Nhanh hộ mai đi thi rồi. Trả lời chi tiết = ctrlhn + cảm ơn + 5*. Cảm ơn

Giải thích các bước giải:

`a)`

`2x-19 vdots x-7`

`=>2x-19 vdots 2x-14`

`=>2x-19-2x+14 vdots x-7`

`=>-5 vdots x-7`

`=>x-7 in Ư(-5)`

`=>x-7 in {1;5;-1;-5}` mà `x in N`

`=>x in {8;12;6;2}`

Vậy `x in {8;12;6;2}`

`b)`

`12 vdots 2x-1`

`=>2x-1 in Ư(12)`

`=>2x-1 in {1;2;3;4;6;12;-1;-2;-3;-4;-6;-12}` mà `x in N`

`=>2x in {2;3;4;5;7;13;0}` mà `x in N`

`=>x in {1;2;0}`

Vậy `x in {1;2;0}`

`c)`

`6 vdots 2x+1`

`=>2x+1 in Ư(6)`

`=>2x+1 in {1;2;3;6}`

`=>2x in {0;1;2;5}` mà `x in N`

`=>x in {0;1}`

Vậy `x in {0;1}`

`d)`

`10 vdots 3x+4`

`=>3x+4 in Ư(10)`

`=>3x+4 in {1;2;5;10}`

`=>3x in {-3;-2;1;6}` mà `x in N`

`=>x in {2}`

Vậy `x in {2}`

b, $\text{Vì 12 $\vdots$ (2x -1)}$

$\Rightarrow$ $\text{(2x - 1) $\in$ Ư(12) = {1,2,3,4,6,12.-1,-2,-3,-4,-6,-12}}$

 Ta có bảng: 

\begin{array}{|1|2|3|}\hline \text{2x -1}&\text{}&\text{}\\\hline \text{}&\text{}&\text{ }\\\hline \text{}&\text{}&\text{}\\\hline \text{}&\text{1}&\text{2}\\\hline