Đáp án:
$\begin{array}{l}
a)A = \left( {x - 3} \right)\left( {x + 4} \right) - 2\left( {3x - 2} \right)\\
= {x^2} + x - 12 - 6x + 4\\
= {x^2} - 5x - 8 = B = {\left( {x - 4} \right)^2}\\
\Rightarrow {x^2} - 5x - 8 = {x^2} - 8x + 16\\
\Rightarrow 3x = 24\\
\Rightarrow x = 8\\
b)A = \left( {x + 2} \right)\left( {x - 2} \right) + 3{x^2}\\
= {x^2} - 4 + 3{x^2}\\
= 4{x^2} - 4\\
B = {\left( {2x + 1} \right)^2} + 2x\\
= 4{x^2} + 4x + 1 + 2x\\
= 4{x^2} + 6x + 1\\
\Rightarrow 4{x^2} - 4 = 4{x^2} + 6x + 1\\
\Rightarrow 6x = - 5\\
\Rightarrow x = - \frac{5}{6}\\
d)A = {\left( {x + 1} \right)^3} - {\left( {x - 2} \right)^3}\\
= 9{x^2} - 9x + 9\\
B = \left( {3x - 1} \right)\left( {3x + 1} \right)\\
= 9{x^2} - 1\\
\Rightarrow 9{x^2} - 9x + 9 = 9{x^2} - 1\\
\Rightarrow - 9x = - 10\\
\Rightarrow x = \frac{{10}}{9}
\end{array}$