\(3x+4⋮x-1\)
Mà \(x-1⋮x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+4⋮x-1\\3x-3⋮x-1\end{matrix}\right.\)
\(\Leftrightarrow7⋮x-1\)
Vì \(x\in N\Leftrightarrow x-1\in N;x-1\inƯ\left(7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)
Vậy ..
\(\left(3x+4\right)⋮\left(x-1\right)\)
vì \(\left(x-1\right)⋮\left(x-1\right)\)
=> \(\left(3x-3\right)⋮\left(x-1\right)\)
=> \(\left(3x+4\right)-\left(3x-3\right)⋮\left(x-1\right)\)
=> \(\left(3x+4-3x+3\right)⋮\left(x-1\right)\)
=> \(7⋮\left(x-1\right)\)
=> \(\left(x-1\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
ta có bảng sau
| x-1 | -7 | -1 | 1 | 7 |
| x | -6 | 0 | 2 | 8 |
vậy x\(\in\left\{-6;0;2;8\right\}\)