Ta có: 2a-3\(⋮\)2a+1
\(\Leftrightarrow2a+1-4⋮2a+1\)
hay -4\(⋮\)2a+1
\(\Rightarrow2a+1\inƯ\left(-4\right)\)
\(\Rightarrow2a+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2a\in\left\{0;-2;1;-3;3;-5\right\}\)
hay \(a\in\left\{0;-1;\frac{1}{2};\frac{-3}{2};\frac{3}{2};\frac{-5}{2}\right\}\)
mà \(a\in Z\)
nên \(a\in\left\{0;-1\right\}\)
Vậy: \(a\in\left\{0;-1\right\}\)
\(\dfrac{2a-3}{2a+1} = \dfrac{2a+1-4}{2a+1}=\dfrac{2a+1}{2a+1}+\dfrac{-4}{2a+1}=1+\dfrac{-4}{2a+1}\) \(\in Z\)
mà \(1 \in Z\)
=> \(\dfrac{4}{2a+1} \in Z\) => \(4 \vdots 2a+1\) <=> \(2a+1 \in \) Ư(-4) = { -4 ; -2 ; -1 ; 1 ; 2 ; 4 }
Ta có bảng sau :
| 2a+1 | -4 | -2 | -1 | 1 | 2 | 4 |
| 2a | -5 | -3 | -2 | 0 | 1 | 3 |
| a | \(\frac{-5}{2}\) | \(\frac{-3}{2}\) | -1 | 0 | \(\frac{1}{2}\) | \(\frac{3}{2}\) |
Vậy \(a \in \) { -1 ; 0 }