\(\Rightarrow a\left(b-1\right)=3\cdot1=\left(-3\right)\left(-1\right)\\ \left\{{}\begin{matrix}a=1\\b-1=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\end{matrix}\right.\\ \left\{{}\begin{matrix}a=3\\b-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\\ \left\{{}\begin{matrix}a=-3\\b-1=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-3\\b=0\end{matrix}\right.\\ \left\{{}\begin{matrix}a=-1\\b-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-1\\b=-2\end{matrix}\right.\)
Vậy PT có nghiệm \(\left(a;b\right)\) là \(\left(1;4\right);\left(3;2\right);\left(-3;0\right);\left(-1;-2\right)\)
\(a.\left(b-1\right)=3\), \(Ta\) \(lập\) \(bảng\):
| \(a\) | \(1\) | \(-1\) | \(3\) | \(-3\) |
| \(b-1\) | \(3\) | \(-3\) | \(1\) | \(-1\) |
| \(b\) | \(4\) | \(-2\) | \(2\) | \(0\) |
\(Vậy\) \(\left\{a;b\right\}\in\left\{1;4\right\};\left\{-1;-2\right\};\left\{3;2\right\};\left\{-3;0\right\}\)
