Đáp án:
$\begin{array}{l}
a)\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - x\left( {x + 4} \right)\left( {x - 4} \right) = 5\\
\Rightarrow {x^3} - {3^3} - x\left( {{x^2} - 16} \right) = 5\\
\Rightarrow {x^3} - 27 - {x^3} + 16x - 5 = 0\\
\Rightarrow 16x = 32\\
\Rightarrow x = 2\\
b)9{x^2} - 4 = \left( {2x - 1} \right)\left( {2 - 5x} \right)\\
\Rightarrow 9{x^2} - 4 = 4x - 10{x^2} - 2 + 5x\\
\Rightarrow 19{x^2} - 9x - 2 = 0\\
\Rightarrow x = \frac{{9 \pm \sqrt {233} }}{{38}}\\
c){\left( { - x - 3} \right)^2} - \left( {x + 3} \right)\left( {x - 5} \right) = 1\\
\Rightarrow {x^2} + 6x + 9 - {x^2} + 2x + 15 - 1 = 0\\
\Rightarrow 8x = - 23\\
\Rightarrow x = - \frac{{23}}{8}
\end{array}$
