Vì 6 ⋮(x -1) nên (x-1) ∈ Ư(6)
Ta có Ư(6) ={1;2;3;6}
Suy ra: x -1 = 1 ⇒ x = 2
x – 1 = 2 ⇒ x = 3
x – 1 = 3 ⇒ x = 4
x – 1 = 6 ⇒ x = 7
Vậy x ∈ { 2; 3; 4; 7}.
=> x - 1 \(\in\) Ư(6) = { \(\pm\)1 ; \(\pm\)2 ; \(\pm\)3 ; \(\pm\)6 }
\(\Leftrightarrow x-1\in\left\{-1;1;2;3;6\right\}\)
hay \(x\in\left\{0;2;3;4;7\right\}\)
\(6⋮\left(x-1\right)\)
\(\Rightarrow\left(x-1\right)\inƯ_{\left(6\right)}=\left\{1;2;3;6\right\}\)
\(Xét.4TH:\\ TH1:x-1=1\Rightarrow x=1+1=2\\TH2:x-1=2\Rightarrow x=2+1=3\\ TH3:x-1=3\Rightarrow x=3+1=4\\ TH4:x-1=6\Rightarrow x=6+1=7\)
Vì 6 ⋮(x -1) nên (x-1) ∈ Ư(6)
Ta có Ư(6) ={1;2;3;6}
Suy ra: x -1 = 1 ⇒ x = 2
x – 1 = 2 ⇒ x = 3
x – 1 = 3 ⇒ x = 4
x – 1 = 6 ⇒ x = 7
Vậy x ∈ { 2; 3; 4; 7}