Để 6 chia hết cho (x-1)
=> x = 7 hoặc -5 nhé bạn
ta có: 6 \(⋮\)x -1
=> x - 1 \(\in\)Ư(6) = { -6;-3;-2;-1;1;2;3;6}
=> x \(\in\){ -5;-2;-1;0;2;3;4;7}
vậy: x \(\in\){ -5;-2;-1;0;2;3;4;7}
bạn ủng hộ nha! đúng 100 %.
Ta có: \(6⋮\left(x-1\right)\)
\(\Rightarrow\left(x-1\right)\inƯ\left(6\right)=\left\{-1;1-2;2;-3;3-6;6\right\}\)
*Nếu \(x-1=-1\Rightarrow x=-1+1=0\)
*Nếu \(x-1=1\Rightarrow x=1+1=2\)
*Nếu \(x-1=-2\Rightarrow x=-2+1=-1\)
*Nếu \(x-1=2\Rightarrow x=2+1=3\)
*Nếu \(x-1=-3\Rightarrow x=-3+1=-2\)
*Nếu \(x-1=3\Rightarrow x=3+1=4\)
*Nếu \(x-1=-6\Rightarrow x=-6+1=-5\)
*Nếu \(x-1=6\Rightarrow x=6+1=7\)
\(\Rightarrow x\in\left\{-5;-2;-1;0;2;3;4;7\right\}\)