Đáp án:
$\begin{array}{l}
B1)\\
a)\left( {{x^2} - 1} \right)\left( {{x^2} + 2x} \right)\\
= {x^4} + 2{x^3} - {x^2} - 2x\\
b)\left( {2x - 1} \right)\left( {3x + 2} \right)\left( {3 - x} \right)\\
= \left( {6{x^2} + x - 2} \right)\left( {3 - x} \right)\\
= 18{x^2} - 6{x^3} + 3x - {x^2} - 6 + 2{x^2}\\
= - 6{x^3} + 19{x^2} + 3x - 6\\
c)\left( {x + 3} \right)\left( {{x^2} + 3x - 5} \right)\\
= {x^3} + 3{x^2} - 5x + 3{x^2} + 9x - 15\\
= {x^3} + 6{x^2} + 4x - 15\\
d)\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\\
= {x^3} + 1\\
e)\left( {2{x^3} - 3x - 1} \right)\left( {5x + 2} \right)\\
= 10{x^4} + 4{x^3} - 15{x^2} - 6x - 5x - 2\\
= 10{x^4} + 4{x^3} - 15{x^2} - 11x - 2\\
f)\left( {{x^2} - 2x + 3} \right)\left( {x - 4} \right)\\
= {x^3} - 4{x^2} - 2{x^2} + 8x + 3x - 12\\
= {x^3} - 6{x^2} + 11x - 12\\
B2)\\
a) - 2{x^3}y\left( {2{x^2} - 3y + 5yz} \right)\\
= - 4{x^5}y + 6{x^3}{y^2} - 10{x^3}{y^2}z\\
b)\left( {x - 2y} \right)\left( {{x^2}{y^2} - xy + 2y} \right)\\
= {x^3}{y^2} - {x^2}y + 2xy - 2{x^2}{y^3} + 2x{y^2} - 4{y^2}\\
c)\dfrac{2}{5}xy\left( {{x^2}y - 5x + 10y} \right)\\
= \dfrac{2}{5}{x^3}{y^2} - 2{x^2}y + 4x{y^2}\\
d)\dfrac{2}{3}{x^2}y\left( {3xy - {x^2} + y} \right)\\
= 2{x^3}{y^2} - \dfrac{2}{3}{x^4}y + \dfrac{2}{3}{x^2}{y^2}\\
e)\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\\
= {x^3} - {y^3}\\
f)\left( {\dfrac{1}{2}xy - 1} \right)\left( {{x^3} - 2x - 6} \right)\\
= \dfrac{1}{2}{x^4}y - {x^2}y - 3xy - {x^3} + 2x + 6
\end{array}$
