Đáp án:

$\begin{array}{l}
1)a){\left( {x + \dfrac{1}{3}} \right)^3} = {x^3} + {x^2} + \dfrac{1}{3}x + \dfrac{1}{{27}}\\
c){\left( {x - 3} \right)^3} = {x^3} - 9{x^2} + 27x - 27\\
e)\left( {x - 2} \right)\left( {{x^2} + 2x + 9} \right)\\
 = {x^3} + 2{x^2} + 9x - 2{x^2} - 4x - 18\\
 = {x^3} + 5x - 18\\
g)\left( {x + 4} \right)\left( {{x^2} - 4x + 16} \right)\\
 = {x^3} + 64\\
b){\left( {2x + y} \right)^3}\\
 = 8{x^3} + 12{x^2}y + 6x{y^2} + {y^3}\\
c){\left( {3x - 2y} \right)^3}\\
 = 27{x^3} - 54{x^2}y + 18x{y^2} - 8{y^3}\\
d)\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\\
 = {x^3} + 1\\
f)\left( {x - 3y} \right)\left( {{x^2} + 3xy + 9{y^2}} \right)\\
 = {x^3} - 27{y^3}\\
h)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
 = {x^3} - 1\\
b2)\\
a){x^2} - 6x + 9\\
 = {\left( {x - 3} \right)^2}\\
b)25 + 10x + {x^2} = {\left( {x + 5} \right)^2}\\
c)\dfrac{1}{4}{a^2} + 2a{b^2} + 4{b^4}\\
 = {\left( {\dfrac{1}{2}a + 2{b^2}} \right)^2}\\
d)4{x^2} - 49 = \left( {2x - 7} \right)\left( {2x + 7} \right)\\
e)4{x^2} - 25{y^2} = \left( {2x - 5y} \right)\left( {2x + 5y} \right)\\
f)8{x^3} + 1 = \left( {2x + 1} \right)\left( {4{x^2} - 4x + 1} \right)\\
g){x^3} - 1 = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
h)8{x^3} - 27 = \left( {2x - 3} \right)\left( {4{x^2} + 6x + 9} \right)\\
i){x^3} - \dfrac{1}{8} = \left( {x - \dfrac{1}{2}} \right)\left( {{x^2} + \dfrac{1}{2}x + \dfrac{1}{4}} \right)\\
k)4{x^2} + 4x + 1 = {\left( {2x + 1} \right)^2}
\end{array}$