Giải thích các bước giải:
\(\begin{array}{l}
a.{x^2}\left( {x - 2} \right) - 9\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {{x^2} - 9} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = \pm 3
\end{array} \right.\\
b.4{x^2} - 1 = 6{x^2} - 7x - 5\\
\to 2{x^2} - 7x - 4 = 0\\
\to \left( {x - 4} \right)\left( {2x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = - \frac{1}{2}
\end{array} \right.\\
c.{x^2} + 2x + 1 - 4{x^2} + 8x - 4 = 0\\
\to - 3{x^2} + 10x - 3 = 0\\
\to \left( {x - 3} \right)\left( {3x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - \frac{1}{3}
\end{array} \right.\\
d.{\left( {2x + 1} \right)^2} = 0\\
\to x = - \frac{1}{2}\\
e.\left( {x - 3} \right)\left( {2x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - \frac{3}{2}
\end{array} \right.\\
f.x\left( {2x + 1} \right)\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - \frac{1}{2}\\
x = - 3
\end{array} \right.
\end{array}\)
