Đáp án:
$\\$
Bổ sung : `x ∈ ZZ`
`a`,
`7 \vdots x-2`
`-> x - 2 ∈ Ư (7) = {±1; ±7} (x ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline x-2& 1 & -1 & 7 & -7 \\\hline x& 3 & 1& 9 & -5\\\hline\end{array}$
Vậy `x ∈ {3;1;9;-5}` để `7 \vdots x-2`
$\\$
`b,`
`x + 2 \vdots x-4`
`-> x - 4 + 6 \vdots x-4`
Vì `x-4 \vdots x-4`
`-> 6 \vdots x-4`
`-> x-4 ∈ Ư (6) = {±1; ±2; ±3; ±6} (x ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline x-4& 1 & -1 & 2 & -2&3&-3&6&-6 \\\hline x& 5 & 3& 6 & 2&7&1&10&-2\\\hline\end{array}$
Vậy `x ∈ {5;3;6;2;7;1;10;-2}` để `x + 2 \vdots x-4`
$\\$
`c,`
`2x + 5 \vdots x + 1`
`-> 2x + 2 + 3 \vdots x + 1`
`-> 2 (x+1) + 3 \vdots x+1`
Vì `x+1 \vdots x+1 -> 2 (x+1) \vdots x+1`
`-> 3 \vdots x+1`
`-> x + 1 ∈ Ư (3) ={±1; ±3} (x ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline x+1& 1 & -1 & 3 & -3 \\\hline x& 0 & -2 & 2 & -4 \\\hline\end{array}$
Vậy `x∈{0;-2;2;-4}` để `2x + 5 \vdots x + 1`
