Đáp án:
$\begin{array}{l}
A = - {x^2}{y^3};B = \dfrac{{ - 2}}{9}x{y^2};C = - 3y + 2x\\
a)A.C + B\\
= \left( { - {x^2}{y^3}} \right).\left( { - 3y + 2x} \right) - \dfrac{2}{9}x{y^2}\\
= 3{x^2}{y^4} - 2{x^3}{y^3} - \dfrac{2}{9}x{y^2}\\
b)B.C - A\\
= \left( { - \dfrac{2}{9}x{y^2}} \right).\left( { - 3y + 2x} \right) + {x^2}{y^3}\\
= \dfrac{2}{3}x{y^3} - \dfrac{4}{9}{x^2}{y^2} + {x^2}{y^3}\\
c)A.B.C\\
= \left( { - {x^2}{y^3}} \right)\left( {\dfrac{{ - 2}}{9}x{y^2}} \right)\left( { - 3y + 2x} \right)\\
= \dfrac{2}{9}.{x^3}{y^5}.\left( {2x - 3y} \right)\\
= \dfrac{4}{9}{x^4}{y^5} - \dfrac{2}{3}{x^3}{y^6}\\
d)\dfrac{A}{B}.C\\
= \dfrac{{ - {x^2}{y^3}}}{{\dfrac{{ - 2}}{9}x{y^2}}}.\left( {2x - 3y} \right)\\
= \dfrac{9}{2}xy.\left( {2x - 3y} \right)\\
= 9{x^2}y - \dfrac{{27}}{2}x{y^2}
\end{array}$
