Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}x^2+xy-4x=-6\\y^2+xy=-1\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy-4x+6=0\\3y^2+3xy+3=0\end{matrix}\right.\)

\(\Rightarrow x^2-3y^2-2xy-4x+3=0\)

\(\Leftrightarrow x^2-2\left(y+2\right)x-3y^2+3=0\)

\(\Delta'=\left(y+2\right)^2+3y^2-3=4y^2+4y+1=\left(2y+1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=y+2+2y+1=3y+3\\x=y+2-2y-1=-y+1\end{matrix}\right.\)

Thế vào pt dưới:

\(\left[{}\begin{matrix}y^2+\left(3y+3\right)y=-1\\y^2+\left(-y+1\right)y=-1\end{matrix}\right.\)