Đáp án:
$\begin{array}{l}
{x^2} + 2x - 2.\left| {x + 1} \right| + m - 1 = 0\\
\Rightarrow {x^2} + 2x + 1 - 2\left| {x + 1} \right| + {m^2} - 2 = 0\\
\Rightarrow {\left( {x + 1} \right)^2} - 2\left| {x + 1} \right| + {m^2} - 2 = 0\\
\Rightarrow {\left( {\left| {x + 1} \right|} \right)^2} - 2\left| {x + 1} \right| + {m^2} - 2 = 0\\
\text{Đặt}:\left| {x + 1} \right| = t\left( {t \ge 0} \right)\\
\Rightarrow {t^2} - 2t + {m^2} - 2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta ' \ge 0\\
\dfrac{{ - b}}{a} \ge 0\\
\dfrac{c}{a} \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
1 - {m^2} + 2 \ge 0\\
2 \ge 0\\
{m^2} - 2 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{m^2} \le 3\\
{m^2} \ge 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt 2 \le m \le \sqrt 3 \\
- \sqrt 3 \le m \le - \sqrt 2
\end{array} \right.
\end{array}$
