№ 1. Вычислите: $$9\cdot(\frac{1}{5})^2-19\frac{1}{9}$$

$$9\cdot(\frac{1}{5})^2-19\frac{1}{9} = 9 \cdot \frac{1}{25} - \frac{19 \cdot 9 + 1}{9} = \frac{9}{25} - \frac{172}{9} = \frac{9 \cdot 9 - 172 \cdot 25}{225} = \frac{81 - 4300}{225} = \frac{-4219}{225} = -18 \frac{169}{225}$$

Ответ: $$-18 \frac{169}{225}$$