Ta có : \(y=\dfrac{x+3}{x+1}\\ \Rightarrow y\left(x+1\right)=x+3\\ \Leftrightarrow xy+y-x-3=0\\ \Leftrightarrow\left(xy+y\right)-\left(x+1\right)-2=0\\ \Leftrightarrow y\left(x+1\right)-\left(x+1\right)=2\\ \Leftrightarrow\left(x+1\right)\left(y-1\right)=2.1=1.2\)
\(\left[{}\begin{matrix}\left(x+1\right)\left(y-1\right)=2.1\\\left(x+1\right)\left(y-1\right)=1.2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(x+1\right)=2\\\left(y-1\right)=1\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x+1\right)=1\\\left(y-1\right)=2\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\end{matrix}\right.\left(TM\right)\)
Vậy x=1 và y=2 hoặc x=0 và y=3
