\(\hept{\begin{cases}x^2+xy+x-y-2y^2=0\\x^2-y^2+x+y=6\end{cases}}\)
\(hpt\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x+2y+1\right)=0\left(1\right)\\\left(x+y\right)\left(x-y+1\right)=6\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x-y=0\\x+2y+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=y\\x=-1-2y\end{cases}}\)
Xét \(x=y\) thay vào \(\left(2\right)\) ta có:\(\left(2\right)\Leftrightarrow2y=6\Leftrightarrow y=3\Leftrightarrow x=y=3\)
Xét \(x=-1-2y\) thay vào \(\left(2\right)\) ta có:\(\left(2\right)\Leftrightarrow3y^2+3y=6\Leftrightarrow3y^2+3y-6=0\)
\(\Leftrightarrow3\left(y-1\right)\left(y+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}y=1\Rightarrow x=-1-2y=-3\\y=-2\Rightarrow x=-1-2y=3\end{cases}}\)
Vậy hpt có nghiệm là \(\left(x;y\right)=\left(3;3\right);\left(-3;1\right);\left(3;-2\right)\)
