\(x^2y-5x^2-3y+15=7\)
\(\Leftrightarrow x^2\left(y-5\right)-3\left(y-5\right)=7\)
\(\Leftrightarrow\left(x^2-3\right)\left(y-5\right)=7=\left(-1\right)\left(-7\right)=1.7=7.1\)
TH1: \(\left\{{}\begin{matrix}x^2-3=-1\\y-5=-7\end{matrix}\right.\) \(\Rightarrow x^2=2\) (ko có x nguyên thỏa mãn)
TH2: \(\left\{{}\begin{matrix}x^2-3=1\\y-5=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2=4\\y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=12\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x^2-3=7\\y-5=1\end{matrix}\right.\) \(\Rightarrow x^2=10\) (ko có x nguyên thỏa mãn)
Trường hợp \(x^2-3=-7\) khỏi xét vì \(x^2-3\ge-3\forall x\)
