2) \(\dfrac{\sqrt{70}-\sqrt{32}}{\sqrt{35}-\sqrt{75}}\)

Упростим корни:

\[\sqrt{32}=4\sqrt{2},\qquad \sqrt{75}=5\sqrt{3}.\]

Домножим дробь на сопряжённое выражение \(\sqrt{35}+\sqrt{75}\):

\[\frac{\sqrt{70}-\sqrt{32}}{\sqrt{35}-\sqrt{75}}\cdot\frac{\sqrt{35}+\sqrt{75}}{\sqrt{35}+\sqrt{75}}\]

\[=\frac{35\sqrt{2}+5\sqrt{210}-4\sqrt{70}-20\sqrt{6}}{35-75}\]

\[=\frac{35\sqrt{2}+5\sqrt{210}-4\sqrt{70}-20\sqrt{6}}{-40}.\]

Ответ: \(\dfrac{4\sqrt{70}+20\sqrt{6}-35\sqrt{2}-5\sqrt{210}}{40}\).