Đáp án:
$\begin{array}{l}
13)a) - \dfrac{1}{3}.{x^2}.y.2x{y^3}\\
= - \dfrac{1}{3}.2.\left( {{x^2}.x} \right).\left( {y.{y^3}} \right)\\
= - \dfrac{2}{3}.{x^3}.{y^4}\\
\Rightarrow \text{Bậc}:3 + 4 = 7\\
b)\dfrac{1}{4}.{x^3}y.\left( { - 2{x^3}{y^5}} \right)\\
= \dfrac{1}{4}.\left( { - 2} \right).{x^3}.{x^3}.y.{y^5}\\
= - \dfrac{1}{2}.{x^6}.{y^6}\\
\Rightarrow \text{Bậc}:6 + 6 = 12\\
14)Khi:x = - 1;y = 1\\
Va:a.x.y = 9\\
\Rightarrow a.\left( { - 1} \right).1 = 9\\
\Rightarrow a = - 9\\
\Rightarrow \text{Biểu thức}: - 9xy
\end{array}$
