a) Ta có:
\(a-1⋮2a+1\)
\(\Rightarrow2\left(a-1\right)⋮2a+1\)
\(\Rightarrow2a-2⋮2a+1\)
\(\Rightarrow\left(2a+1\right)-3⋮2a+1\)
\(\Rightarrow-3⋮2a+1\)
\(\Rightarrow2a+1\in\left\{-1;1;-3;3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}2a+1=-1\Rightarrow a=-1\\2a+1=1\Rightarrow a=0\\2a+1=-3\Rightarrow a=-2\\2a+1=3\Rightarrow a=1\end{matrix}\right.\)
Vậy \(a\in\left\{-1;0;-2;1\right\}\)
b) Ta có:
\(2a-1⋮3a+2\)
\(\Rightarrow3\left(2a-1\right)⋮3a+2\)
\(\Rightarrow6a-3⋮3a+2\)
\(\Rightarrow\left(6a+4\right)-7⋮3a+2\)
\(\Rightarrow2\left(3a+2\right)-7⋮3a+2\)
\(\Rightarrow-7⋮3a+2\)
\(\Rightarrow3a+2\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}3a+2=-1\Rightarrow a=-1\left(thoa\right)\\3a+2=1\Rightarrow a=\dfrac{-1}{3}\left(loai\right)\\3a+2=-7\Rightarrow a=-3\left(thoa\right)\\3a+2=7\Rightarrow a=\dfrac{5}{3}\left(loai\right)\end{matrix}\right.\)
Vì \(a\in Z\) nên \(a\in\left\{-1;-3\right\}\)
