Ta có (3x+8)⋮(x-1)
⇒3(x-1)+11⋮(x-1)
Mà 3(x-1)⋮(x-1)
⇒11⋮(x-1)
⇒x-1∈Ư (11)
⇒x-1∈{1;11}
⇒x∈{2;12}
Vậy x∈{2;12}
Ta có :
\(3x+8⋮x-1\)
Mà \(x-1⋮x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+8⋮x-1\\3x-3⋮x-1\end{matrix}\right.\)
\(\Leftrightarrow11⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=11\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy...