1. Là các PT bậc 2 hai ẩn x,y
2. Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}a^2-b=4\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+a-6=0\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=2\\a=-3\end{matrix}\right.\\a+b=2\end{matrix}\right.\)
Với \(a=2\Leftrightarrow b=0\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=2\\y=0\Rightarrow x=2\end{matrix}\right.\)
Với \(a=-3\Leftrightarrow b=5\Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\xy=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y-3\\-y\left(y+3\right)=5\end{matrix}\right.\)
\(\Leftrightarrow y^2+3y+5=0\left(\text{vô nghiệm}\right)\\ \Leftrightarrow x,y\in\varnothing\)
Vậy hệ có nghiệm \(\left(x;y\right)\) là \(\left(0;2\right);\left(2;0\right)\)
