\(\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\dfrac{2x_1-1}{x_1+3}-\dfrac{2x_2-1}{x_2+3}}{x_1-x_2}\)
\(=\left(\dfrac{2x_1x_2+6x_1-x_2-3-\left(2x_1x_2+6x_2-x_1-3\right)}{\left(x_1+3\right)\left(x_2+3\right)}\right):\left(x_1-x_2\right)\)
\(=\dfrac{6x_1-6x_2+x_1-x_2}{\left(x_1+3\right)\left(x_2+3\right)}\cdot\dfrac{1}{x_1-x_2}\)
\(=\dfrac{7}{\left(x_1+3\right)\left(x_2+3\right)}\)
Vì \(x_1;x_2\in\left(-\infty;-3\right)\)
nên \(\left\{{}\begin{matrix}x_1< -3\\x_2< -3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1+3< 0\\x_2+3< 0\end{matrix}\right.\Leftrightarrow\left(x_1+3\right)\left(x_2+3\right)>0\)
\(\Leftrightarrow\dfrac{7}{\left(x_1+3\right)\left(x_2+3\right)}>0\)
Vậy: Hàm số đồng biến trên khoảng \(\left(-\infty;-3\right)\)
