Answer :
[tex]\frac{8}{2-\sqrt{12}}=\frac{8}{2-\sqrt{4\cdot3}}=\frac{8}{2-2\sqrt3}=\frac{8}{2(1-\sqrt3)}=\frac{4}{1-\sqrt3}\\\\\frac{4}{1-\sqrt3}\cdot\frac{1+\sqrt3}{1+\sqrt3}=\frac{4(1+\sqrt3)}{1^2-(\sqrt3)^2}=\frac{4(1+\sqrt3)}{1-3}=\frac{4(1+\sqrt3)}{-2}=-2(1+\sqrt3)}\\\\=-2-2\sqrt3[/tex]
[tex]if\ \frac{8}{2}-\sqrt{12}=4-\sqrt{4\cdot3}=4-2\sqrt3[/tex]
[tex]if\ \frac{8}{2}-\sqrt{12}=4-\sqrt{4\cdot3}=4-2\sqrt3[/tex]
[tex]\frac{8}{2-\sqrt{12}}\\
\frac{8(2+\sqrt{12})}{4-12}=\\
\frac{8(2+\sqrt{12})}{-8}=\\
-(2+\sqrt{12}})=\\
-2-\sqrt{12}=\\-2-2\sqrt3
[/tex]
