Đáp án + Giải thích các bước giải:
`a//2x+3` $\vdots$ `x`
`⇒3` $\vdots$ `x` . Do `2x` $\vdots$ `x`
`⇒x∈Ư(3)={±1;±3}`
`b//8x+4` $\vdots$ `2x-1`
`⇒4(2x-1)+8` $\vdots$ `2x-1`
`⇒8` $\vdots$ `2x-1` . Do `4(2x-1)` $\vdots$ `2x-1`
`⇒2x-1∈Ư(8)={±1;±2;±4;±8}`
`⇒2x∈{0;-1;-3;-7;2;3;5;9}`
Mà `x∈Z`
`⇒x∈{0;1}`
`c//x^{2}-4x+4` $\vdots$ `x-4`
`⇒x(x-4)+4` $\vdots$ `x-4`
`⇒4` $\vdots$ `x-4` . Do `x(x-4)` $\vdots$ `x-4`
`⇒x-4∈Ư(4)={±1;±2;±4}`
`⇒x∈{3;2;0;5;6;8}`