\(P=(\dfrac{-2}{3xy^2})^3(\dfrac{-1}{2x^2y^3})^2\) (1)
b, Thay x=-1 ; y =1 vào đơn thức (1) ta có :
\(P=(\dfrac{-2}{3(-1)1^2})^3(\dfrac{-1}{2(-1)^2.1^3})^2\)
\(P=(\dfrac{2}{3})^3 . (\dfrac{-1}{2})^2\)
\(P=\dfrac{8}{27}.\dfrac{1}{4}\)
\(P= \dfrac{2}{27}\)