Trừ vế cho vế:
\(\Rightarrow x^2+2y^2+3xy-x-3y=2\)
\(\Leftrightarrow\left(x^2+xy-2x\right)+\left(2xy+2y^2-4y\right)+x+y-2=0\)
\(\Leftrightarrow x\left(x+y-2\right)+2y\left(x+y-2\right)+x+y-2=0\)
\(\Leftrightarrow\left(x+y-2\right)\left(x+2y+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y-2=0\\x+2y+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-y+2\\x=-2y-1\end{matrix}\right.\)
- Với \(x=-y+2\) thế vào pt đầu:
\(2\left(-y+2\right)^2+y^2+5y\left(-y+2\right)-y+2=0\)
\(\Leftrightarrow-2y^2+y+10=0\)
\(\Rightarrow\left[{}\begin{matrix}y=-2\Rightarrow x=4\\y=\dfrac{5}{2}\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
- Với \(x=-2y-1\) thế vào pt đầu:
\(2\left(-2y-1\right)^2+y^2+5y\left(-2y-1\right)-y+2=0\)
\(\Leftrightarrow-y^2+2y+4=0\)
\(\Rightarrow\left[{}\begin{matrix}y=1-\sqrt{5}\Rightarrow x=-3+2\sqrt{5}\\y=1+\sqrt{5}\Rightarrow x=-3-2\sqrt{5}\end{matrix}\right.\)
