Ta có: \(6x+1=2\left(3x-1\right)+3\)
Vì \(2\left(3x-1\right)⋮\left(3x-1\right)\Rightarrow3⋮\left(3x-1\right)\)
\(\Rightarrow3x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow3x=\left\{2;0;4;-2\right\}\)
\(\Rightarrow x=\left\{\frac{2}{3};0;\frac{4}{3};\frac{-2}{3}\right\}\)
Vì biểu thức là số nguyên
Vậy x = 0
\(6x+1=6x-2+3.\)
\(\left(6x-2\right)+3=2\left(3x-1\right)+3⋮3x-1\)
\(6x+1⋮3x-1\)
ta có \(3x-1⋮3x-1\)
\(\Rightarrow2\left(3x-1\right)⋮3x-1\)
\(\Rightarrow6x-2⋮3x-1\)
mà \(6x+1⋮3x-1\)
\(\Rightarrow6x+1-\left(6x-2\right)⋮3x-1\)
\(\Rightarrow6x+1-6x+2\) \(⋮3x-1\)
\(\Rightarrow3\) \(⋮3x-1\)
\(\Rightarrow3x-1\in\text{Ư}_{\left(3\right)}=\text{ }\left\{1;3\right\}\)
nếu \(3x-1=1\Rightarrow\) ko tìm đc \(x\) thỏa mãn
nếu \(3x-1=3\Rightarrow\) ko tìm đc \(x\) thỏa mãn
vậy ko tìm đc \(x\) thỏa mãn để \(6x+1⋮3x-1\)
