Ta có:
\(x^2-xy=6x-5y-8\)
\(\Leftrightarrow x\left(x-y\right)-5\left(x-y\right)=x-8\)
\(\Leftrightarrow\left(x-5\right)\left(x-y\right)-\left(x-5\right)=-3\)
\(\Leftrightarrow\left(x-5\right)\left(x-y-1\right)=-3\)
Ta có bảng sau:
| x - 5 | -1 | -3 | 1 | 3 |
| x - y - 1 | 3 | 1 | -3 | -1 |
| x | 4 | 2 | 6 | 8 |
| y | 0 | 0 | 8 | 8 |
Vậy...
\(x^2-xy=6x-5y-8\\ \Leftrightarrow\left(x^2-5x\right)-\left(xy-5y\right)-\left(x-5\right)=-3\\ \Leftrightarrow x\left(x-5\right)-y\left(x-5\right)-\left(x-5\right)=-3\\ \Leftrightarrow\left(x-y-1\right)\left(x-5\right)=-3\\ =\left(-1\right)\cdot3=3\cdot\left(-1\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1\)
Do \(x;y\in Z\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y-1=-1\\x-5=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=3\\x-5=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=1\\x-5=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=-3\\x-5=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}8-y-1=-1\\x=8\end{matrix}\right.\\\left\{{}\begin{matrix}4-y-1=3\\x=4\end{matrix}\right.\\\left\{{}\begin{matrix}2-y-1=1\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}6-y-1=-3\\x=6\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=8\\y=8\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\end{matrix}\right.\)
Vậy pt có tập nghiệm nguyên \(\left\{x;y\right\}=\left\{8;8\right\};\left\{4;0\right\};\left\{2;0\right\};\left\{6;8\right\}\)
