\(2\sqrt{27}-\sqrt{\dfrac{16}{3}}-\sqrt{48}-\sqrt{8\dfrac{1}{3}}\)
\(=2\cdot3\sqrt{3}-\dfrac{4}{\sqrt{3}}-4\sqrt{3}-\sqrt{\dfrac{25}{3}}\)
\(=6\sqrt{3}-4\sqrt{3}-\dfrac{4}{3}\sqrt{3}-\dfrac{5\sqrt{3}}{3}\)
\(=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
\[
6\sqrt{3} - \frac{4}{3} - 4\sqrt{3} - \frac{2\sqrt{2}}{\sqrt{3}}
\]
\[
(6\sqrt{3} - 4\sqrt{3}) - \frac{4}{3} - \frac{2\sqrt{2}}{\sqrt{3}} = 2\sqrt{3} - \frac{4}{3} - \frac{2\sqrt{2}}{\sqrt{3}}
\]
\[
\frac{4\sqrt{3}}{3\sqrt{3}} - \frac{2\sqrt{2} \cdot 3}{3\sqrt{3}} = \frac{4\sqrt{3} - 6\sqrt{2}}{3\sqrt{3}}
\]
\[
2\sqrt{3} - \frac{4\sqrt{3} - 6\sqrt{2}}{3\sqrt{3}} = \frac{6\sqrt{3} - (4\sqrt{3} - 6\sqrt{2})}{3} = \frac{2\sqrt{3} + 6\sqrt{2}}{3}
\]
\[ =>
\frac{2\sqrt{3} + 6\sqrt{2}}{3}
\]
