Đáp án:
$\begin{array}{l}
c)\left| {2x + 1} \right| = \left| {{x^2} - 3x - 4} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = {x^2} - 3x - 4\\
2x + 1 = - {x^2} + 3x + 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 5x - 5 = 0\\
{x^2} - x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{5 \pm 3\sqrt 5 }}{2}\\
x = \frac{{1 \pm \sqrt {13} }}{2}
\end{array} \right.\\
d){\left( {x + 1} \right)^2} - 3\left| {x + 1} \right| + 2 = 0\\
\Leftrightarrow {\left( {\left| {x + 1} \right|} \right)^2} - 3\left| {x + 1} \right| + 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left| {x + 1} \right| = 2\\
\left| {x + 1} \right| = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}$
