Вычислите: а) \(\dfrac{\left(6\dfrac45-4\dfrac12\right)\cdot\dfrac35}{\left(2\dfrac56+1\dfrac14\right)\cdot\dfrac35}\); б) \(\dfrac{\left(1\dfrac{22}{25}+2\dfrac45\right)\cdot\dfrac74}{2\dfrac14+3\dfrac15}\); в) \(\dfrac{5\dfrac34-2\dfrac34}{5\dfrac18:4\dfrac13}\).

1. \(6\dfrac45-4\dfrac12=\dfrac{34}{5}-\dfrac92=\dfrac{23}{10}\), \(2\dfrac56+1\dfrac14=\dfrac{17}{6}+\dfrac54=\dfrac{49}{12}\). Поэтому \(\dfrac{\frac{23}{10}\cdot\frac35}{\frac{49}{12}\cdot\frac35}=\dfrac{23}{10}\cdot\dfrac{12}{49}=\dfrac{138}{245}\).
2. \(1\dfrac{22}{25}+2\dfrac45=\dfrac{47}{25}+\dfrac{14}{5}=\dfrac{117}{25}\), \(2\dfrac14+3\dfrac15=\dfrac94+\dfrac{16}{5}=\dfrac{109}{20}\). Поэтому \(\dfrac{\frac{117}{25}\cdot\frac74}{\frac{109}{20}}=\dfrac{117}{25}\cdot\dfrac74\cdot\dfrac{20}{109}=\dfrac{819}{545}\).
3. \(5\dfrac34-2\dfrac34=3\), а \(5\dfrac18:4\dfrac13=\dfrac{41}{8}:\dfrac{13}{3}=\dfrac{123}{104}\). Поэтому \(3:\dfrac{123}{104}=\dfrac{104}{41}\).

Ответ: а) \(\dfrac{138}{245}\); б) \(\dfrac{819}{545}\); в) \(\dfrac{104}{41}\).