\(\Leftrightarrow2x+1\in\left\{1;15\right\}\)
hay \(x\in\left\{0;7\right\}\)
3x + 9 ⋮ 2x + 1
=> (2x + 1) + (x + 8) ⋮ 2x + 1
=> x + 8 ⋮ 2x + 1
=> 2. (x + 8) ⋮ 2x + 1
=> 2x + 16 ⋮ 2x + 1
=> 15 ⋮ 2x + 1
=> 2x + 1 ∈ Ư (15)
=> 2x + 1 ∈ {1; 3; 5; 15}
=> 2x ∈ {0; 2; 4; 14}
=> x ∈ {0; 1; 2 ; 7}