X) [(\(\frac{5}{6}-\frac{3}{8}\)):\(\frac{3}{4}\)-(\(\frac{3}{8}+\frac{7}{20}\))\(\cdot\)\(\frac{9}{20}\)]:\(\frac{1}{50}\)

Решение:

Выражение в квадратных скобках:

1. \( \frac{5}{6} - \frac{3}{8} = \frac{20-9}{24} = \frac{11}{24} \)
2. \( \frac{11}{24} : \frac{3}{4} = \frac{11}{24} \cdot \frac{4}{3} = \frac{44}{72} = \frac{11}{18} \)
3. \( \frac{3}{8} + \frac{7}{20} = \frac{15+14}{40} = \frac{29}{40} \)
4. \( \frac{29}{40} \cdot \frac{9}{20} = \frac{261}{800} \)
5. \( \frac{11}{18} - \frac{261}{800} = \frac{11\cdot400 - 261\cdot9}{7200} = \frac{4400 - 2349}{7200} = \frac{2051}{7200} \)

Основное вычисление:

1. \( \frac{2051}{7200} : \frac{1}{50} = \frac{2051}{7200} \cdot 50 = \frac{2051\cdot50}{7200} = \frac{2051}{144} \)

Ответ: \(\frac{2051}{144}\)