a) \(A=\dfrac{8}{3}x^2y^2\cdot\left(-\dfrac{1}{4}x^2y\right)=\left(\dfrac{8}{3}\cdot\dfrac{-1}{4}\right)\left(x^2y^2\cdot x^2y\right)=-\dfrac{2}{3}x^4y^3\)
Hệ số: \(-\dfrac{2}{3}\); bậc: 7.
b) Thay \(x=1;y=1:\)
\(A=-\dfrac{2}{3}\cdot1^4\cdot1^3=-\dfrac{2}{3}\)
Vậy \(A=-\dfrac{2}{3}\) khi \(x=1;y=1.\)