4. Освободитесь от иррациональности

a) $$\frac{1}{3\sqrt{2}} = \frac{1 \cdot \sqrt{2}}{3\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{2}}{3 \cdot 2} = \frac{\sqrt{2}}{6}$$

Ответ: $$\frac{\sqrt{2}}{6}$$

б) $$\frac{6}{5-\sqrt{2}} = \frac{6 \cdot (5+\sqrt{2})}{(5-\sqrt{2}) \cdot (5+\sqrt{2})} = \frac{6(5+\sqrt{2})}{5^2 - (\sqrt{2})^2} = \frac{6(5+\sqrt{2})}{25-2} = \frac{6(5+\sqrt{2})}{23} = \frac{30+6\sqrt{2}}{23}$$

Ответ: $$\frac{30+6\sqrt{2}}{23}$$