Đáp án:
$\begin{array}{l}
B1) - 3x\left( {2x{y^2} - \dfrac{5}{3}{x^2}y + z} \right)\\
= - 6{x^2}{y^2} + 5{x^3}y - 3xz\\
\left( {3{x^2} - 2x} \right)\left( {4{x^3} + 5x + 1} \right)\\
= 12{x^5} + 15{x^3} + 3{x^2} - 8{x^4} - 10{x^2} - 2x\\
= 12{x^5} - 8{x^4} + 15{x^3} - 7{x^2} - 2x\\
\dfrac{1}{3}{x^3}{y^2}\left( {2x + y} \right)\left( {2x - y} \right)\\
= \dfrac{1}{3}{x^3}{y^2}\left( {4{x^2} - {y^2}} \right)\\
= \dfrac{4}{3}{x^5}{y^2} - \dfrac{1}{3}{x^3}{y^4}\\
B2)\\
a)12{x^2}y - 18x{y^2}\\
= 6xy\left( {2x - 3y} \right)\\
b)\dfrac{3}{5}{x^3}{y^2} - 6{x^2}y - x{y^3}\\
= xy\left( {\dfrac{3}{5}{x^2}y - 6x - {y^2}} \right)\\
c)\dfrac{5}{3}x\left( {2x + y} \right) - \dfrac{5}{3}y\left( {2x + y} \right)\\
= \dfrac{5}{3}\left( {2x + y} \right)\left( {x - y} \right)\\
d)\dfrac{3}{7}x\left( {y - 2x} \right) + y\left( {2x - y} \right)\\
= \left( {2x - y} \right)\left( {y - \dfrac{3}{7}x} \right)
\end{array}$
