Решение:
Используем распределительное свойство умножения: (a \(\cdot\) b + a \(\cdot\) c = a \(\cdot\) (b + c))
(8\(\frac{5}{11}\) \(\cdot\) 4\(\frac{2}{9}\) + 8\(\frac{7}{11}\) \(\cdot\) 6\(\frac{5}{9}\) = 8\(\frac{5}{11}\) \(\cdot\) 4\(\frac{2}{9}\) + 8\(\frac{7}{11}\) \(\cdot\) 6\(\frac{5}{9}\) = 8\(\frac{5}{11}\) \(\cdot\) 4\(\frac{2}{9}\) + 8\(\frac{7}{11}\) \(\cdot\) 6\(\frac{5}{9}\) = 8\(\frac{5}{11}\)\(4\frac{2}{9}+6\frac{5}{9}\) = 8\(\frac{5}{11}\) \(\cdot\) \(4\frac{2}{9} + 6\frac{5}{9}\) = 8\(\frac{5}{11}\) \(\cdot\) \(10 + \frac{2}{9} + \frac{5}{9}\) = 8\(\frac{5}{11}\) \(\cdot\) \(10 + \frac{7}{9}\) = 8\(\frac{5}{11}\) \(\cdot\) 10\(\frac{7}{9}\) = \(\frac{93}{11}\) \(\cdot\) \(\frac{97}{9}\) = \(\frac{9021}{99}\) = 91\(\frac{32}{99}\))
Ответ: \(91\frac{32}{99}\\)