Có: \(12-x⋮12-x\)
Mà theo đề bài: \(x+3⋮12-x\)
\(\Rightarrow\left(12-x\right)+\left(x+3\right)⋮12-x\)
\(\Rightarrow12-x+x+3⋮12-x\)
\(\Rightarrow15⋮12-x\)
Mà \(12-x\le12\) do \(x\in N\)
\(\Rightarrow12-x\in\left\{1;-1;3;-3;5;-5;-15\right\}\)
\(\Rightarrow x\in\left\{11;13;9;15;7;17;27\right\}\)
Vậy \(x\in\left\{11;13;9;15;7;17;27\right\}\)
\(x+3⋮12-x\\ \Rightarrow12+x-9⋮12-x\\ \Rightarrow-\left(12-x\right)-9⋮12-x\\ \Rightarrow9⋮12-x\\ \Rightarrow12-x\in\text{Ư}\left(9\right)=\left\{1;3;9\right\}\\ \Rightarrow x\in\left\{11;9;3\right\}\)
x+3\(⋮\)12-x
12-x\(⋮\)12-x
x+3-(12-x)\(⋮\)12-x
x+3-12+x\(⋮\)12-x
15\(⋮\)12-x
\(\Rightarrow\)12-x={1,3,5,15 }
\(\Rightarrow\)x={11,9,7,-2}
vi x\(\in\)n nen x={11,9,7}
ta có : (x+3) chia hết cho (12-x)
(12 -x) + 15 chia hết cho (12-x)
vì (12-x) chia hết cho (12-x)
\(\Rightarrow\) 15 chia hết cho 12 - x
\(\Rightarrow\) 12 - x \(\in\) { 1 ; 3 ; 5 ; 15 }
từ đây ta có bảng
| 12-x | 1 | 3 | 5 | 15 |
| x | 11 | 9 | 7 | \(\varnothing\) |
Vậy x = 11 ; 9 ; 7
