a) 4+\(\frac{7}{3}\)
Приведем дроби к общему знаменателю 3:
$$ 4+\frac{7}{3} = \frac{4 \cdot 3}{3} + \frac{7}{3} = \frac{12}{3} + \frac{7}{3} = \frac{12+7}{3} = \frac{19}{3} $$
б) (2+\(\frac{7}{24}\)) - (\(\frac{11}{24}\)-\(\frac{5}{12}\))
Приведем дроби к общему знаменателю 24:
$$ (2+\frac{7}{24}) - (\frac{11}{24} - \frac{5}{12}) = (\frac{2 \cdot 24}{24} + \frac{7}{24}) - (\frac{11}{24} - \frac{5 \cdot 2}{24}) = (\frac{48}{24} + \frac{7}{24}) - (\frac{11}{24} - \frac{10}{24}) = \frac{55}{24} - \frac{1}{24} = \frac{55-1}{24} = \frac{54}{24} = \frac{9}{4} $$
в) \(\frac{13}{12} \cdot \frac{12}{13} - \frac{25}{156}.\)
$$ \frac{13}{12} \cdot \frac{12}{13} - \frac{25}{156} = \frac{13 \cdot 12}{12 \cdot 13} - \frac{25}{156} = 1 - \frac{25}{156} = \frac{156}{156} - \frac{25}{156} = \frac{156-25}{156} = \frac{131}{156} $$
Ответ: a) \(\frac{19}{3}\); б) \(\frac{9}{4}\); в) \(\frac{131}{156}\)