e) x + 13 chia hết cho x + 1.
Ta có: x + 13 = (x + 1) + 12 \(\Rightarrow\) (x + 1) + 12 \(⋮\) (x + 1) khi 12 \(⋮\) (x + 1)
\(\Rightarrow\) x + 1 \(\in\) Ư(12) = {1; 2; 3; 4; 6; 12}
\(\Rightarrow\) x \(\in\) {0; 1; 2; 3; 5; 11}
Vì x \(\in\) N*
\(\Rightarrow\) x \(\in\) {1; 2; 3; 5; 11}
f) 2x + 108 chia hết cho 2x + 3.
Vì (2x + 108) \(⋮\) (2x + 3) và (2x + 3) \(⋮\) (2x + 3)
\(\Rightarrow\) (2x + 108) - (2x + 3) \(⋮\) (2x + 3)
\(\Rightarrow\) 105 \(⋮\) (2x + 3)
\(\Rightarrow\) 2x + 3 \(\in\) Ư(105) = {1; 3; 5; 7; 15; 21; 35; 105}
\(\Rightarrow\) 2x \(\in\) {-2; 0; 2; 4; 12; 18; 32; 102}
\(\Rightarrow\) x \(\in\) {-1; 0; 1; 2; 6; 9; 16; 51}
Vì x \(\in\) N*
\(\Rightarrow\) x \(\in\) {1; 2; 6; 9; 16; 51}
