Đáp án:
$\begin{array}{l}
4)\\
- 3{x^5}{y^7}\left( {\dfrac{2}{3}{x^4}y - {y^3} + \dfrac{1}{2}} \right)\\
= - 2{x^9}{y^8} + 3{x^5}{y^{10}} - \dfrac{3}{2}{x^5}{y^7}\\
5)\dfrac{1}{2}{x^3}y\left( {2{x^4}{y^3} - 4xy - 6} \right)\\
= {x^7}{y^4} - 2{x^4}{y^2} - 3{x^3}y\\
6) - \dfrac{5}{6}x\left( {\dfrac{2}{3}{x^2}y + \dfrac{3}{4}x{y^2} - \dfrac{1}{2}xy} \right)\\
= - \dfrac{5}{9}{x^3}y - \dfrac{5}{8}{x^2}{y^2} + \dfrac{5}{{12}}{x^2}y\\
14)\left( {\dfrac{1}{2}x + 3} \right)\left( {2{x^2} - 4x - 6} \right)\\
= {x^3} - 2{x^2} - 3x + 6{x^2} - 12x - 18\\
= {x^3} + 4{x^2} - 15x - 18\\
15)\left( {\dfrac{3}{2}x - 1} \right)\left( {3{x^2} - 6x + 9} \right)\\
= \dfrac{9}{2}{x^3} - 9{x^2} + \dfrac{{27}}{2}x - 3{x^2} + 6x - 9\\
= \dfrac{9}{2}{x^3} - 12{x^2} + \dfrac{{39}}{2}x - 9\\
16)4{x^2} - \left( {x + 3} \right)\left( {x - 5} \right) + x\\
= 4{x^2} - \left( {{x^2} - 2x - 15} \right) + x\\
= 3{x^2} + 3x + 15\\
17)4x\left( {{x^2} - x - 1} \right) - \left( {{x^2} - 2} \right)\left( {x + 3} \right)\\
= 4{x^3} - 4{x^2} - 4x - {x^3} - 3{x^2} + 2x + 6\\
= 3{x^3} - 7{x^2} - 2x + 6\\
18)\\
\left( {x - 5} \right)\left( {x + 7} \right) - x\left( {x + 3} \right)\left( {x - 3} \right)\\
= {x^2} + 2x - 35 - x\left( {{x^2} - 9} \right)\\
= {x^2} + 2x - 35 - {x^3} + 9x\\
= - {x^3} + {x^2} + 11x - 35\\
20)\left( {\dfrac{1}{2}x - \dfrac{1}{3}{y^2}} \right)\left( {\dfrac{1}{2}x + \dfrac{1}{3}{y^2}} \right)\\
= {\left( {\dfrac{1}{2}x} \right)^2} - {\left( {\dfrac{1}{3}{y^2}} \right)^2}\\
= \dfrac{1}{4}{x^2} - \dfrac{1}{9}{y^4}
\end{array}$
