ta có 4x-1 \(⋮x+1\)
x+1\(⋮x+1\)=> \(4\left(x+1\right)⋮x+1\)
=>4(x+1) -(4x-1)\(⋮x+1\)
=> 4x+4-4x+1\(⋮x+1\)
=>\(5⋮x+1\)
=>\(x+1\inƯ\left(5\right)\)
=> \(x+1\in\left\{\pm1,\pm5\right\}\)=> \(x\in\left\{-6,-2,0,4\right\}\)
Vậy..........
các phần sau bn lm tương tự nhé