\(\Leftrightarrow x^2y^2\left(x+y\right)+x+y=xy+2\)
\(\Leftrightarrow\left(x+y\right)\left(x^2y^2+1\right)=xy+2\)
\(\Rightarrow xy+2⋮x^2y^2+1\)
\(\Rightarrow\left(xy-2\right)\left(xy+2\right)⋮x^2y^2+1\)
\(\Rightarrow x^2y^2-4⋮x^2y^2+1\)
\(\Rightarrow5⋮x^2y^2+1\)
\(\Rightarrow\left[{}\begin{matrix}x^2y^2=4\\x^2y^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}xy=2\\xy=-2\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\end{matrix}\right.\)
Xét \(xy=2\)\(\Rightarrow\)\(5\left(x+y\right)=6\)(pt vô nghiệm nguyên)
Xét xy=-2\(\Rightarrow5\left(x+y\right)=0\)
\(\Rightarrow x=-y\)
\(\Rightarrow y^2=2\)(pt vô nghiệm nguyên)
Xét x=0\(\Rightarrow y=2\)
Xét y=0\(\Rightarrow x=2\)
Thử lại ta thấy cặp số (x;y)=(0;2);(2;0) thỏa mãn
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