a) \(\left(3x+4\right)⋮\left(x-1\right)\)
vì \(\left(x-1\right)⋮\left(x-1\right)\)
=> \(3\left(x-1\right)⋮\left(x-1\right)\)
=> \(\left(3x-3\right)⋮\left(x-1\right)\)
=> \(\left(3x+4\right)-\left(3x-3\right)⋮\left(x-1\right)\)
=> \(\left(3x+4-3x+3\right)⋮\left(x-1\right)\)
=> \(7⋮\left(x-1\right)\)
=> \(x-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
ta có bảng sau
| x-1 | -7 | -1 | 1 | 7 |
| x | -6 | 0 | 2 | 8 |
vậy x\(\in\left\{-6;0;2;8\right\}\)
b) \(\left(2x+7\right)⋮\left(x-1\right)\)
vì \(\left(x-1\right)⋮\left(x-1\right)\)
=> \(2\left(x-1\right)⋮\left(x-1\right)\)
=> \(\left(2x-2\right)⋮\left(x-1\right)\)
=> \(\left(2x+7\right)-\left(2x-2\right)⋮\left(x-1\right)\)
=> \(\left(2x+7-2x+2\right)⋮\left(x-1\right)\)
=> \(9⋮\left(x-1\right)\)
=> \(x-1\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
ta có bảng sau
| x-1 | -9 | -3 | -1 | 1 | 3 | 9 |
| x | -8 | -2 | 0 | 2 | 4 |
10 |
vậy x\(\in\left\{-8;-2;0;2;4;10\right\}\)
