\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3+2}{3-2}+m\ge0\\\dfrac{4+2}{4-2}+m\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5+m\ge0\\3+m\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5\ge-m\\3\ge-m\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5.\left(-1\right)\le-m.\left(-1\right)\\3.\left(-1\right)\le-m.\left(-1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m\ge-5\\m\ge-3\end{matrix}\right.\)
\(\Leftrightarrow m\ge\dfrac{x+2}{-x+2}\)
\(\Leftrightarrow m\ge\max\limits_{\left[3;4\right]}\dfrac{x+2}{-x+2}\)
Xét hàm \(f\left(x\right)=\dfrac{x+2}{-x+2}\) trên \(\left[3;4\right]\)
\(f'\left(x\right)=\dfrac{4}{\left(x+2\right)^2}< 0\Rightarrow f\left(x\right)\) đồng biến trên mỗi khoảng xác định
\(\Rightarrow\max\limits_{\left[3;4\right]}f\left(x\right)=f\left(4\right)=3\)
\(\Rightarrow m\ge3\)
