6\(⋮\left(x-1\right)\)
=>x-1ϵƯ(6)ϵ{\(\pm1;\pm2;\pm3;\pm6\)}
| x-1 | 1 | 2 | 3 | 6 | -1 | -2 | -3 | -6 |
| x | 2 | 3 | 4 | 7 | 0 | -1 | -2 | -5 |
Ta có: \(6⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
\(\Rightarrow x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)
6\(⋮\)x-1=>x-1ϵƯ(6)={1;2;3;6}
Với x-1=1=>x=2
x-1=2=>x=3
x-1=3=>x=4
x-1=6=>x=7
Vậy xϵ{2;3;4;7}
6 chia hết cho (x-1)
\(=>x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(x-1=1\\ =>x=2\\ x-1=-1\\ =>x=0\\ x-1=2\\ =>x=3\\ x-1=-2\\ =>x=-1\)
\(x-1=3\\ =>x=4\\ x-1=-3\\ =>x=-2\\ x-1=6\\ =>x=7\\ x-1=-6\\ =>x=-5\)
Vậy: \(x\in\left\{-5;-2;-1;0;2;3;4;7\right\}\)
Vì 6\(⋮\)x-1=>x-1ϵƯ(6)={1;2;3;6}
Với x-1=1=>x=2
x-1=2=>x=3
x-1=3=>x=4
x-1=6=>x=7
Vậy xϵ{2;3;4;7}
