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