\(HPT\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+1\right)+y\left(x+y-4\right)=0\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+1\right)+\left(x^2+1\right)\left(x+y-2\right)\left(x+y-4\right)=0\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-3\right)^2\left(x^2+1\right)=0\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x^2+1=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3-x\\x^2+x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1;y=2\\x=-2;y=5\end{matrix}\right.\)