Đáp án:
$\begin{array}{l}
a){\left( {3x + 5} \right)^2} = 9{x^2} + 30x + 25\\
b){\left( {6{x^2} + \dfrac{1}{3}} \right)^2} = 36{x^4} + 4{x^2} + \dfrac{1}{9}\\
c){\left( {5x - 4y} \right)^2} = 25{x^2} - 40xy + 16{y^2}\\
d){\left( {2x - \dfrac{1}{2}} \right)^2} = 4{x^2} - 2x + \dfrac{1}{4}\\
e)\left( {6x + 5y} \right)\left( {6x - 5y} \right)\\
= 36{x^2} - 25{y^2}\\
f){\left( {4 - x} \right)^2} = 16 - 8x + {x^2}\\
g){\left( {x - 4} \right)^2} = {x^2} - 8x + 16\\
h){x^4} - 1 = \left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right) = \left( {{x^2} + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)\\
i)\left( {1 + 4x} \right)\left( {4x - 1} \right) = 16{x^2} - 1\\
l)\dfrac{1}{4} - 36{x^2} = \left( {\dfrac{1}{2} - 6x} \right)\left( {\dfrac{1}{2} + 6x} \right)\\
B3)\\
a){x^2} - 2x + 1 = {\left( {x - 1} \right)^2}\\
b){x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\\
c)4{x^2} - 4x + 1 = {\left( {2x - 1} \right)^2}\\
d)25{x^2} - 20xy + 4{y^2}\\
= {\left( {5x - 2y} \right)^2}\\
e){a^2} - a + \dfrac{1}{4} = {\left( {a - \dfrac{1}{2}} \right)^2}\\
f){\left( {3x + 2y} \right)^2} - 2x\left( {3x + 2y} \right) + {x^2}\\
= {\left( {3x + 2y - x} \right)^2}\\
= {\left( {2x + 2y} \right)^2}
\end{array}$
