Giải:
\(\dfrac{4x-1}{3-x}\in Z\Rightarrow4x-1⋮3-x\)
\(\Rightarrow4x-1⋮-x+3\)
\(\Rightarrow4x-12+11⋮-x+3\)
\(\Rightarrow-4\left(-x+3\right)+11⋮-x+3\)
\(\Rightarrow11⋮-x+3\)
\(\Rightarrow-x+3\in\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;4;-8;14\right\}\)
Vậy...
Đặt \(A=\dfrac{4x-1}{3-x}\left(x\in Z;x\ne3\right)\)\(A=\dfrac{4x-1}{3-x}=\dfrac{4x-1}{-x+3}=\dfrac{4x-12+11}{-x+3}=\dfrac{-4\left(-x+3\right)+11}{-x+3}=\dfrac{-4\left(-x+3\right)}{-x+3}+\dfrac{11}{-x+3}=-4+\dfrac{11}{-x+3}\)
Để phân số A là số nguyên thì \(11⋮-x+3\).
\(\Rightarrow-x+3\inƯ\left(11\right)\)
Ta lập bảng sau:
| -x+3 | 1 | -1 | 11 | -11 |
| x | 2 | 4 | -8 | 14 |
Vậy \(x\in\left\{-8;2;4;14\right\}\).
Để \(\dfrac{4x-1}{3-x}\) là số nguyên thì \(4x-1⋮\left(3-x\right)\)
\(\Leftrightarrow4x-1+\left(3-x\right)⋮3-x\)
\(\Leftrightarrow4x-1+4\left(3-x\right)⋮3-x\)
\(\Leftrightarrow4x-1+12-4x⋮3-x\)
\(\Rightarrow11⋮3-x\)
\(\Rightarrow3-x\inƯ_{\left(11\right)}=\left\{-11;-1;1;11\right\}\)
Ta có bảng sau
| 3-x | -11 | -1 | 1 | 11 |
| x | 14 | 4 | 2 | -8 |
Vậy \(x\in\left\{14;4;2;-8\right\}\) thoả mãn điều kiện đề bài
