Đáp án:
$\begin{array}{l}
1a){\left( {x + 2} \right)^2} = {x^2} + 4x + 4\\
b){\left( {3x - 2} \right)^2} = 9{x^2} - 12x + 4\\
c){\left( {2x - 3y} \right)^2} = 4{x^2} - 12xy + 9{y^2}\\
d){x^2} - 9 = \left( {x - 3} \right)\left( {x + 3} \right)\\
e)4{x^2} - 16 = \left( {2x - 4} \right)\left( {2x + 4} \right)\\
f){\left( {3x - \dfrac{1}{3}} \right)^2} = 9{x^2} - 2x + \dfrac{1}{9}\\
B2)\\
a){\left( {x - 3} \right)^2} - x\left( {x + 1} \right) = 5\\
\Leftrightarrow {x^2} - 6x + 9 - {x^2} - x = 5\\
\Leftrightarrow - 7x = - 4\\
\Leftrightarrow x = \dfrac{4}{7}\\
Vậy\,x = \dfrac{4}{7}\\
b)\left( {x - 1} \right)\left( {x + 4} \right) - {\left( {x - 2} \right)^2} = 6\\
\Leftrightarrow {x^2} + 3x - 4 - {x^2} + 4x - 4 = 6\\
\Leftrightarrow 7x = 14\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
c)\left( {3x - 2} \right)\left( {x + 1} \right) - 3{\left( {x - 1} \right)^2} = 2\\
\Leftrightarrow 3{x^2} + x - 2 - 3\left( {{x^2} - 2x + 1} \right) = 2\\
\Leftrightarrow 3{x^2} + x - 2 - 3{x^2} + 6x - 3 - 2 = 0\\
\Leftrightarrow 7x - 7 = 0\\
\Leftrightarrow 7x = 7\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
d){\left( {x - 2} \right)^2} + \left( {x - 3} \right)\left( {x + 3} \right)\\
- \left( {2x - 1} \right)\left( {x - 4} \right) = 5\\
\Leftrightarrow {x^2} - 4x + 4 + {x^2} - 9\\
- \left( {2{x^2} - 9x + 4} \right) = 5\\
\Leftrightarrow 2{x^2} - 4x - 5 - 2{x^2} + 9x - 4 = 5\\
\Leftrightarrow 5x = 14\\
\Leftrightarrow x = \dfrac{{14}}{5}\\
Vậy\,x = \dfrac{{14}}{5}
\end{array}$
