\(\left\{{}\begin{matrix}x^2+y^2+xy=2\\x^3+y^3=2x+4y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2+xy\right)\left(x-y\right)=2x-2y\\x^3+y^3=2x+4y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=2x-2y\\x^3+y^3=2x+4y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y^3=6y\\x^2+xy+y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y\left(y^2-3\right)=0\\x^2+y^2+xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=0\\x^2+xy+y^2=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\sqrt{3}\\x^2+xy+y^2=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=\sqrt{3}\\x^2+xy+y^2=2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=0\\x^2=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\sqrt{3}\\x^2-\sqrt{3}x+3=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=\sqrt{3}\\x^2+\sqrt{3}x+3=2\end{matrix}\right.\end{matrix}\right.\)
* TH1: \(\left\{{}\begin{matrix}y=0\\x^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}\\y=0\end{matrix}\right.\)
* TH2: \(\left\{{}\begin{matrix}y=-\sqrt{3}\\x^2-\sqrt{3}x+3=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\sqrt{3}\\x^2-\sqrt{3}x+1=0\left(1\right)\end{matrix}\right.\)
Giải (1):\(\Delta=3-4\cdot1=-1< 0\) => phương trình vô nghiệm
* TH3: \(\left\{{}\begin{matrix}y=\sqrt{3}\\x^2+\sqrt{3}x+3=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}\\x^2+\sqrt{3}x+1=0\left(2\right)\end{matrix}\right.\)
Giải (2): \(\Delta=3-4\cdot1=-1< 0\) => phương trình vô nghiệm
Vậy hệ phương trình có 2 cặp nghiệm \(\left(x,y\right)=\left(-\sqrt{2};0\right);\left(\sqrt{2};0\right)\)
