Задание 06. Вычислите: а) \((\frac{1}{9})^2-20:\frac{1}{9}\); б) \((\frac{19}{9}+\frac{11}{12}):\frac{5}{48}\); в) \((2\frac{3}{4}+2\frac{1}{5})\cdot16\).

1. а) \(\left(\frac{1}{9}\right)^2-20:\frac{1}{9}=\frac{1}{81}-20\cdot9=\frac{1}{81}-180=-\frac{14579}{81}\).
2. б) \(\left(\frac{19}{9}+\frac{11}{12}\right):\frac{5}{48}=\left(\frac{76}{36}+\frac{33}{36}\right)\cdot\frac{48}{5}=\frac{109}{36}\cdot\frac{48}{5}=\frac{436}{15}=29\frac{1}{15}\).
3. в) \(\left(2\frac{3}{4}+2\frac{1}{5}\right)\cdot16=\left(\frac{11}{4}+\frac{11}{5}\right)\cdot16=\frac{99}{20}\cdot16=\frac{396}{5}=79\frac{1}{5}\).

Ответ: а) \(-\frac{14579}{81}\); б) \(29\frac{1}{15}\); в) \(79\frac{1}{5}\).