a) \(\left( {{x^2} - 4} \right) + x\left( {x - 2} \right) = 0;\)
\(\begin{array}{l}\left( {{x^2} - 4} \right) + x\left( {x - 2} \right) = 0\\\left( {x - 2} \right)\left( {x + 2} \right) + x\left( {x - 2} \right) = 0\\\left( {x - 2} \right)\left( {x + 2 + x} \right) = 0\end{array}\)
\(\begin{array}{l}\left( {x - 2} \right)\left( {2x + 2} \right) = 0\\TH1:x - 2 = 0\\x = 2\\TH2:2x + 2 = 0\\2x = - 2\\x = - 1\end{array}\)
Vậy \(x \in \left\{ { - 1;2} \right\}.\)
b) \({\left( {2x + 1} \right)^2} - 9{x^2} = 0.\)
\(\begin{array}{l}{\left( {2x + 1} \right)^2} - {\left( {3x} \right)^2} = 0\\\left( {2x + 1 - 2x} \right)\left( {2x + 1 + 3x} \right) = 0\\1.\left( {5x + 1} \right) = 0\\5x = - 1\\x = \frac{{ - 1}}{5}\end{array}\)
Vậy \(x = \frac{{ - 1}}{5}.\)