Решение:
$(7\frac{1}{9} - 2\frac{14}{15}) : (2\frac{2}{3} + 1\frac{3}{5}) - (\frac{3}{4} - \frac{1}{20}) \cdot (\frac{5}{7} - \frac{5}{14}) = \frac{35}{48}.$

$1) 7\frac{1}{9} - 2\frac{14}{15} = 7\frac{5}{45} - 2\frac{42}{45} = 6 + 1 + \frac{5}{45} - 2 - \frac{42}{45} = 6 + \frac{45}{45} + \frac{5}{45} - 2 - \frac{42}{45} = 6 - 2 + \frac{45 + 5 - 42}{45} =$

$= 4 + \frac{8}{45} = 4\frac{8}{45};$

$2) 2\frac{2}{3} + 1\frac{3}{5} = 2\frac{10}{15} + 1\frac{9}{15} = 2 + 1 + \frac{10 + 9}{15} = 3 + \frac{19}{15} = 3 + 1 + \frac{4}{15} = 4 + \frac{4}{15} = 4\frac{4}{15};$

$3) 4\frac{8}{45} : 4\frac{4}{15} = \frac{188}{45} \cdot \frac{15}{64} = \frac{188 \cdot 15}{45 \cdot 64} = \frac{47 \cdot 1}{3 \cdot 16} = \frac{47}{48};$

$4) \frac{3}{4} - \frac{1}{20} = \frac{15}{20} - \frac{1}{20} = \frac{15 - 1}{20} = \frac{14}{20} = \frac{7}{10};$

$5) \frac{5}{7} - \frac{5}{14} = \frac{10}{14} - \frac{5}{14} = \frac{10 - 5}{14} = \frac{5}{14};$

$6) \frac{7}{10} \cdot \frac{5}{14} = \frac{7 \cdot 5}{10 \cdot 14} = \frac{1 \cdot 1}{2 \cdot 2} = \frac{1}{4};$

$7) \frac{47}{48} - \frac{1}{4} = \frac{47}{48} - \frac{12}{48} = \frac{47 - 12}{48} = \frac{35}{48}.$
Ответ: $\frac{35}{48}$.

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