\(A=\dfrac{2x-1}{x+2}\)
\(\Rightarrow A=\dfrac{2x+4-5}{x+2}\)
\(\Rightarrow A=\dfrac{2\left(x+2\right)-5}{x+2}\)
\(\Rightarrow A=\dfrac{2\left(x+2\right)}{x+2}-\dfrac{5}{x+2}\)
\(\Rightarrow A=2-\dfrac{5}{x+2}\)
\( Để.A\in Z.mà.2\in Z\Rightarrow\dfrac{5}{x+2}\in Z\Rightarrow5⋮\left(x+2\right)\Rightarrow x+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng:
| x+2 | -1 | -5 | 1 | 5 |
| x | -3 | -7 | -1 | 3 |
Vậy \(x\in\left\{-3;-7;-1;3\right\}\)
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