Câu hỏi 23

\(y=\frac{mx+6}{2x+m+1}\Rightarrow y'=\frac{m\left( m+1 \right)-6.2}{{{\left( 2x+m+1 \right)}^{2}}}=\frac{{{m}^{2}}+m-12}{{{\left( 2x+m+1 \right)}^{2}}}\)

Hàm số nghịch biến trên 

\(\left( { - 1;1} \right) \Leftrightarrow \left\{ \begin{array}{l}
y' < 0\,\,\\
\frac{{ - m - 1}}{2} \notin \left( { - 1;1} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{m^2} + m - 12 < 0\,\,\\
\left[ \begin{array}{l}
\frac{{ - m - 1}}{2} \le - 1\\
\frac{{ - m - 1}}{2} \ge 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 4 < m < 3\,\,\\
\left[ \begin{array}{l}
- m + 1 \le 0\\
- m - 3 \ge 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 4 < m < 3\,\,\\
\left[ \begin{array}{l}
m \ge 1\\
m \le - 3
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
- 4 < m \le - 3\\
1 \le m < 3
\end{array} \right.\)

Chọn D.