Đáp án:
$\begin{array}{l}
B2)\\
a)3{x^2}y\left( {2{x^2} - y} \right) - 2{x^2}\left( {2{x^2}y - {y^2}} \right)\\
= 6{x^4}y - 3{x^2}{y^2} - 4{x^4}y + 2{x^2}{y^2}\\
= 2{x^4}y - {x^2}{y^2}\\
b)3{x^2}\left( {2y - 1} \right) - \left[ {2{x^2}\left( {5y - 3} \right) - 2x\left( {x - 1} \right)} \right]\\
= 6{x^2}y - 3{x^2} - 10{x^2}y + 6{x^2} + 2{x^2} - 2x\\
= - 4{x^2}y + 5{x^3} - 2x\\
B3)\\
a)3\left( {2x - 1} \right) - x\left( {3x - 2} \right) = 3x\left( {1 - x} \right) + 2\\
\Leftrightarrow 6x - 3 - 3{x^2} + 2x = 3x - 3{x^2} + 2\\
\Leftrightarrow 5x = 5\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
b)\dfrac{1}{4}{x^2} - \left( {\dfrac{1}{2}x - 4} \right).\dfrac{1}{2}x = - 14 - \dfrac{3}{2}\left( {x - \dfrac{8}{3}} \right)\\
\Leftrightarrow \dfrac{1}{4}{x^2} - \dfrac{1}{4}{x^2} + 2x = - 14 - \dfrac{3}{2}x + 4\\
\Leftrightarrow 2x + \dfrac{3}{2}x = - 10\\
\Leftrightarrow \dfrac{7}{2}x = - 10\\
\Leftrightarrow x = - \dfrac{{20}}{7}\\
Vậy\,x = \dfrac{{ - 20}}{7}\\
c)2{x^3}\left( {2x - 3} \right) - {x^2}\left( {4{x^2} - 6x + 2} \right) = 0\\
\Leftrightarrow {x^2}\left( {4{x^2} - 6x - 4{x^2} + 6x - 2} \right) = 0\\
\Leftrightarrow {x^2}.\left( { - 2} \right) = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$
