Là \(2xy^2+x+x+y+1=x^2+2y^2+xy\) nhé :v
\(2xy^2+x+x+y+1=x^2+2y^2+xy\)
\(\Leftrightarrow2y^2x-2y^2-x^2+x-xy+y=-1\)
\(\Leftrightarrow2y^2.\left(x-1\right)-x.\left(x-1\right)-y.\left(x-1\right)=-1\)
\(\Leftrightarrow\left(x-1\right).\left(2y^2-x-y\right)=-1\)
. Xét TH1:
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\2y^2-x-y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(y-1\right).\left(2y+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\) \(\left(y\in Z\right)\)
. Xét TH2:
\(\Leftrightarrow\left[{}\begin{matrix}x-1=-1\\2y^2-x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(\left(2;1\right);\left(0;1\right).\)
