\(\left(x+1\right)\left(xy-1\right)^2=3=1.3=3.1\)
có \(\left(xy-1\right)^2\ge0\)nên \(\left(xy-1\right)^2=1\Rightarrow x+1=3\Leftrightarrow x=2\)
\(\left(xy-1\right)^2=1\Leftrightarrow\orbr{\begin{cases}2y-1=1\\2y-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}y=1\\y=0\end{cases}}}\)
Vậy có các nghiệm \(\left(x,y\right)=\left\{\left(2,1\right),\left(2,0\right)\right\}\)