Đáp án:
$\begin{array}{l}
1)a)\left( {3{x^2}y - 6xy + 9x} \right).\left( {\frac{{ - 4}}{{3xy}}} \right)\\
= - 4x + 24 - \frac{{12}}{y}\\
b)\left( {\frac{1}{3}x + 2y} \right)\left( {\frac{1}{9}{x^2} - \frac{2}{3}xy + 4{y^2}} \right)\\
= \left( {\frac{1}{3}x + 2y} \right)\left[ {{{\left( {\frac{1}{3}x} \right)}^2} - \frac{1}{3}x.2y + {{\left( {2y} \right)}^2}} \right]\\
= {\left( {\frac{1}{3}x} \right)^3} + {\left( {2y} \right)^3}\\
= \frac{1}{{27}}{x^3} + 8{y^3}\\
c)\left( {x - 2} \right)\left( {{x^2} - 5x + 1} \right) + x\left( {{x^2} + 11} \right)\\
= {x^3} - 5{x^2} + x - 2{x^2} + 10x - 2 + {x^3} + 11x\\
= 2{x^3} - 7{x^2} + 22x - 2\\
d)\left( {x - 3y} \right)\left( {{x^2} + 3xy + 9{y^2}} \right)\\
= {x^3} - {\left( {3y} \right)^3}\\
= {x^3} - 27{y^3}\\
e)\left( {3 + x} \right)\left( {{x^2} + 3x - 5} \right)\\
= 3{x^2} + 9x - 15 + {x^3} + 3{x^2} - 5x\\
= {x^3} + 6{x^2} + 4x - 15\\
f)\left( {x + 2} \right)\left( {x - 2} \right) - \left( {2x + 1} \right)\\
= {x^2} - 4 - 2x - 1\\
= {x^2} - 2x - 5\\
2)a)\\
{\left( {3x + 2} \right)^2} + 2\left( {2 + 3x} \right)\left( {1 - 2y} \right) + {\left( {2y - 1} \right)^2}\\
= {\left( {3x + 2 + 1 - 2y} \right)^2}\\
= {\left( {3x - 2y + 3} \right)^2}\\
= 9{x^2} + 4{y^2} + 9 - 12xy + 18x - 12y\\
b){\left( {3x + 1} \right)^2} - 2\left( {3x + 1} \right)\left( {3x + 5} \right) + {\left( {3x + 5} \right)^2}\\
= {\left( {3x + 1 - 3x - 5} \right)^2}\\
= {\left( { - 4} \right)^2}\\
= 16\\
c)2x{\left( {2x - 1} \right)^2} - 3x\left( {x + 3} \right)\left( {x - 3} \right) - 4x{\left( {x + 1} \right)^2}\\
= 2x\left( {4{x^2} - 4x + 1} \right) - 3x\left( {{x^2} - 9} \right)\\
- 4x\left( {{x^2} + 2x + 1} \right)\\
= 8{x^3} - 8{x^2} + 2x - 3{x^3} + 27x\\
- 4{x^3} - 8{x^2} - 4x\\
= {x^3} - 16{x^2} + 25x
\end{array}$
