\(\Leftrightarrow a\left(b-1\right)=5.1=\left(-5\right)\left(-1\right)\)
\(TH_1:\left\{{}\begin{matrix}a=5\\b-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\\ TH_2:\left\{{}\begin{matrix}a=1\\b-1=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=6\end{matrix}\right.\\ TH_3:\left\{{}\begin{matrix}a=-1\\b-1=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-4\end{matrix}\right.\\ TH_4:\left\{{}\begin{matrix}a=-5\\b-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-5\\b=0\end{matrix}\right.\)
Vậy \(\left(a;b\right)\in\left\{\left(5;2\right);\left(1;6\right);\left(-1;-4\right);\left(-5;0\right)\right\}\)
