Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}\,\,\,\,\,\,1 + {\log _2}\left( {2 - x} \right) - 2{\log _2}\left( {m - \dfrac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)} \right) \le - {\log _2}\left( {x + 1} \right)\,\,\left( { - 1 < x < 2} \right)\\ \Leftrightarrow 1 + {\log _2}\left( {x + 1} \right) + {\log _2}\left( {2 - x} \right) - 2{\log _2}\left( {m - \dfrac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)} \right) \le 0\\ \Leftrightarrow {\log _2}\left( {2x + 2} \right) + {\log _2}\left( {2 - x} \right) - 2{\log _2}\left( {m - \dfrac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)} \right) \le 0\\ \Leftrightarrow 2{\log _2}\sqrt {2x + 2} + 2{\log _2}\sqrt {2 - x} - 2{\log _2}\left( {m - \frac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)} \right) \le 0\\ \Leftrightarrow {\log _2}\sqrt {2x + 2} + {\log _2}\sqrt {2 - x} - {\log _2}\left( {m - \dfrac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)} \right) \le 0\\ \Leftrightarrow {\log _2}\dfrac{{\sqrt {2x + 2} \sqrt {2 - x} }}{{m - \frac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)}} \le 0\\ \Leftrightarrow \dfrac{{\sqrt {2x + 2} \sqrt {2 - x} }}{{m - \dfrac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)}} \le 1\\ \Leftrightarrow \sqrt {2x + 2} \sqrt {2 - x} \le m - \dfrac{x}{2} + 4\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)\\ \Leftrightarrow 2\sqrt {2x + 2} \sqrt {2 - x} \le 2m - x + 8\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right)\\ \Leftrightarrow 2\sqrt {2x + 2} \sqrt {2 - x} + x - 8\left( {\sqrt {2 - x} + \sqrt {2x + 2} } \right) \le 2m\end{array}\)
Đặt \(t = \sqrt {2 - x} + \sqrt {2x + 2} \) ta có:
\(\begin{array}{l}t'\left( x \right) = \dfrac{{ - 1}}{{2\sqrt {2 - x} }} + \dfrac{2}{{2\sqrt {2x + 2} }} = 0\\ \Leftrightarrow \dfrac{2}{{2\sqrt {2x + 2} }} = \dfrac{1}{{2\sqrt {2 - x} }}\\ \Leftrightarrow 2\sqrt {2 - x} = \sqrt {2x + 2} \\ \Leftrightarrow 4\left( {2 - x} \right) = 2x + 2\\ \Leftrightarrow 8 - 4x = 2x + 2\\ \Leftrightarrow x = 1\,\,\left( {tm} \right)\end{array}\)
\(t\left( { - 1} \right) = \sqrt 3 ,\,\,\,\,t\left( 2 \right) = \sqrt 6 ,\,\,\,t\left( 1 \right) = 3\).
Do đó \(t \in \left( {\sqrt 3 ;3} \right)\).
Ta có:
\(\begin{array}{l}{t^2} = 2 - x + 2x + 2 + 2\sqrt {2 - x} \sqrt {2x + 2} \\{t^2} = 4 + x + 2\sqrt {2 - x} \sqrt {2x + 2} \\ \Leftrightarrow x + 2\sqrt {2 - x} \sqrt {2x + 2} = {t^2} - 4\end{array}\)
Bất phương trình trở thành \(g\left( t \right) = {t^2} - 4 - 8t \le 2m\) (*) với \(t \in \left( {\sqrt 3 ;3} \right)\).
Để bất phương trình ban đầu có nghiệm thì (*) có nghiệm \(t \in \left( {\sqrt 3 ;3} \right)\)\( \Rightarrow 2m \ge \mathop {\min }\limits_{\left[ { \sqrt 3 ;3} \right]} g\left( t \right)\).
Ta có: \(g'\left( t \right) = 2t - 8 = 0 \Leftrightarrow t = 4\,\,\left( {ktm} \right)\).
\(g\left( {\sqrt 3 } \right) = - 1 - 8\sqrt 3 ,\,\,g\left( 3 \right) = - 19\).
\( \Rightarrow \mathop {\min }\limits_{\left[ { \sqrt 3 ;3} \right]} g\left( t \right) = - 19 \Leftrightarrow 2m \ge - 19 \Leftrightarrow m \ge - 9,5\).
Vậy \({m_0} = - 9,5 \in \left( { - 10; - 9} \right)\).
Chọn D.
