Đáp án:
$\begin{array}{l}
a)2x\left( {{x^2} - 7x - 3} \right)\\
= 2{x^3} - 14{x^2} - 6x\\
b)\left( { - 2{x^3} + \dfrac{3}{4}{y^2} - 7xy} \right).4x{y^2}\\
= - 8{x^4}{y^2} + 3x{y^4} - 28{x^2}{y^3}\\
c)\left( { - 5{x^3}} \right)\left( {2{x^2} + 3x - 5} \right)\\
= - 10{x^5} - 15{x^4} + 25{x^3}\\
d)\left( {2{x^2} - \dfrac{1}{3}xy + {y^2}} \right).\left( { - 3{x^3}} \right)\\
= - 6{x^5} + {x^4}y - 3{x^3}{y^2}\\
e)\left( {{x^2} - 2x + 3} \right)\left( {x - 4} \right)\\
= {x^3} - 4{x^2} - 2{x^2} + 8x + 3x - 12\\
= {x^3} - 6{x^2} + 11x - 12\\
f)\left( {2{x^3} - 3x - 1} \right)\left( {5x + 2} \right)\\
= 10{x^4} + 4{x^3} - 15{x^2} - 6x - 5x - 2\\
= 10{x^4} + 4{x^3} - 15{x^2} - 11x - 2\\
B2)\\
a){\left( {2x + 3y} \right)^2} = 4{x^2} + 12xy + 9{y^2}\\
b){\left( {5x - y} \right)^2} = 25{x^2} - 10xy + {y^2}\\
c){\left( {x + \dfrac{1}{4}} \right)^2} = {x^2} + \dfrac{1}{2}x + \dfrac{1}{{16}}\\
d)\left( {{x^2} + \dfrac{2}{5}y} \right)\left( {{x^2} - \dfrac{2}{5}y} \right)\\
= {\left( {{x^2}} \right)^2} - {\left( {\dfrac{2}{5}y} \right)^2}\\
= {x^4} - \dfrac{4}{{25}}{y^2}
\end{array}$
