Đáp án: B
Giải thích các bước giải:
$\begin{array}{l}
y = \frac{{{2^{x + 1}} + 1}}{{{2^x} - m}} = \frac{{{{2.2}^x} + 1}}{{{2^x} - m}}\\
+ )x \in \left( { - 1;1} \right)\\
\Rightarrow {2^x} \in (\frac{1}{2};2)\,\left( {do\,{2^x}\,đồng\,biến} \right)\\
Đặt\,{2^x} = t\,\left( {\frac{1}{2} < t < 2} \right)\\
y = \frac{{2t + 1}}{{t - m}}\\
\Rightarrow y' = \frac{{ - 2m - 1}}{{{{\left( {t - m} \right)}^2}}} < 0\forall \frac{1}{2} < t < 2\\
\Rightarrow \left\{ \begin{array}{l}
- 2m - 1 < 0\\
\left[ \begin{array}{l}
m \ge 2\\
m \le \frac{1}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
m > \frac{{ - 1}}{2}\\
\left[ \begin{array}{l}
m \ge 2\\
m \le \frac{1}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
m \ge 2\\
\frac{{ - 1}}{2} < m \le \frac{1}{2}
\end{array} \right.\\
\Rightarrow m \in {\rm{\{ 0;2;3;4}}...{\rm{;49\} }}
\end{array}$
