Đáp án:
$a \in \left\{ {\dfrac{{ - 5}}{3}; - 1} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
{12.1^3} + 2.a.1 + {a^2} = 2.{\left( { - 2} \right)^2} - \left| {2a + 3} \right|.\left( { - 2} \right) + {a^2}\\
\Leftrightarrow 12 + 2a + {a^2} = 8 + 2\left| {2a + 3} \right| + {a^2}\\
\Leftrightarrow 2\left| {2a + 3} \right| = 2a + 4\\
\Leftrightarrow \left| {2a + 3} \right| = a + 2\\
\Leftrightarrow \left\{ \begin{array}{l}
a + 2 \ge 0\\
\left[ \begin{array}{l}
2a + 3 = a + 2\\
2a + 3 = - a - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a \ge - 2\\
\left[ \begin{array}{l}
a = - 1\\
a = \dfrac{{ - 5}}{3}
\end{array} \right.\left( c \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = - 1\\
a = \dfrac{{ - 5}}{3}
\end{array} \right.
\end{array}$
Vậy $a \in \left\{ {\dfrac{{ - 5}}{3}; - 1} \right\}$
