\(\left(3x+4\right)\)⋮\(\left(x-1\right)\)
⇔ \(\left[3\left(x-1\right)+7\right]\text{⋮}\left(x-1\right)\)
Vì \(3\left(x-1\right)\text{⋮}\left(x-1\right)\)
⇒ \(7\text{⋮}\left(x-1\right)\)
⇒ \(\left(x-1\right)\) ∈\(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)
Còn lại em tự làm tiếp nha
3x+4=3x-3+7\(⋮\)(x-1)=>3(x-1)+7\(⋮\)(x-1)
mà 3(x-1)\(⋮\)(x-1)
=> 7\(⋮\)(x-1)
=> x-1 là ước của 7
=>\(\left[{}\begin{matrix}x-1=7\\x-1=-7\\x-1=1\\x-1=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=8\\x=-6\\x=2\\x=0\end{matrix}\right.\)
\(3x+4=3\left(x-1\right)+7⋮x-1\)
\(\Rightarrow x-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
| x - 1 | 1 | -1 | 7 | -7 |
| x | 2 | 0 | 8 | -6 |
