Đáp án:
Ta có: `(5x + 1) ⋮ (x – 3)` `(1)`
mà `(x – 3) ⋮ (x – 3) => (5x – 15) ⋮ (x – 3)` `(2)`
Từ `(1)` và `(2)` suy ra: `(5x + 1) – (5x – 15) ⋮ (x – 3)`
`=> (5x + 1 – 5x + 15) ⋮ (x – 3)`
`=> 16 ⋮ (x – 3)`
`=> (x – 3) ∈ Ư(16) = {1; 2; 4; 8; 16}`
`=> x ∈ {4; 5; 7; 11; 19}`