \(=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
A = \(\dfrac{4\sqrt{x}}{x+4}=\dfrac{4\left(\sqrt{5}-1\right)}{10-2\sqrt{5}}=\dfrac{4\left(\sqrt{5}-1\right)}{2\sqrt{5}\left(\sqrt{5}-1\right)}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)