**Решение:**
1) $(\frac{13}{48} + 2\frac{5}{12}) : (-\frac{3}{4} + \frac{39}{4} : (-13)) = (\frac{13}{48} + \frac{29}{12}) : (-\frac{3}{4} + \frac{3}{4} : (-1)) = (\frac{13 + 116}{48}) : (-\frac{3}{4} - \frac{3}{4}) = \frac{129}{48} : (-\frac{6}{4}) = \frac{129}{48} \cdot (-\frac{4}{6}) = \frac{43}{16} \cdot (-\frac{1}{2}) = -\frac{43}{32}$;
2) $(1\frac{2}{3} - 3,6) : (-2\frac{7}{9} + 4\frac{1}{15}) \cdot (-2,6) = (\frac{5}{3} - \frac{18}{5}) : (-\frac{25}{9} + \frac{61}{15}) \cdot (-\frac{13}{5}) = (\frac{25 - 54}{15}) : (\frac{-125 + 183}{45}) \cdot (-\frac{13}{5}) = (\frac{-29}{15}) : (\frac{58}{45}) \cdot (-\frac{13}{5}) = (\frac{-29}{15}) \cdot (\frac{45}{58}) \cdot (-\frac{13}{5}) = (\frac{-1}{1}) \cdot (\frac{3}{2}) \cdot (-\frac{13}{5}) = \frac{39}{10} = 3,9$;
**Ответ:**
1) $-\frac{43}{32}$;
2) $3,9$.
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