\(B=\dfrac{2x-1}{x+5}=\dfrac{2\left(x+5\right)-11}{x+5}=\dfrac{2\left(x+5\right)}{x+5}-\dfrac{11}{x+5}=2-\dfrac{11}{x+5}\)
Để `B ∈ Z <=> 2 - 11/(x+5) ∈ Z`
Mà `2∈ Z`
`=> 11/(x+5) ∈ Z`
`=> 11 ⋮ x+5`
`=> x+5 ∈ Ư(11)`
`=> x + 5∈ {1 ; 11 ; -1; -11}`
`=> x ∈ {-4 ; 6 ; -6 ; -16}`
Vậy .......
Để b là số nguyên thì \(2x-1⋮x+5\)
=>\(2x+10-11⋮x+5\)
=>\(-11⋮x+5\)
=>\(x+5\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-4;-6;6;-16\right\}\)