\(a,\) Vì \(2x⋮x\Rightarrow3⋮x\Rightarrow x\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(b,\left(8x+4\right)⋮\left(2x-1\right)\\ \Rightarrow\left[\left(8x-4\right)+8\right]⋮\left(2x-1\right)\\ \Rightarrow\left[4\left(2x-1\right)+8\right]⋮\left(2x-1\right)\)
\(Vì.4\left(2x-1\right)⋮\left(2x-1\right)\Rightarrow8⋮\left(2x-1\right)\Rightarrow\left(2x-1\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
Ta có bảng:
| 2x-1 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| x | -3,5(loại) | -1,5(loại) | -0,5(loại) | 0 | 1 | 1,5(loại) | 2,5(loại) | 4,5(loại) |
Vậy \(x\in\left\{0;1\right\}\)
\(c,\left(x^2-x+7\right)⋮\left(x-1\right)\\ \Rightarrow\left[x\left(x-1\right)+7\right]⋮\left(x-1\right)\)
\(Vì.x\left(x-1\right)⋮\left(x-1\right)\Rightarrow7⋮\left(x-1\right)\Rightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Ta có bảng:
| x-1 | -7 | -1 | 1 | 7 |
| x | -6 | 0 | 2 | 8 |
Vậy \(x\in\left\{-6;0;2;8\right\}\)
