Đáp án: A
Giải thích các bước giải:
$\begin{array}{l}
\log _2^2\left( {4x} \right) - m.{\log _{\sqrt 2 }}.x - 2m - 4 = 0\\
\Rightarrow {\left( {{{\log }_2}4 + {{\log }_2}x} \right)^2} - 2m{\log _2}x - 2m - 4 = 0\\
\Rightarrow {\left( {{{\log }_2}x + 2} \right)^2} - 2m.{\log _2}x - 2m - 4 = 0\\
\text{Đặt}:{\log _2}x = t\\
Khi:x \in \left[ {1;8} \right] \Rightarrow t \in \left[ {0;3} \right]\\
\Rightarrow {\left( {t + 2} \right)^2} - 2m.t - 2m - 4 = 0\\
\Rightarrow {t^2} + 4t + 2 - 2mt - 4 = 0\\
\Rightarrow {t^2} + 4t - 2 = 2mt\\
+ Khi:t = 0 \Rightarrow - 2 = 0\left( {vo\,nghiem} \right)\\
+ Khi:t \ne 0\\
\Rightarrow m = \dfrac{{{t^2} + 4t - 2}}{{2t}}\\
\text{Đặt}:f\left( t \right) = \dfrac{{{t^2} + 4t - 2}}{{2t}}\left( {t \in \left( {0;3} \right]} \right)\\
\Rightarrow f'\left( t \right) = \dfrac{{\left( {2t + 4} \right).2t - 2\left( {{t^2} + 4t - 2} \right)}}{{4{t^2}}}\\
= \dfrac{{4{t^2} + 8t - 2{t^2} - 8t + 4}}{{4{t^2}}}\\
= \dfrac{{2{t^2} + 4}}{{4{t^2}}} > 0\\
\Rightarrow Khi:t \in \left( {0;3} \right]\\
\Rightarrow 0 < f\left( t \right) \le f\left( 3 \right)\\
\Rightarrow 0 < f\left( t \right) \le \dfrac{{19}}{6}\\
\Rightarrow 0 < m \le \dfrac{{19}}{6}\\
\Rightarrow \left[ \begin{array}{l}
m = 1\\
m = 2\\
m = 3
\end{array} \right.
\end{array}$
