\(y=\dfrac{2x+3}{x+2}\)
\(Txd:\left\{{}\begin{matrix}x+2\ne0\\-1\le x\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\-1\le x\le1\end{matrix}\right.\) \(\Leftrightarrow-1\le x\le1\)
\(\Rightarrow D=\left[-1;1\right]\)
\(y'=\dfrac{2\left(x+2\right)-\left(2x+3\right)}{\left(x+2\right)^2}=\dfrac{2x+4-2x-3}{\left(x+2\right)^2}=\dfrac{1}{\left(x+2\right)^2}>0,\forall x\in D\)
Nên hàm số cho đồng biến trên \(D=\left[-1;1\right]\)
\(\Rightarrow GTLN\left(y\right)=y\left(1\right)=\dfrac{2.1+3}{1+2}=\dfrac{5}{3}\)
\(GTNN\left(y\right)=y\left(-1\right)=\dfrac{2.\left(-1\right)+3}{-1+2}=1\)
Vậy \(y\left(max\right)=\dfrac{5}{3}\left(x=1\right);y\left(min\right)=1\left(x=-1\right)\)
