Đáp án:
$\begin{array}{l}
a)3x.\left( {5{x^2} - 2x - 1} \right)\\
= 15{x^3} - 6{x^2} - 3x\\
b)\left( {{x^2} - 2xy + 3} \right).\left( { - xy} \right)\\
= - {x^3}y + 2{x^2}{y^2} - 3xy\\
c)\\
\dfrac{1}{2}{x^2}y\left( {2{x^3} - \dfrac{2}{5}x{y^2} - 1} \right)\\
= {x^5}y - \dfrac{1}{5}{x^3}{y^3} - \dfrac{1}{2}{x^2}y\\
d)\dfrac{2}{7}x\left( {1,4x - 3,5y} \right)\\
= \dfrac{2}{7}x.\dfrac{7}{5}x - \dfrac{2}{7}x.\dfrac{7}{2}y\\
= \dfrac{2}{5}{x^2} - xy\\
e)\dfrac{1}{2}xy\left( {\dfrac{2}{3}{x^2} - \dfrac{3}{4}xy + \dfrac{4}{5}{y^2}} \right)\\
= \dfrac{1}{2}xy.\dfrac{2}{3}{x^2} - \dfrac{1}{2}xy.\dfrac{3}{4}xy + \dfrac{1}{2}xy.\dfrac{4}{5}{y^2}\\
= \dfrac{1}{3}{x^3}y - \dfrac{3}{8}{x^2}{y^2} + \dfrac{2}{5}x{y^3}\\
f)\left( {1 + 2x - {x^2}} \right).5x\\
= - 5{x^3} + 10{x^2} + 5x\\
g)\left( {{x^2}y - xy + x{y^2} + {y^3}} \right).3x{y^2}\\
= 3{x^3}{y^3} - 3{x^2}{y^3} + 3{x^2}{y^5} + 3x{y^5}\\
h)\dfrac{2}{3}{x^2}y\left( {15x - 0,9y + 6} \right)\\
= 10{x^3}y - \dfrac{3}{5}{x^2}{y^2} + 4{x^2}y\\
i) - \dfrac{3}{7}{x^4}\left( {2,1{y^2} - 0,7x + 35} \right)\\
= - \dfrac{3}{7}.{x^4}.\dfrac{{21}}{{10}}{y^2} + \dfrac{3}{7}{x^4}.\dfrac{7}{{10}}x - \dfrac{3}{7}{x^4}.35\\
= - \dfrac{9}{{10}}{x^4}{y^2} + \dfrac{3}{{10}}{x^5} - 15{x^4}
\end{array}$
