TA có: \(\frac{10}{3\sqrt6-7}-\frac{3\sqrt2-2\sqrt3}{\sqrt2-\sqrt3}-21\cdot\sqrt{\frac23}\)
\(=\frac{10\left(3\sqrt6+7\right)}{\left(3\sqrt6-7\right)\left(3\sqrt6+7\right)}+\frac{\sqrt6\left(\sqrt3-\sqrt2\right)}{\sqrt3-\sqrt2}-21\cdot\frac{\sqrt2}{\sqrt3}\)
\(=2\left(3\sqrt6+7\right)+\sqrt6-7\sqrt6=6\sqrt6+14+\sqrt6-7\sqrt6=14\)

