Thay \(x=-\dfrac{2}{3}\) vào A ta có:
\(A=6\cdot\left(-\dfrac{2}{3}\right)^3-3\cdot\left(-\dfrac{2}{3}\right)^2+2\cdot\left|-\dfrac{2}{3}\right|+4\)
\(A=6\cdot-\dfrac{8}{27}-3\cdot\dfrac{4}{9}+2\cdot\dfrac{2}{3}+4\)
\(A=-\dfrac{16}{9}-\dfrac{4}{3}+\dfrac{4}{3}+4\)
\(A=\dfrac{-16}{9}+4\)
\(A=\dfrac{20}{9}\)
Vậy: ....