5xy - 5x + y = 5
<=> 5xy = 5 + 5x - y
<=> \(\left\{{}\begin{matrix}x=\dfrac{5+5x-y}{5y}\\y=\dfrac{5+5x-y}{5x}\end{matrix}\right.\)
\(5xy-5x+y=5\)
\(\Rightarrow5x\left(y-1\right)+\left(y-1\right)=4\)
\(\Rightarrow\left(y-1\right)\left(5x+1\right)=4\)
Do \(x,y\in Z\)
TH1: \(\left\{{}\begin{matrix}y-1=1\\5x+1=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=2\left(tm\right)\\x=-\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}y-1=4\\5x+1=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=5\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}y-1=2\\5x+1=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}y-1=-2\\5x+1=-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\left(tm\right)\\x=-\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\)
TH5: \(\left\{{}\begin{matrix}y-1=-1\\5x+1=-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
TH6: \(\left\{{}\begin{matrix}y-1=-4\\5x+1=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\left(tm\right)\\x=-\dfrac{2}{5}\left(ktm\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(0;5\right);\left(-1;0\right)\right\}\)
