Đáp án: A
Giải thích các bước giải:
$\begin{array}{l}
{2^{a.{x^2} - 4x - 2a}} = \dfrac{1}{{{{\left( {\sqrt 2 } \right)}^{ - 4}}}} = {\left( {\sqrt 2 } \right)^4} = {2^2}\\
\Rightarrow a.{x^2} - 4x - 2a = 2\\
\Rightarrow a.{x^2} - 4x - 2a - 2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
a \ne 0\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a \ne 0\\
{2^2} - a.\left( { - 2a - 2} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a \ne 0\\
4 + 2{a^2} + 2a > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a \ne 0\\
{a^2} + a + 2 > 0\left( {tm} \right)
\end{array} \right.\\
\Rightarrow a \ne 0
\end{array}$
