Để \(\frac{3}{2a-5}\) thuộc Z
Thì 3 chia hết cho 2a - 5
=> 2a - 5 thuộc Ư(3) = {-3;-1;1;3}
Ta có bảng :
| 2a - 5 | -3 | -1 | 1 | 3 |
| 2a | 2 | 4 | 6 | 8 |
| a | 1 | 2 | 3 | 4 |
Ta có: \(\frac{3}{2a-5}\in Z\)
\(\Rightarrow3⋮2a-5\Rightarrow\left(2a-5\right)\inƯ\left(3\right)\)
\(Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow2a-5\in\left\{-3;-1;1;3\right\}\)
\(\cdot2a-5=-3\Rightarrow a=\left(-3+5\right):2=1\)
\(\cdot2a-5=-1\Rightarrow a=2\)
\(\cdot2a-5=1\Rightarrow a=3\)
\(\cdot2a-5=3\Rightarrow a=4\)
Tất cả đều thỏa mãn a \(\in z\)
\(\Rightarrow a\in\left\{1;2;3;4\right\}\)
Để \(\frac{3}{2a-5}\in Z\)
\(\Rightarrow2a-5\inƯ\left(3\right)\\ \Rightarrow2a-5\in\left\{1;-1;3;-3\right\}\\2a\in\left\{6;4;8;2\right\}\\ a\in\left\{3;2;4;1\right\} \)
Vậy \(a\in\left\{3;2;4;1\right\}\)
