1) (20\frac{8}{15} \cdot 7\frac{1}{2} - 54\frac{3}{5} : 2\frac{1}{2}) : (3\frac{13}{21} \cdot 8\frac{2}{5} - 29\frac{2}{5}) - \frac{5}{6} \cdot 1\frac{1}{5} + \frac{1}{25}

Решение:

1. Первая скобка:
   • \(20\frac{8}{15} = \frac{20 \cdot 15 + 8}{15} = \frac{308}{15}\)
   • \(7\frac{1}{2} = \frac{7 \cdot 2 + 1}{2} = \frac{15}{2}\)
   • \(20\frac{8}{15} \cdot 7\frac{1}{2} = \frac{308}{15} \cdot \frac{15}{2} = \frac{308}{2} = 154\)
   • \(54\frac{3}{5} = \frac{54 \cdot 5 + 3}{5} = \frac{273}{5}\)
   • \(2\frac{1}{2} = \frac{2 \cdot 2 + 1}{2} = \frac{5}{2}\)
   • \(54\frac{3}{5} : 2\frac{1}{2} = \frac{273}{5} : \frac{5}{2} = \frac{273}{5} \cdot \frac{2}{5} = \frac{546}{25}\)
   • \(154 - \frac{546}{25} = \frac{154 \cdot 25 - 546}{25} = \frac{3850 - 546}{25} = \frac{3304}{25}\)
2. Вторая скобка:
   • \(3\frac{13}{21} = \frac{3 \cdot 21 + 13}{21} = \frac{76}{21}\)
   • \(8\frac{2}{5} = \frac{8 \cdot 5 + 2}{5} = \frac{42}{5}\)
   • \(3\frac{13}{21} \cdot 8\frac{2}{5} = \frac{76}{21} \cdot \frac{42}{5} = \frac{76 \cdot 2}{5} = \frac{152}{5}\)
   • \(29\frac{2}{5} = \frac{29 \cdot 5 + 2}{5} = \frac{147}{5}\)
   • \(\frac{152}{5} - 29\frac{2}{5} = \frac{152}{5} - \frac{147}{5} = \frac{5}{5} = 1\)
3. Деление скобок:
   • \(\frac{3304}{25} : 1 = \frac{3304}{25}\)
4. Оставшиеся действия:
   • \(\frac{5}{6} \cdot 1\frac{1}{5} = \frac{5}{6} \cdot \frac{6}{5} = 1\)
   • \(\frac{3304}{25} - 1 + \frac{1}{25} = \frac{3304 - 25 + 1}{25} = \frac{3280}{25}\)
   • \(\frac{3280}{25} = \frac{656}{5} = 131\frac{1}{5}\)

Ответ: 131\frac{1}{5}