Đáp án:
$\begin{array}{l}
Dkxd:y \ge - 6\\
\left| {3{{\left( {{x^2} - 1} \right)}^2} + 1} \right| + \sqrt {y + 6} = 1\\
Do:{\left( {{x^2} - 1} \right)^2} \ge 0\\
\Rightarrow 3{\left( {{x^2} - 1} \right)^2} + 1 \ge 1\\
\Rightarrow \left| {3{{\left( {{x^2} - 1} \right)}^2} + 1} \right| \ge 1\\
\sqrt {y + 6} \ge 0\\
\Rightarrow \left| {3{{\left( {{x^2} - 1} \right)}^2} + 1} \right| + \sqrt {y + 6} \ge 1 + 0\\
\Rightarrow \left| {3{{\left( {{x^2} - 1} \right)}^2} + 1} \right| + \sqrt {y + 6} \ge 1\\
\text{Dấu = xảy ra}\\
\Rightarrow \left\{ \begin{array}{l}
\left| {3{{\left( {{x^2} - 1} \right)}^2} + 1} \right| = 0\\
\sqrt {y + 6} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} - 1 = 0\\
y + 6 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} = 1\\
y = - 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 1;x = - 1\\
y = - 6
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {1; - 6} \right);\left( { - 1; - 6} \right)} \right\}
\end{array}$
