Вычислите 1/(√7-√6)-3/(√6-√3)-4/(√7+√3)

Решите задачу:

$\frac{1}{\sqrt7-\sqrt6}-\frac{3}{\sqrt6-\sqrt3}-\frac{4}{\sqrt7+\sqrt3}=\frac{\sqrt7+\sqrt6}{7-6}-\frac{3\, (\sqrt6+\sqrt3)}{6-3}-\frac{4\, (\sqrt7-\sqrt3)}{7-3}=\\\\=\frac{\sqrt7+\sqrt6}{1}-\frac{3\sqrt6+3\sqrt3}{3}-\frac{4\sqrt7-4\sqrt3}{4}=\frac{12\sqrt7+12\sqrt6-12\sqrt6-12\sqrt3-12\sqrt7+12\sqrt3}{12}=\\\\=\frac{0}{12}=0$