\(=\dfrac{\left(\sqrt{3}+1\right)^2+\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{2}\\ =\dfrac{8}{2}=4\)
\(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
\(=\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{2}\)
=4