Đáp án: + Giải thích các bước giải:
$a$) $2x+3$ $\vdots$ $x$
$⇔$ $3$ $\vdots$ $x$ (vì $2x \vdots x$)
$⇒$ $x$ $∈$ `Ư(3)={±1;±3}` ($x ∈ Z$)
Vậy $x$ $∈$ `{±1;±3}`.
$b$) $8x+4$ $\vdots 2x-1$ ($x \neq \dfrac{1}{2}$)
$⇔ 8x - 4 + 8$ $\vdots$ $2x-1$
$⇔ 4(2x-1) + 8$ $\vdots$ $2x-1$
$⇔ 8$ $\vdots$ $2x-1$ ($4(2x-1)$ $\vdots $2x-1$)
$⇒$ $2x-1$ $∈$ `Ư(8)={±1;±2;±4;±8}` ($x ∈ Z$)
Mặt khác : $2x-1$ là số lẻ $⇔$ $2x-1$ $∈$ `{±1}`
$⇔$ $x$ $∈$ `{0;1}`
Vậy $x$ $∈$ `{0;1}` .
$c$) $x^2 - x + 5x + 1$ $\vdots x-1$ ($x \neq 1$)
$⇔ x(x-1) + 5x - 5 + 6$ $\vdots$ $x-1$
$⇔ x(x-1) + 5(x-1) + 6$ $\vdots$ $x-1$
$⇔ (x+5)(x-1) + 6$ $\vdots$ $x-1$
$⇔6 \vdots x-1$ (vì $(x+5)(x-1)$ $\vdots $ $x-1$)
$⇒$ $x-1$ $∈$ `Ư(6)={±1;±2;±3;±6}`
$⇔$ $x$ $∈$ `{-5;-2;-1;0;2;3;4;7}`
Vậy $x$ $∈$ `{-5;-2;-1;0;2;3;4;7}`.
