Đáp án:
$\begin{array}{l}
a){\left( {2x + 3y} \right)^2} = 4{x^2} + 2.2x.3y + 9{y^2}\\
= 4{x^2} + 12xy + 9{y^2}\\
b){\left( {5x - y} \right)^2} = 25{x^2} - 2.5x.y + {y^2}\\
= 25{x^2} - 10xy + {y^2}\\
c){\left( {2x + {y^2}} \right)^3}\\
= {\left( {2x} \right)^3} + 3.{\left( {2x} \right)^2}.{y^2} + 3.2x.{\left( {{y^2}} \right)^2} + {\left( {{y^2}} \right)^3}\\
= 8{x^3} + 12{x^2}{y^2} + 6x{y^4} + {y^6}\\
d)\left( {{x^2} + \frac{2}{5}y} \right)\left( {{x^2} - \frac{2}{5}y} \right)\\
= {\left( {{x^2}} \right)^2} - {\left( {\frac{2}{5}y} \right)^2}\\
= {x^4} - \frac{4}{{25}}{y^2}\\
e){\left( {x + \frac{1}{4}} \right)^2} = {x^2} + 2.x.\frac{1}{4} + \frac{1}{{16}}\\
= {x^2} + \frac{1}{2}x + \frac{1}{{16}}\\
f){\left( {\frac{2}{3}{x^2} - \frac{1}{2}y} \right)^3}\\
= {\left( {\frac{2}{3}{x^2}} \right)^3} - 3.{\left( {\frac{2}{3}{x^2}} \right)^2}.\frac{1}{2}y\\
+ 3.\frac{2}{3}{x^2}.{\left( {\frac{1}{2}y} \right)^2} - {\left( {\frac{1}{2}y} \right)^3}\\
= \frac{8}{{27}}{x^6} - \frac{2}{3}{x^4}y + \frac{1}{2}{x^2}y - \frac{1}{8}{y^3}\\
e){\left( {3{x^2} - 2y} \right)^3}\\
= {\left( {3{x^2}} \right)^3} - 3.{\left( {3{x^2}} \right)^2}.2y + 3.3{x^2}.{\left( {2y} \right)^2}\\
- {\left( {2y} \right)^3}\\
= 27{x^6} - 54{x^4}y + 36{x^2}{y^2} - 8{y^3}\\
h)\left( {x - 3y} \right)\left( {{x^2} + 3xy + 9{y^2}} \right)\\
= {x^3} - {\left( {3y} \right)^3}\\
= {x^3} - 27{y^3}\\
i)\left( {{x^2} - 3} \right)\left( {{x^4} + 3{x^2} + 9} \right)\\
= {\left( {{x^2}} \right)^3} - {3^3}\\
= {x^6} - 27
\end{array}$
