\(\left\{{}\begin{matrix}x^2+y^2+x+y=4\\3xy-x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2+\left(x+y\right)-2xy=4\\3xy-\left(x+y\right)=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)
Hpt \(\Leftrightarrow\left\{{}\begin{matrix}a^2+a-2b=4\\3b-a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+a-2b=4\\a=3b-1\end{matrix}\right.\)
\(\Rightarrow\left(3b-1\right)^2+\left(3b-1\right)-2b=4\)
\(\Leftrightarrow9b^2-6b+1+3b-1-2b-4=0\)
\(\Leftrightarrow9b^2-5b-4=0\)
\(\Leftrightarrow9b^2-9b+4b-4=0\)
\(\Leftrightarrow\left(b-1\right)\left(9b+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=1\Rightarrow a=2\\b=\frac{-4}{9}\Rightarrow a=\frac{-7}{3}\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\Leftrightarrow x=y=1\)
+) \(\left\{{}\begin{matrix}a=\frac{-7}{3}\\b=\frac{-4}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=\frac{-7}{3}\\xy=\frac{-4}{9}\end{matrix}\right.\)
\(\Rightarrow y\left(\frac{-7}{3}-y\right)=\frac{-4}{9}\)
\(\Leftrightarrow y=\frac{-7\pm\sqrt{65}}{6}\)
\(\Leftrightarrow x=\frac{-7\pm\sqrt{65}}{6}\)
Vậy \(\left[{}\begin{matrix}x=y=1\\x=y=\frac{-7\pm\sqrt{65}}{6}\end{matrix}\right.\)
