Đáp án:
$\\$
`a,`
`8x + 4 \vdots 2x - 1`
`-> 8x - 4 + 8 \vdots 2x - 1`
`-> 4 (2x - 1) + 8 \vdots 2x - 1`
Vì `4 (2x - 1) \vdots 2x-1`
`-> 8 \vdots 2x - 1`
`-> 2x - 1 ∈ Ư (8) = {±1; ±2; ±4; ±8} (x ∈ ZZ)`
Vì `1` là số lẻ
`-> 2x-1` là số lẻ
`-> 2x-1 ∈ {±1} (x ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline 2x-1& 1 & -1 \\\hline x& 1 & 0 \\\hline\end{array}$
Vậy `x ∈ {1;0}` để `8x + 4 \vdots 2x - 1`
$\\$
`b,`
`x^2 - 5x + 7 \vdots x-5`
`-> x (x - 5) + 7 \vdots x-5`
Vì `x (x-5) \vdots x-5`
`-> 7 \vdots x-5`
`-> x-5∈ Ư (7) = {±1; ±7} (x ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline x-5& 1 & -1&7&-7 \\\hline x& 6 & 4&12&-2 \\\hline\end{array}$
Vậy `x ∈ {6;4;12;-2}` để `x^2 - 5x + 7 \vdots x-5`
$\\$
`c,`
`6x + 4 \vdots 2x - 1`
`-> 6x - 3 + 7 \vdots 2x - 1`
`-> 3 (2x - 1) + 7 \vdots 2x - 1`
Vì `3 (2x-1) \vdots 2x-1`
`-> 7 \vdots 2x-1`
`-> 2x-1 ∈ Ư (7) = {±1;±7} (x ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline 2x-1& 1 & -1&7&-7 \\\hline x& 1 & 0&4&-3 \\\hline\end{array}$
Vậy `x ∈ {1;0;4;-3}` để `6x + 4 \vdots 2x - 1`
$\\$
`c,`
`x^2 -x + 7 \vdots x-1`
`-> x (x-1) + 7 \vdots x-1`
Vì `x (x-1) \vdots x-1`
`-> 7 \vdots x-1`
`-> x-1∈ Ư (7) = {±1; ±7} (x ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline x-1& 1 & -1&7&-7 \\\hline x& 2 & 0&8&-6\\\hline\end{array}$
Vậy `x ∈ {2;0;8;-6}` để `x^2 -x + 7 \vdots x-1`
