Question

(x+5) chia hết cho (x-1) 

Answer 1

\(x+5\) ⋮ \(x-1\)

\(\Rightarrow x-1+6\) ⋮ \(x-1\)

\(\Rightarrow4\) ⋮ \(x-1\)

\(\Rightarrow x-1\inƯ\left(6\right)\)

Mà: \(Ư\left(6\right)=\left\{1;-1;-2;2;3;-3;6;-6\right\}\)

Ta có bảng:

\(x-1\)	1	-1	2	-2	3	-3	6	-6
\(x\)	2	0	3	-1	4	-2	7	-5

Vậy \(x+5\) ⋮ \(x-1\) khi

\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)

Answer 2

\(\dfrac{\text{x - 5}}{\text{x - 1}}=\dfrac{\text{x + 1 + 4}}{\text{x - 1}}=\text{1 +}\dfrac{4}{\text{x + 1}}\)

để  \(\text{x + 5 ⋮ x + 1}\) thì \(\text{4 ⋮ x + 1}\)

 \(\rightarrow\) \(\text{x + 1 ∈ Ư(4)}\) \(=\left\{0;\pm1;\pm2;\pm3\right\}\)

 \(\rightarrow\) \(\text{x ∈}\left\{0;1;-1;2;-2;3;-3\right\}\)