Ta có: \(\displaystyle {2^{{x^2} - 6x - \frac{5}{2}}} = 16.\sqrt 2 \)\(\displaystyle \Leftrightarrow {2^{{x^2} - 6x - \frac{5}{2}}} = {2^4}{.2^{\frac{1}{2}}}\) \(\displaystyle \Leftrightarrow {2^{{x^2} - 6x - \frac{5}{2}}} = {2^{\frac{9}{2}}}\) \(\displaystyle \Leftrightarrow {x^2} - 6x - \frac{5}{2} = \frac{9}{2}\)
\(\displaystyle \Leftrightarrow {x^2} - 6x - 7 = 0 \Leftrightarrow \left[ \begin{array}{l}x = - 1\\x = 7\end{array} \right.\).
Vậy phương trình có tập nghiệm \(\displaystyle \left\{ { - 1;7} \right\}\).
Chọn B.
