Đặt \(\left\{{}\begin{matrix}x-y=a\\xy=b\end{matrix}\right.\) : Hệ trở thành ;
\(\left\{{}\begin{matrix}a-b=7\\ab=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+7\\b^2+7b+12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+7\\\left(b+3\right)\left(b+4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=-3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=-4\end{matrix}\right.\end{matrix}\right.\)
Với \(a=4;b=-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\xy=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y^2+4y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+4\\\left(y+1\right)\left(y+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Với \(a=3;b=-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=3\\xy=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\y^2+3y+4=0\end{matrix}\right.\) ( Vô nghiệm )
Vậy \(\left(x;y\right)=\left(3;-1\right)\) \(\left(x;y\right)=\left(1;-3\right)\)
