Решение:
- \(3\frac{4}{15} = \frac{3 \cdot 15 + 4}{15} = \frac{49}{15}\)
- \(\frac{29}{35} + \frac{4}{7} = \frac{29 + 4 \cdot 5}{35} = \frac{29 + 20}{35} = \frac{49}{35} = \frac{7}{5}\)
- \(\frac{7}{5} : \frac{49}{15} = \frac{7 \cdot 15}{5 \cdot 49} = \frac{105}{245} = \frac{3}{7}\)
- \(-1 + \frac{3}{7} = \frac{-7 + 3}{7} = \frac{-4}{7}\)
- \(\frac{-4}{7} - \frac{1}{3} = \frac{-4 \cdot 3 - 1 \cdot 7}{7 \cdot 3} = \frac{-12 - 7}{21} = \frac{-19}{21}\)
Ответ: \(\frac{-19}{21}\)