Răspuns :
[tex]\it -6\cdot\dfrac{36}{\sqrt3}+\dfrac {8}{\sqrt3}+6\sqrt3-\dfrac {4}{\sqrt3}=\dfrac {-216+8-4}{\sqrt3}+6\sqrt3=\dfrac {-212}{\sqrt3}+6\sqrt3=\\ \\ \\ \dfrac {-212+18}{\sqrt3}=\dfrac {-194}{\sqrt3}=-\dfrac {194\sqrt3}{3}[/tex]
[tex]\it -6\cdot\dfrac{36}{\sqrt3}+\dfrac {8}{\sqrt3}+6\sqrt3-\dfrac {4}{\sqrt3}=\dfrac {-216+8-4}{\sqrt3}+6\sqrt3=\dfrac {-212}{\sqrt3}+6\sqrt3=\\ \\ \\ \dfrac {-212+18}{\sqrt3}=\dfrac {-194}{\sqrt3}=-\dfrac {194\sqrt3}{3}[/tex]