\(\left\{{}\begin{matrix}\left(xy+x\right)\left(x^2+xy+x\right)=\left(x-1\right)\left(3x-1\right)\\xy+x=x^2-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-1\right)\left(2x^2-1\right)=\left(x-1\right)\left(3x-1\right)\\xy+x=x^2-1\end{matrix}\right.\)(1)
Nếu x=1 thì thay vào hệ dưới, tìm được y=-1
Nếu x\(\ne\)1 thì hệ (1) trở thành:\(\left\{{}\begin{matrix}\left(2x^2-1\right)\left(x+1\right)=3x-1\\xy+x=x^2-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x^3+2x^2-x-1=3x-1\\xy+x=x^2-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x\left(x^2+x-2\right)=0\\xy+x=x^2-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x\left(x-1\right)\left(x+2\right)=0\\xy+x=x^2-1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\xy+x=x^2-1\end{matrix}\right.\)( vì x\(\ne\)1) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\xy+x=x^2-1\end{matrix}\right.\\\left[{}\begin{matrix}x=-2\\xy+x=x^2-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-\dfrac{5}{2}\end{matrix}\right.\)( vì với x=0 thì ko tìm đc y)
Vậy, hệ pt đã cho có các cặp nghiệm (x;y) là:(1;-1);(-2;\(-\dfrac{2}{5}\))
