Đáp án:
$\begin{array}{l}
4\sqrt {\left( {4 - x} \right)\left( {2 + x} \right)} = {x^2} - 2x - 12\\
Dkxd:\left\{ \begin{array}{l}
\left( {4 - x} \right)\left( {2 + x} \right) \ge 0\\
{x^2} - 2x - 12 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 2 \le x \le 4\\
\left[ \begin{array}{l}
x \ge 1 + \sqrt {13} \\
x \le 1 - \sqrt {13}
\end{array} \right.
\end{array} \right.\\
\Rightarrow 1 + \sqrt {13} \le x \le 4\\
Pt:\\
4\sqrt {\left( {4 - x} \right)\left( {2 + x} \right)} = {x^2} - 2x - 12\\
\Rightarrow 4.\sqrt { - \left( {{x^2} - 2x - 8} \right)} - \left( {{x^2} - 2x - 8} \right) + 4 = 0\\
Dat:\sqrt { - \left( {{x^2} - 2x - 8} \right)} = a\\
\Rightarrow 4.a + {a^2} + 4 = 0\\
\Rightarrow {a^2} + 4a + 4 = 0\\
\Rightarrow {\left( {a + 2} \right)^2} = 0\\
\Rightarrow a = - 2\\
\Rightarrow \sqrt { - \left( {{x^2} - 2x - 8} \right)} = - 2\left( {vo\,nghiem} \right)
\end{array}$
Vậy pt vô nghiệm.
