Question

Tìm nghiệm nguyên của PT: \(x^2+xy-2y^2-3x-6y+2=0\)

Answer 1

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

TH1: \(\left\{{}\begin{matrix}x+2y=1\\x-y-3=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+2y=-1\\x-y-3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=2\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}x+2y=2\\x-y-3=-1\end{matrix}\right.\)

TH4: \(\left\{{}\begin{matrix}x+2y=-2\\x-y-3=1\end{matrix}\right.\)