Đề bài chắc là : Tìm \(x\in Z\) sao cho 3x-1⋮x+2
Ta có:
x+2⋮x+2
\(\Rightarrow\) 3(x+2)⋮x+2
\(\Rightarrow\) 3x+6⋮x+2
Mà 3x-1⋮x+2
\(\Rightarrow\) (3x+6)-(3x-1)⋮x+2
\(\Rightarrow\) 7⋮x+2
\(\Rightarrow x+2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x\in\left\{-9;-3;-1;5\right\}\)
Vậy \(x\in\left\{-9;-3;-1;5\right\}\)
Ta co: 3x+2=3.(x-1)+5
Vi 3.(x-1) chia het cho x-1
=>De 3x+2 chia het cho x-1 thi 5 chia het cho x-1
=>x-1 thuoc uoc cua 5
=>x-1 thuoc tap hop -1;1;5;-5
=>x thuoc tap hop 0;2;6;-4
3x - 1 chia hết x + 2
=> 3( x + 2 ) + (-7) chia hết cho x + 2
vì 3( x + 2 ) chia hết cho x + 2 => (-7) cũng chia hết cho x + 2
=> x + 2 thuộc Ư( -7) ={ -1; -7; 1; 7}
=> x + 2 = 1 hay x + 2 = -1 hay x + 2 = 7 hay x + 2 = -7
x = 1 - 2 x = -1 - 2 x = 7 - 2 x = -7 - 2
x = -1 x = -3 x = 5 x = -9
=> x = -1 ; x = -3 ; x = 5 ; x = -9