Задание 4.

1. $$\frac{1}{\frac{1}{28}+\frac{1}{12}} = \frac{1}{\frac{1\cdot3}{28\cdot3}+\frac{1\cdot7}{12\cdot7}} = \frac{1}{\frac{3}{84}+\frac{7}{84}} = \frac{1}{\frac{3+7}{84}} = \frac{1}{\frac{10}{84}} = \frac{84}{10} = 8,4$$
2. $$\frac{1}{\frac{1}{14}-\frac{1}{63}} = \frac{1}{\frac{1\cdot9}{14\cdot9}-\frac{1\cdot2}{63\cdot2}} = \frac{1}{\frac{9}{126}-\frac{2}{126}} = \frac{1}{\frac{9-2}{126}} = \frac{1}{\frac{7}{126}} = \frac{126}{7} = 18$$