\(A=\dfrac{-6}{2x-3}\)
Để \(A\in Z\) thì \(-6⋮\left(2x-3\right)\)
=> \(\left(2x-3\right)\in U\left(-6\right)=\left\{-1,-2,-3,-6,1,2,3,6\right\}\)
=> \(2x\in\left\{2,1,0,-3,4,5,6,9\right\}\)
=> \(x\in\left\{1,\dfrac{1}{2},0,\dfrac{-3}{2},2,\dfrac{5}{2},3,\dfrac{9}{2}\right\}\)
Mà \(x\in Z\Rightarrow x\in\left\{1,0,2,3\right\}\)
Vậy \(x\in\left\{1,0,2,3\right\}\) thì A thuộc Z
\(A=\dfrac{-6}{2x-3}\inℤ\left(x\ne\dfrac{3}{2}\right)\)
\(\Rightarrow2x-3\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
\(\Rightarrow x\in\left\{1;2;\dfrac{1}{2};\dfrac{5}{2};0;3;-\dfrac{3}{2};\dfrac{9}{2}\right\}\)
\(\Rightarrow x\in\left\{1;2;0;3\right\}\left(x\inℤ\right)\)
