Đặt \(A=2x^2y+3xy^2-2x^2y-2xy^2+3\)
\(=\left(2x^2y-2x^2y\right)+\left(3xy^2-2xy^2\right)+3=xy^2+3\)
Thay \(x=-\dfrac{2}{3};y=\dfrac{1}{2}\) vào A, ta được:
\(A=-\dfrac{2}{3}\cdot\left(\dfrac{1}{2}\right)^2+3=-\dfrac{2}{3}\cdot\dfrac{1}{4}+3=-\dfrac{1}{6}+3=\dfrac{17}{6}\)