Đáp án:
$\begin{array}{l}
b) - 5{x^3}{y^2} + 10{x^3}{y^2} + \left( { - \dfrac{3}{4}{x^3}{y^2}} \right) - {x^3}{y^2}\\
= {x^3}.{y^2}.\left( { - 5 + 10 - \dfrac{3}{4} - 1} \right)\\
= {x^3}{y^2}.\left( {4 - \dfrac{3}{4}} \right)\\
= {x^3}{y^2}.\dfrac{{13}}{4}\\
= \dfrac{{13}}{4}.{x^3}{y^2}\\
c)\dfrac{1}{3}{x^2}y + x{y^2} - xy - \dfrac{1}{2}x{y^2} - 3xy - \dfrac{1}{3}{x^2}y\\
= \left( {\dfrac{1}{3}{x^2}y - \dfrac{1}{3}{x^2}y} \right) + \left( {x{y^2} - \dfrac{1}{2}x{y^2}} \right) + \left( { - xy - 3xy} \right)\\
= 0 + \dfrac{1}{2}x{y^2} - 4xy\\
= \dfrac{1}{2}x{y^2} - 4xy
\end{array}$
