Ta có: \(\left\{{}\begin{matrix}3x-\dfrac{2}{y+1}=\dfrac{-1}{2}\\2x+\dfrac{1}{y+1}=2\end{matrix}\right.\left(ĐKXĐ:y\ne-1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-\dfrac{4}{y+1}=-1\\6x+\dfrac{3}{y+1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7}{y+1}=-7\\3x-\dfrac{2}{y+1}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=1\\3x-2=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\3x=-\dfrac{1}{2}+2=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=0\left(nhận\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(\dfrac{1}{2};0\right)\)