giúp mình với cảm ơn trước

`3)`

`a)x(y+2)=8`

`=>x;y+2∈Ư(8)={1;2;4;8}`

Ta có bảng sau :

$\begin{array}{|c|c|c|}\hline \text{x}&\text{1}&\text{8} &\text{2}&\text{4}\\\hline \text{y-2}&\text{8}&\text{1}&\text{4}&\text{2}\\\hline \end{array}$

`=>`$\begin{array}{|c|c|c|}\hline \text{x}&\text{1}&\text{8} &\text{2}&\text{4}\\\hline \text{y}&\text{10}&\text{3}&\text{6}&\text{4}\\\hline \end{array}$

Vậy `x;y∈{10;1};{8;3};{2;6};{4;4}`

`b)(x-2)(2y+3)=26`

`=>26.1=>{(x-2=26),(2y+3=1):}(`loại`)`

`=>1.26=>{(x-2=1),(2y+3=26):}(`loại`)`

`=>13.2=>{(x-2=13),(2y+3=2):}(`loại`)`

`=>2.13=>{(x-2=2),(2y+3=13):}=>{(x=4),(y=5):}`

`4)`

`a) 5 vdots (x-2)`

`=> x-2 in Ư(5)={1,-1,5,-5}`

Ta có bảng sau :

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

`=>x in { 3,1,7,-3}`

`b) (x+13) vdots (x-5)`

`=> x+13 = (x-5)+18 vdots (x-5)`

Vì `x-5 vdots x-5`

`=>18 vdots x-5`

`=>x-5 in Ư(18)={1,-1,2,-2,3,-3,6,-6,9,-9,18,-18}`

`=>x in {6,4,7,3,8,2,11,-1,14,-4,23,-13}`