[tex]\dfrac{cos150^{o}\cdot tg120^{o}}{\sqrt{3}\cdot sin120^{o}-tg135^{o}}=\dfrac{cos(180^{o}-30^{o})\cdot tg(180^{o}-60^{o})}{\sqrt{3}\cdot sin(180^{o}-60^{o})-tg(180^{o}-45^{o})}=\\\\\\=\dfrac{-cos30^{o}\cdot(-tg60^{o})}{\sqrt{3}\cdot sin60^{o}-(-tg45^{o})}=\dfrac{-\frac{\sqrt{3}}{2}\cdot(-\sqrt{3})}{\sqrt{3}\cdot\frac{\sqrt{3}}{2}-(-1)}=\dfrac{\frac{3}{2}}{\frac{3}{2}+1} = \dfrac{\frac{3}{2}}{\frac{3}{2}+\frac{2}{2}} = \dfrac{\frac{3}{2}}{\frac{5}{2}} =\boxed{ \frac{3}{5}}[/tex]
