Ta có hệ phương trình \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\left(1\right)\\x-3y=3\end{matrix}\right.\)
Ta có x-3y=3\(\Leftrightarrow x=3y+3\)
Thay x=3y+3 vào (1)\(\Leftrightarrow\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-23=0\Leftrightarrow9y^2+18y+9+y^2-6y-6-2y-23=0\Leftrightarrow10y^2+10y-20=0\Leftrightarrow10\left(y^2+y-2\right)=0\Leftrightarrow y^2+y-2=0\Leftrightarrow y^2+2y-y-2=0\Leftrightarrow y\left(y+2\right)-\left(y+2\right)=0\Leftrightarrow\left(y+2\right)\left(y-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y+2=0\\y-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}y=-2\\y=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
Vậy (x;y)={(-3;-2);(6;1)}
