Question

Giải pt nghiệm nguyên sau

\(x^2y^2+xy+1=x^2\)

Answer 1

\(x^2y^2+xy+1=x^2\)

\(\Leftrightarrow4x^2y^2+4xy+4=4x^2\)

\(\Leftrightarrow\left(2xy+1\right)^2+3=4x^2\)

\(\Leftrightarrow\left(2x-2xy-1\right)\left(2x+2xy+1\right)=3=1.3=\left(-1\right).\left(-3\right)\)

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

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

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

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