Đáp án:
$\begin{array}{l}
d) - xy\left( {2x + 3xy - 8y} \right)\\
= - 2{x^2}y - 3{x^2}{y^2} + 8x{y^2}\\
B2)\\
a)3\left( {3{x^2} + 2} \right) - {x^2}\left( {4x + 3} \right) + {x^3}\\
= 9{x^2} + 6 - 4{x^3} - 3{x^2} + {x^3}\\
= - 3{x^3} + 6{x^2} + 6\\
= - {3.5^3} + {6.5^2} + 6\\
= - 3.125 + 6.25 + 6\\
= - 375 + 156\\
= - 219\\
b)\\
7xy\left( {xy - {x^2}} \right) - 3{y^2}\left( {{x^2} - y} \right) - 4{x^2}\left( {{y^2} - xy} \right)\\
= 7{x^2}{y^2} - 7{x^3}y - 3{x^2}{y^2} + 3{y^3}\\
- 4{x^2}{y^2} + 4{x^3}y\\
= - 3{x^3}y + 3{y^3}\\
= - {3.1^3}.\dfrac{1}{2} + 3.{\left( {\dfrac{1}{2}} \right)^3}\\
= - \dfrac{3}{2} + \dfrac{3}{8}\\
= \dfrac{{ - 9}}{8}\\
B3)\\
a)3x\left( {x + 1} \right) - x\left( {3x + 2} \right) = 12\\
\Leftrightarrow 3{x^2} + 3x - 3{x^2} - 2x = 12\\
\Leftrightarrow x = 12\\
Vậy\,x = 12\\
b){x^2}\left( {4 - 3x} \right) + x\left( {3{x^2} - 4x + 2} \right) = 8\\
\Leftrightarrow 4{x^2} - 3{x^3} + 3{x^3} - 4{x^2} + 2x = 8\\
\Leftrightarrow 2x = 8\\
\Leftrightarrow x = 4\\
Vậy\,x = 4
\end{array}$
