5) \((52,25-49\frac{1}{7})\cdot(40,01-36,81):(6\frac{1}{6}-2\frac{1}{42})\);

1. \(52,25-49\frac{1}{7}=\frac{209}{4}-\frac{344}{7}=\frac{77}{28}=\frac{11}{4}\).
2. \(40,01-36,81=3,2=\frac{16}{5}\).
3. \(6\frac{1}{6}-2\frac{1}{42}=\frac{37}{6}-\frac{85}{42}=\frac{259-85}{42}=\frac{29}{7}\).
4. \(\frac{11}{4}\cdot\frac{16}{5}:\frac{29}{7}=\frac{11}{4}\cdot\frac{16}{5}\cdot\frac{7}{29}=\frac{308}{145}=2\frac{18}{145}\).

Ответ: \(\frac{308}{145}=2\frac{18}{145}\).