Câu 1:
\(\dfrac{14\cdot34-21\cdot10}{28\cdot10-28\cdot16}=\dfrac{7\cdot2\cdot34-7\cdot3\cdot10}{28\left(10-16\right)}\)
\(=\dfrac{7\left(2\cdot34-3\cdot10\right)}{28\cdot\left(-6\right)}=\dfrac{1}{4}\cdot\dfrac{68-30}{-6}=\dfrac{1}{4}\cdot\dfrac{38}{-6}=\dfrac{38}{-24}=-\dfrac{19}{12}\)