Trừ vế cho vế:
\(\Rightarrow x^2+2y^2+3xy-x-3y-2=0\)
\(\Leftrightarrow x^2+\left(3y-1\right)x+2y^2-3y-2=0\)
Coi đây là pt bậc 2 ẩn x tham số y
\(\Delta=\left(3y-1\right)^2-4\left(2y^2-3y-2\right)=\left(y+3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3y+1-y-3}{2}=-2y-1\\x=\dfrac{-3y+1+y+3}{2}=-y+2\end{matrix}\right.\)
Thế vào pt đầu:
\(\Rightarrow\left[{}\begin{matrix}2\left(-2y-1\right)^2+y^2+5y\left(-2y-1\right)-y+2=0\\2\left(-y+2\right)^2+y^2+5y\left(-y+2\right)-y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-y^2+2y+4=0\\-2y^2+y+10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=1-\sqrt{5}\Rightarrow x=-3+2\sqrt{5}\\y=1+\sqrt{5}\Rightarrow x=-3-2\sqrt{5}\\y=-2\Rightarrow x=4\\y=\dfrac{5}{2}\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
