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