\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\left(1\right)\\x^2+y^2+x+y-4=0\left(2\right)\end{cases}}\)
pt(1)\(\Leftrightarrow2x^2+2xy-4x-x-y+2-xy-y^2+2y=0\)
\(\Leftrightarrow2x\left(x+y-2\right)-\left(x+y-2\right)-y\left(x+y-2\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left(2x-y-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y-2=0\\2x-y-1=0\end{cases}}\)
Thế xuống pt 2 rồi giải