\(\left\(6\frac{7}{20}-15\frac{2}{5}\cdot\frac{2}{4}\right\):\(\frac{1}{10}\)+\(\frac{7}{20}\)

Решение:

1. \( 15\frac{2}{5} = \frac{77}{5} \)
2. \( \frac{77}{5} \cdot \frac{2}{4} = \frac{77}{5} \cdot \frac{1}{2} = \frac{77}{10} \)
3. \( 6\frac{7}{20} = \frac{127}{20} \)
4. \( \frac{127}{20} - \frac{77}{10} = \frac{127}{20} - \frac{154}{20} = \frac{-27}{20} \)
5. \( \frac{-27}{20} : \frac{1}{10} = \frac{-27}{20} \cdot 10 = \frac{-27}{2} \)
6. \( \frac{-27}{2} + \frac{7}{20} = \frac{-270}{20} + \frac{7}{20} = \frac{-263}{20} \)

Ответ: \(\frac{-263}{20}\)