\(x-1\in U\left(8\right)=\left\{1;2;4;8\right\}\)
\(\Rightarrow x\in\left\{2;3;5;9\right\}\)
\(x-1\inƯ\left(8\right)=\left\{1;2;4;8\right\}\)
x - 1 = 1 => x = 2
x - 1 = 2 => x = 3
x - 1 = 4 => x = 5
x - 1 = 8 => x = 9
Vậy x \(\in\){2;3;5;9}
Để 8 chia hết cho x - 1 thì x - 1 phải là Ư(8) mà Ư(8) = -1 ; - 2; -4 ; -8; 1 ; 2 ; 4 ; 8
x -1 = -1 \(\Rightarrow\)x = 0
x - 1 = -2 \(\Rightarrow\) x = -1
x - 1 = - 4 \(\Rightarrow\)x = -3
x - 1 = - 8 \(\Rightarrow\)x = -7
x - 1 = 1 \(\Rightarrow\) x = 2
x - 1 = 2 \(\Rightarrow\) x = 3
x - 1 = 4 \(\Rightarrow\)x = 5
x - 1 = 8 \(\Rightarrow\)x = 9
Vậy x \(\varepsilon\)\(\left\{....\right\}\)