Đáp án + Giải thích các bước giải:
`a//a+5` $\vdots$ `a-1`
`→(a-1)+6` $\vdots$ `a-1`
`→6` $\vdots$ `a-1` . Do `(a-1)` $\vdots$ `a-1`
`→a-1∈Ư(6)={±1;±2;±3;±6}`
`→a∈{0;-1;-2;-5;2;3;4;7}`
Vậy để `a+5` $\vdots$ `a-1` thì `a∈{0;-1;-2;-5;2;3;4;7}`
`b//2a` $\vdots$ `a-1`
`→2(a-1)+2` $\vdots$ `a-1`
`→2` $\vdots$ `a-1` . Do `2(a-1)` $\vdots$ `a-1`
`→a-1∈Ư(2)={±1;±2}`
`→a∈{0;-1;2;3}`
Vậy để `2a` $\vdots$ `a-1` thì `a∈{0;-1;2;3}`
`c//3a-8` $\vdots$ `a-4`
`→3(a-4)+4` $\vdots$ `a-4`
`→4` $\vdots$ `a-4` . Do `3(a-4)` $\vdots$ `a-4`
`→a-4∈Ư(4)={±1;±2;±4}`
`→a∈{3;2;0;5;6;8}`
Vậy để `3a-8` $\vdots$ `a-4` thì `a∈{3;2;0;5;6;8}`
`d//a^{2}+a+1` $\vdots$ `a+1`
`→a(a+1)-a+a+1` $\vdots$ `a+1`
`→a(a+1)+1` $\vdots$ `a+1`
`→1` $\vdots$ `a+1` . Do `a(a+1)` $\vdots$ `a+1`
`→a+1∈Ư(1)={±1}`
`→a∈{-2;0}`
Vậy để `a^{2}+a+1` $\vdots$ `a+1` thì `a∈{-2;0}`
