Ta có:
(x2 + 5)⋮(x + 1)
=> (x2 + 1x - 1x + 5)⋮(x + 1)
=> [x(x + 1) - (1x + 1) + 4)⋮(x + 1)
=> [x(x + 1) - 1(x + 1) + 4]⋮(x + 1)
Vì x(x + 1)⋮(x + 1) và 1(x + 1)⋮(x + 1)
nên để [x(x + 1) - 1(x + 1) + 4]⋮(x + 1) thì 4⋮(x + 1)
=> x + 1 ∈ Ư(4)
=> x + 1 ∈ {1; 2; 4}
=> x ∈ {0; 1; 3}
Vậy x ∈ {0; 1; 3}