Đáp án:
$\begin{array}{l}
B1)\\
\dfrac{1}{2}x.\dfrac{1}{4}{x^2}\dfrac{1}{8}{x^3}.2y.4{y^2}.8{y^3}\\
= \dfrac{1}{2}.\dfrac{1}{4}.\dfrac{1}{8}.2.4.8.{x^6}.{y^6}\\
= {x^6}.{y^6}\\
B2)\\
{2^2}.{y^2}.{z^2} = \left( {3{x^3}.{y^2}.z} \right).A\\
\Leftrightarrow A = {2^2}.{y^2}.{z^2}:\left( {3{x^3}.{y^2}.z} \right)\\
= 4{y^2}{z^2}:\left( {3{x^3}.{y^2}.z} \right)\\
= \dfrac{{4z}}{{3{x^3}}}
\end{array}$