Đáp án:
$\begin{array}{l}
a)\left( {2x + 3y} \right)\left( {\dfrac{1}{2}xy - \dfrac{2}{3}} \right)\\
= {x^2}y - \dfrac{4}{3}x + \dfrac{3}{2}x{y^2} - 2y\\
b)\left( {{x^2} - 2x} \right)\left( {{x^3} + 4x - 1} \right)\\
= {x^5} + 4{x^3} - {x^2} - 2{x^4} - 8{x^2} + 2x\\
= {x^5} - 2{x^4} + {4^3} - 9{x^2} + 2x\\
c)x\left( {x - 1} \right) - x + 1\\
= {x^2} - x - x + 1\\
= {x^2} - 2x + 1\\
d)8{x^3} + 12{x^2}y + 6x{y^2} + {y^3}\\
= {\left( {2x + y} \right)^3}\\
e)\dfrac{1}{{25}}{x^2} - 64{y^2}\\
= \left( {\dfrac{1}{5}x - 8y} \right)\left( {\dfrac{1}{5}x + 8y} \right)\\
f){x^2} + 42 - {y^2} + 4\\
= {x^2} - {y^2} + 46
\end{array}$
