a, \(\sqrt{\dfrac{\sqrt{5}-2}{\sqrt{5}+2}}+\sqrt{\dfrac{\sqrt{5}+2}{\sqrt{5}-2}}\\ =\sqrt{\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}+\sqrt{\dfrac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2+\sqrt{5}+2\\ =2\sqrt{5}\)
b: Ta có: \(\dfrac{2\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\dfrac{3}{3-\sqrt{6}}\)
\(=\dfrac{-\sqrt{6}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{3\left(3+\sqrt{6}\right)}{3}\)
\(=-\sqrt{6}+3+\sqrt{6}\)
=3
