Giải  hệ:
$\left\{ \begin{array}{l}
{2^{\left| {{x^2} - 2x - 3} \right| - {{\log }_2}3}} = {3^{ - y - 4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
4\left| y \right| - \left| {y - 1} \right| + {\left( {y + 3} \right)^2} \le 8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{array} \right.$
$(1)\,\,\, \Leftrightarrow \,\,{2^{\left| {{x^2} - 2x - 3} \right|}} = {3^{ - y - 3}}$
* Với $y \ge 1$ ta có   (2)  $ \Leftrightarrow 4y - y + 1 + {\left( {y + 3} \right)^2} \le 8$
   $ \Leftrightarrow \,\,{y^2} + 9y + 2 \le 0$ vô nghiệm vì $y \ge 1$ thì vế trái dương.
* Với $0 \le y < 1:\,\,\,\,(2)\,\,\, \Leftrightarrow \,\,4y + y - 1 + {\left( {y + 3} \right)^2} \le 8$
   $ \Leftrightarrow {y^2}11y \le 0\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow  - 11 \le y \le 0$
Suy ra     $y = 0\,\,\, \Rightarrow \,\,\,{2^{\left| {{x^2} - 2x - 3} \right|}} = {3^{ - 3}} < 1$
        $ \Rightarrow \,\,\,\,\left| {{x^2} - 2x - 3} \right| < 0 \Rightarrow $vô nghiệm
* Với $y \le 0\,\,\,\,\,\,:\,\,\,\,(2)\,\,\,\, \Leftrightarrow \,\,\, - 4y + y - 1 + {\left( {y + 3} \right)^2} \le 8$
   $ \Leftrightarrow {y^2} + 3y \le 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\, - 3 \le y \le 0\,\,\,\,\, \Leftrightarrow \,\,\, - y - 3 \le 0$
   $ \Leftrightarrow \,\,\,\,{3^{ - y - 3}} < 1$   Mặt khác $\,\,\,\,\left| {{x^2} - 2x - 3} \right| \ge 0 \Leftrightarrow $
   $ \Leftrightarrow \,\,\,{2^{\,\,\,\left| {{x^2} - 2x - 3} \right|}} \ge 1$
Từ đó ta có hệ:   $\left\{ \begin{array}{l}
{3^{ - y - 3}} = 1\\
\,\,\,{2^{\,\,\,\left| {{x^2} - 2x - 3} \right|}} = 1
\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\left\{ \begin{array}{l}
 - y - 3 = 0\\
{x^2} - 2x - 3 = 0
\end{array} \right.$

$ \Leftrightarrow \,\,\left\{ \begin{array}{l}
x =  - 1\\
y =  - 3
\end{array} \right.$   hoặc    $\left\{ \begin{array}{l}
x = 3\\
y =  - 3
\end{array} \right.$
Hệ có $2$ nghiệm là : $\left( { - 1,\, - 3} \right)\,\,;\,\,\left( {3,\, - 3} \right)$

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