Вычислить:
$$41. [41\frac{29}{72} - (18\frac{7}{8} - 5\frac{1}{4})(10\frac{1}{2} - 7\frac{2}{3})] : 22\frac{7}{18}.$$
Решение:
\(41. [41\frac{29}{72} - (18\frac{7}{8} - 5\frac{1}{4})(10\frac{1}{2} - 7\frac{2}{3})] : 22\frac{7}{18} = \frac{1}{8}.\)

\(
1) 18\frac{7}{8} - 5\frac{1}{4} = 18 + \frac{7}{8} - 5 - \frac{2}{8} = 18 - 5 + \frac{7 - 2}{8} = 13 + \frac{5}{8} = 13\frac{5}{8};
\)

\(
2) 10\frac{1}{2} - 7\frac{2}{3} = 10 + \frac{3}{6} - 7 - \frac{4}{6} = 9 + 1 + \frac{3}{6} - 7 - \frac{4}{6} = 9 + \frac{6}{6} + \frac{3}{6} - 7 - \frac{4}{6} = 9 - 7 + \frac{6 + 3 - 4}{6} =
\)

\(
= 2 + \frac{5}{6} = 2\frac{5}{6};
\)

\(
3) 13\frac{5}{8} \cdot 2\frac{5}{6} = \frac{109}{8} \cdot \frac{17}{6} = \frac{109 \cdot 17}{8 \cdot 6} = \frac{1853}{48} = 38\frac{29}{48};
\)

\(
4) 41\frac{29}{72} - 38\frac{29}{48} = 41\frac{58}{144} - 38\frac{87}{144} = 40 + 1 + \frac{58}{144} - 38 - \frac{87}{144} = 40 + \frac{144}{144} + \frac{58}{144} - 38 - \frac{87}{144} =
\)

\(
= 40 - 38 + \frac{144 + 58 - 87}{144} = 2 + \frac{115}{144} = 2\frac{115}{144};
\)

\(
5) 2\frac{115}{144} : 22\frac{7}{18} = \frac{403}{144} \cdot \frac{18}{403} = \frac{403 \cdot 18}{144 \cdot 403} = \frac{1 \cdot 1}{8 \cdot 1} = \frac{1}{8}.
\)
Ответ: \(\frac{1}{8}\).