Làm giúp mình với a) 6 ⋮ (x-1) b) 12 ⋮ (x+3) c) x+16 ⋮ x+1 d) x+11⋮ x+1

Giải thích các bước giải:

 $a)6\vdots(x-1)$

$⇒x-1∈Ư(6)$

$⇔x-1∈\{±1;±2;±3;±6\}$

$\text{Ta có bảng sau:}$

$\begin{array}{|c|c|}\hline x-1&-6&-3&-2&-1&1&2&3&6\\\hline x&-5&-2&-1&0&2&3&4&7\\\hline\end{array}$

$\text{Vậy với $x∈\{-5;-2;-1;0;2;3;4;7\}$ thì $6\vdots(x-1)$}$

$b)12\vdots(x+3)$

$⇒x+3∈Ư(12)$

$⇔x+3∈\{±1;±2;±3;±4;±6;±12\}$

$\text{Ta có bảng sau:}$

$\begin{array}{|c|c|}\hline x+3&-12&-6&-4&-3&-2&-1&1&2&3&4&6&12\\\hline x&-15&-9&-7&-6&-5&-4&-2&-1&0&1&3&9\\\hline\end{array}$

$\text{Vậy với $x∈\{-15;-9;-7;-6;-5;-4;-2;-1;0;1;3;9\}$ thì $12\vdots(x+3)$}$

$c)(x+16)\vdots(x+1)$

$\text{Ta có:}$

$\dfrac{x+16}{x+1}$

$=\dfrac{x+1+15}{x+1}$

$=1+\dfrac{15}{x+1}$

$\text{Để $(x+16)\vdots(x+1)$ thì $15\vdots(x+1)$}$

$⇒x+1∈Ư(15)$

$⇔x+1∈\{±1;±3;±5;±15\}$

$\text{Ta có bảng sau:}$

$\begin{array}{|c|c|}\hline x+1&-15&-5&-3&-1&1&3&5&15\\\hline x&-16&-6&-4&-2&0&2&4&14\\\hline\end{array}$

$\text{Vậy với $x∈\{-16;-6;-4;-2;0;2;4;14\}$ thì $(x+16)\vdots(x+1)$}$

$d)(x+11)\vdots(x+1)$

$\text{Ta có:}$

$\dfrac{x+11}{x+1}$

$=\dfrac{x+1+10}{x+1}$

$=1+\dfrac{10}{x+1}$

$\text{Để $(x+11)\vdots(x+1)$ thì $10\vdots(x+1)$}$

$⇒x+1∈Ư(10)$

$⇔x+1∈\{±1;±2;±5;±10\}$

$\text{Ta có bảng sau:}$

$\begin{array}{|c|c|}\hline x+1&-10&-5&-2&-1&1&2&5&10\\\hline x&-11&-6&-3&-2&0&1&4&9\\\hline\end{array}$

$\text{Vậy với $x∈\{-11;-6;-3;-2;0;1;4;9\}$ thì $(x+11)\vdots(x+1)$}$

Học tốt!!!

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