Câu hỏi của Nguyễn Sự - Toán lớp 6 | Học trực tuyến
\(xy-3x-y+3=23\)
\(\Rightarrow x\left(y-3\right)-1\left(y-3\right)=23\)
\(\Rightarrow\left(x-1\right)\left(y-3\right)=23\)
\(\Rightarrow x-1;y-3\in U\left(23\right)\)
\(U\left(23\right)=\left\{\pm1;\pm23\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=1\Rightarrow x=2\\y-3=23\Rightarrow y=26\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-1\Rightarrow x=0\\y-3=-23\Rightarrow y=-20\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=23\Rightarrow x=24\\y-3=1\Rightarrow y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-23\Rightarrow x=-22\\y-3=-1\Rightarrow y=2\end{matrix}\right.\end{matrix}\right.\)
