a) (x + 16) ⋮ (x + 1)
Ta có: x + 16 = (x + 1) +15 nên (x + 1) + 15 ⋮ (x + 1) khi 15 \(⋮\) (x + 1)
\(\Rightarrow\) x + 1 \(\in\) Ư(15) = {1; 3; 5;15}
\(\Rightarrow\) x x \(\in\) {0; 2; 4; 14}
Vậy x \(\in\) {0; 2; 4; 14}.
b) (x + 11) ⋮ (x + 1)
Ta có: x + 11 = (x + 1) + 10 nên (x + 1) + 10 ⋮ (x + 1) khi 10 \(⋮\) (x + 1)
\(\Rightarrow\) x + 1 \(\in\) Ư(10) = {1; 2; 5; 10}
\(\Rightarrow\) x \(\in\) {0; 1; 4; 9}
Vậy x \(\in\) {0; 1; 4; 9}.
