Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
1,2x\left( {x - 1} \right) + 3 = 3x\\
\to 2{x^2} - 2x + 3 - 3x = 0\\
\to \left[ \begin{array}{l}
2x - 3 = 0\\
x - 1 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = \frac{3}{2}\\
x = 1
\end{array} \right.\\
2,\left| {x - 2} \right| = \left| {3x - 5} \right|\\
\to {x^2} - 4x + 4 = 9{x^2} - 30x + 25\\
\to 8{x^2} - 26x + 21 = 0\\
\to \left[ \begin{array}{l}
x = \frac{7}{4}\\
x = \frac{3}{2}
\end{array} \right.\\
B3:\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
9\left( {{m^2} + 2m + 1} \right) - \left( {m + 1} \right)\left( {2m - 3} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
9{m^2} + 18m + 9 - 2{m^2} + 3m - 2m + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
7{m^2} + 19m + 12 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
\left[ \begin{array}{l}
m = - 1\left( l \right)\\
m = - \frac{{12}}{7}
\end{array} \right.
\end{array} \right.
\end{array}\)
