Для решения данного задания необходимо:
Тогда:
$$\frac{\sqrt{21}-\sqrt{14}}{\sqrt{6}}=\frac{\sqrt{7\cdot 3}-\sqrt{7\cdot 2}}{\sqrt{6}}=\frac{\sqrt{7}\cdot \sqrt{3}-\sqrt{7}\cdot \sqrt{2}}{\sqrt{6}}=\frac{\sqrt{7}(\sqrt{3}-\sqrt{2})}{\sqrt{6}}=\frac{\sqrt{7}(\sqrt{3}-\sqrt{2})}{\sqrt{2}\cdot \sqrt{3}}=\sqrt{\frac{7}{6}}(\sqrt{3}-\sqrt{2})=\sqrt{\frac{7}{6}}\sqrt{3}-\sqrt{\frac{7}{6}}\sqrt{2}=\sqrt{\frac{7}{6}\cdot 3}-\sqrt{\frac{7}{6}\cdot 2}=\sqrt{\frac{7}{2}}-\sqrt{\frac{7}{3}}=\frac{\sqrt{7}}{\sqrt{2}}-\frac{\sqrt{7}}{\sqrt{3}}=\frac{\sqrt{7}\cdot \sqrt{2}}{2}-\frac{\sqrt{7}\cdot \sqrt{3}}{3}=\frac{\sqrt{14}}{2}-\frac{\sqrt{21}}{3}=\frac{3\sqrt{14}-2\sqrt{21}}{6}=\frac{\sqrt{7}(3\sqrt{2}-2\sqrt{3})}{6}$$.
Ответ: $$\frac{\sqrt{7}(3\sqrt{2}-2\sqrt{3})}{6}$$