[tex]\dfrac{3}{\sqrt6-1}\cdot\dfrac{\sqrt6+1}{\sqrt6+1}=\dfrac{3(\sqrt6+1)}{(\sqrt6-1)(\sqrt6+1)}=\dfrac{3\sqrt3+3}{(\sqrt6)^2-1^2}\\\\=\dfrac{3\sqrt3+3}{6-1}=\dfrac{3\sqrt6+3}{5}\to\boxed{3.}\\\\Used:\\\\(a-b)(a+b)=a^2-b^2\\\\(\sqrt{a})^2=a\\\\distributive\ property:\ a(b+c)=ab+ac[/tex]
