Đáp án: $ - \frac{{17}}{4} < m < \frac{{17}}{4};m \ne 0$
Giải thích các bước giải:
$\begin{array}{l}
\left| {{x^2} - 3x - 2} \right| = m\\
\Rightarrow \left[ \begin{array}{l}
{x^2} - 3x - 2 = m\\
{x^2} - 3x - 2 = - m
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} - 3x - 2 - m = 0\left( 1 \right)\\
{x^2} - 3x - 2 + m = 0\left( 2 \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\Delta _1} > 0\\
{\Delta _2} > 0\\
m \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
9 - 4\left( { - 2 - m} \right) > 0\\
9 - 4\left( { - 2 + m} \right) > 0\\
m \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4m + 17 > 0\\
- 4m + 17 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > \frac{{ - 17}}{4}\\
m < \frac{{17}}{4}\\
m \ne 0
\end{array} \right.\\
\Rightarrow - \frac{{17}}{4} < m < \frac{{17}}{4};m \ne 0
\end{array}$
