HPT<=> \(\left\{{}\begin{matrix}x+y=\sqrt{x+3y}\\x^2+y^2+xy=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2+2xy+y^2=x+3y\\x^2+y^2+xy=3\end{matrix}\right.\)
<=> \(xy+3=x+3y\)
<=> \(x\left(1-y\right)-3\left(1-y\right)=0\)
<=> \(\left(x-3\right)\left(y-1\right)=0\)
<=> \(\left[{}\begin{matrix}x=3\\y=1\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y^2+3y+6=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=1\\x^2+x-2=0\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=1\\\left(x+1\right)\left(x-2\right)=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=1\\x=1;-2\end{matrix}\right.\) (TM)
Vậy cặp ( x;y) cần tìm là ( 1;1) , ( -2;1)
