Ta có:
\(x^2+x-xy-2y^2-2xy=0\)
\(\Leftrightarrow x^2-2xy+xy-2y^2+x-2y=0\)
\(\Leftrightarrow x\left(x-2y\right)+y\left(x-2y\right)+\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x+y+1\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+1=0\\x-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=-1\\x=2y\end{matrix}\right.\)
\(\cdot x=2y\) thay vào pt \(x^2+y^2=1\) ta được:
\(4y^2+y^2=1\)
\(\Leftrightarrow5y^2=1\)
\(\Leftrightarrow y^2=\dfrac{1}{5}\)
\(\Leftrightarrow y=\dfrac{\sqrt{5}}{5}\)
\(\Rightarrow x=2.\dfrac{\sqrt{5}}{5}=\dfrac{2\sqrt{5}}{5}\)
\(\cdot x+y=-1\)
\(\Leftrightarrow\left(x+y\right)^2-2xy=1\)
\(\Leftrightarrow1-2xy=1\)
\(\Leftrightarrow xy=0\)
Lại có: \(x+y=-1\Rightarrow x=-1-y\)
\(\Rightarrow\left(-1-y\right).y=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\-1-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1-0\\x=-1+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Vậy nghiệm của hệ pt là:
\(\left(x;y\right)=\left(\dfrac{\sqrt{5}}{5};\dfrac{2\sqrt{5}}{5}\right)\)
\(\left(x;y\right)=\left(0;-1\right)\)
\(\left(x;y\right)=\left(-1;0\right)\)
