Đáp án:
421.
$\begin{array}{l}
{x^2} - 2x + 8 = 4\sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} \\
Đkxđ: - 2 \le x \le 4\\
Pt \Rightarrow \left( {x - 4} \right)\left( {x + 2} \right) = 4\sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} \\
\Rightarrow 4\sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} + \left( {4 - x} \right)\left( {x + 2} \right) = 0\\
\Rightarrow \sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} .\left( {4 + \sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} } \right) = 0\\
\Rightarrow \sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} = 0\\
\Rightarrow \left[ \begin{array}{l}
4 - x = 0\\
x + 2 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tmdk} \right)\\
x = - 2\left( {tmdk} \right)
\end{array} \right.\\
\Rightarrow D
\end{array}$
