A=\(\dfrac{28}{15}\)(\(\dfrac{1}{2}\))\(^2\).3+(\(\dfrac{8}{15}-\dfrac{79}{60}\)):\(\dfrac{47}{24}\) A=28\(\dfrac{28}{15}.\dfrac{1}{4}.3+\dfrac{\left(-47\right)}{60}.\dfrac{24}{47}\) A=\(\dfrac{7}{5}+\dfrac{\left(-2\right)}{5}=\dfrac{5}{5}=1\) Vậy A=1