a)A=\(\dfrac{4x-7}{x-2}=\dfrac{4x-2-5}{x-2}\)=\(\dfrac{4x-2}{x-2}-\dfrac{5}{x-2}=4-\dfrac{5}{x-2}\)
Do \(4\in Z\Rightarrow\dfrac{5}{x-2}\in Z\)
mà \(5\in Z\Rightarrow\left(x-2\right)\in Z\Rightarrow5⋮\left(x-2\right)\)
\(\Rightarrow\left(x-2\right)\inƯ_{\left(5\right)}\Rightarrow\left(x-2\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng:
| x-2 | 1 | -1 | 5 | -5 |
| x | 3 | 1 | 7 | -3 |
Vậy \(x=\left\{3;-1;7;-3\right\}\)để \(A\in Z\)
B=
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