Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2x\left( {3x - 5} \right) - \left( {5 - 3x} \right) = 0\\
\Leftrightarrow 2x\left( {3x - 5} \right) + \left( {3x - 5} \right) = 0\\
\Leftrightarrow \left( {3x - 5} \right)\left( {2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 5 = 0\\
2x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{5}{3}\\
x = - \frac{1}{2}
\end{array} \right.\\
b,\\
9\left( {3x - 2} \right) = x\left( {2 - 3x} \right)\\
\Leftrightarrow 9\left( {3x - 2} \right) - x.\left( {2 - 3x} \right) = 0\\
\Leftrightarrow 9\left( {3x - 2} \right) + x.\left( {3x - 2} \right) = 0\\
\Leftrightarrow \left( {3x - 2} \right)\left( {9 + x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 2 = 0\\
9 + x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{2}{3}\\
x = - 9
\end{array} \right.\\
c,\\
{\left( {2x - 1} \right)^2} - 25 = 0\\
\Leftrightarrow {\left( {2x - 1} \right)^2} = 25\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 5\\
2x - 1 = - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 6\\
2x = - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.\\
d,\\
{x^2} - 25 = 6x - 9\\
\Leftrightarrow {x^2} - 25 - 6x + 9 = 0\\
\Leftrightarrow {x^2} - 6x - 16 = 0\\
\Leftrightarrow \left( {{x^2} - 8x} \right) + \left( {2x - 16} \right) = 0\\
\Leftrightarrow x\left( {x - 8} \right) + 2\left( {x - 8} \right) = 0\\
\Leftrightarrow \left( {x - 8} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 8 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = - 2
\end{array} \right.
\end{array}\)
