К) [3\(\frac{1}{4}\)\(\cdot\)\(14\frac{4}{5}+4\frac{4}{15}\)-47]:5\(\frac{9}{10}\)

Решение:

1. \( 14\frac{4}{5} = \frac{74}{5} \), \( 4\frac{4}{15} = \frac{64}{15} \)
2. \( \frac{74}{5} + \frac{64}{15} = \frac{74\cdot3 + 64}{15} = \frac{222+64}{15} = \frac{286}{15} \)
3. \( 3\frac{1}{4} = \frac{13}{4} \)
4. \( \frac{13}{4} \cdot \frac{286}{15} = \frac{13\cdot286}{4\cdot15} = \frac{3718}{60} = \frac{1859}{30} \)
5. \( \frac{1859}{30} - 47 = \frac{1859 - 47\cdot30}{30} = \frac{1859 - 1410}{30} = \frac{449}{30} \)
6. \( 5\frac{9}{10} = \frac{59}{10} \)
7. \( \frac{449}{30} : \frac{59}{10} = \frac{449}{30} \cdot \frac{10}{59} = \frac{449\cdot10}{30\cdot59} = \frac{449}{3\cdot59} = \frac{449}{177} \)

Ответ: \(\frac{449}{177}\)