\(\hept{\begin{cases}x^2+y^2+xy=3\\xy+3x^2=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4\left(x^2+y^2+xy\right)=3\left(3x^2+xy\right)\text{ }\left(\text{1}\right)\\3x^2+xy=4\end{cases}}\)
\(\left(1\right)\Leftrightarrow5x^2-xy-4y^2=0\Leftrightarrow\left(x-y\right)\left(5x+4y\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-y=0\\5x+4y=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=x\\y=-\frac{5}{4}x\end{cases}}\)
\(\text{TH1:}y=x\), ta được hệ \(\hept{\begin{cases}x=y\\3x^2+xy=4\end{cases}}\)
TH2: \(y=-\frac{5}{4}x\), ta có hệ \(\hept{\begin{cases}y=-\frac{5}{4}x\\3x^2+xy=4\end{cases}}\)
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