Ta có : \(\left\{{}\begin{matrix}x-xy+3y=3\\x^2+xy+y^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(1-y\right)=0\\x^2+xy+y^2=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\x^2+xy+y^2=7\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x=3\\x^2+xy+y^2=7\end{matrix}\right.\)
\(\Rightarrow9+3y+y^2=7\) \(\Leftrightarrow y^2+3y+2=0\Leftrightarrow\left(y+2\right)\left(y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-1\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}y=1\\x^2+xy+y^2=7\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=7\) \(\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
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