`a,`
`2x + 3 \vdots x`
Vì `x \vdots x`
`-> 2x \vdots x`
`-> 3 \vdots x`
`-> x ∈ Ư (3) = {1;-1;3;-3}`
`-> x ∈ {1;-1;3;-3}` (tm)
Vậy `x ∈ {1;-1;3;-3}` để `2x+3 \vdots x`
`b,`
`8x + 4 \vdots 2x-1`
`-> 8x -4 + 8 \vdots 2x-1`
`-> 4 (2x-1) + 8 \vdots 2x-1`
Vì `2x-1 \vdots 2x-1`
`-> 4 (2x-1) \vdots 2x-1`
`-> 8 \vdots 2x-1`
`->2x-1 ∈ Ư (8)={1;-1;2;-2;4;-4;8;-8}`
`-> 2x ∈ {2;0;3;-1; 5;-3; 9;-7}`
`-> x ∈ {1;0;3/2; (-1)/2; 5/2; (-3)/2; 9/2;(-7)/2}`
Vì `x ∈ ZZ`
`-> x ∈ {1;0}`
Vậy `x ∈ {1;0}` để `8x+4 \vdots 2x-1`
`c,`
`x^2 -x + 5x+1 \vdots x-1`
`-> x^2 - x + 5x-5 + 6 \vdots x-1`
`-> (x^2 - x) + (5x-5) + 6 \vdots x-1`
`-> x (x-1) + 5 (x-1) + 6 \vdots x-1`
`-> (x-1) (x+5) + 6 \vdots x-1`
Vì `x-1 \vdots x-1`
`-> (x-1) (x+5) \vdots x-1`
`-> 6 \vdots x-1`
`->x-1 ∈ Ư (6)={1;-1;2;-2;3;-3;6;-6}`
`-> x∈{2;0;3;-1;4;-2;7;-5}` (tm)
Vậy `x ∈ {2;0;3;-1;4;-2;7;-6}` để `x^2 -x+5x+1 \vdots x-1`
