x+6 chia hết cho x+2

Đáp án:

`x \in {-1 ; -3 ; 0 ; -4 ; 2 ; -6}` 

Giải thích các bước giải:

`x+6 \vdots x+2`

`-> (x+2) +4 \vdots x+2`

`-> 4 \vdots x+2` ( vì `(x+2) \vdots (x+2)` )

`-> x+2 \in Ư(4) ={\pm 1; \pm 2 ; \pm 4}`

`-> x \in {-1 ; -3 ; 0 ; -4 ; 2 ; -6}`

Vậy `x \in {-1 ; -3 ; 0 ; -4 ; 2 ; -6}`

Ta có: `x+6 \vdots x+2`

         `x+2 \vdots x+2`

`=> (x+6)-(x+2) \vdots x+2`

`=> x+6-x-2 \vdots x+2`

`=> 4 \vdots x+2`

`=> x+2 in Ư (4) = {+-1; +-2; +-4}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{n+2}&\text{1}&\text{2}&\text{4}&\text{-1}&\text{-2}&\text{-4}\\\hline \text{n}&\text{-1}&\text{0}&\text{2}&\text{-3}&\text{-4}&\text{-6}\\\hline\end{array}

Vậy `x in {-6; -4; -3; -1; 0; 2}`