a,\(x+2\inƯ\left(7\right)\Rightarrow x+2\in\left\{\pm7;\pm1\right\}\)
Ta có bảng sau:
x+2 | 7 | -7 | 1 | -1 |
x | 5 | -9 | -1 | -3 |
Vậy \(x\in\left\{5;-9;-1;-3\right\}\)
a/Ta có:x+2 \(\in\) Ư(7)
=>Ư(7) \(\in\){-7;-1;1;7)
=>x\(\in\){-9;-3;-1;5}
Vậy x\(\in\){-9;-3;-1;5}
b/Ta có:x-4\(\in\) B(x-1)
=>(x-4) \(⋮\) (x-1) ; (x-1) \(⋮\) (x-1)
=>(x-4)-(x-1) \(⋮\) (x-1)
=>(-3) \(⋮\) (x-1)
=>x-1 \(\in\) {-3;-1;1;3)
=>x \(\in\) {-2;0;2;4}
Vậy x \(\in\) {-2;0;2;4}