Ta có:\(\frac{3x-12}{x-3}=\frac{3x-9-3}{x-3}=\frac{3.\left(x-3\right)-3}{x-3}=1-\frac{3}{x-3}\)
Suy ra:\(3⋮\left(x-3\right)\)
Hoặc \(\left(x-3\right)\inƯ\left(3\right)\)
Vậy Ư(3) là:[1,-1,3,-3]ư
Do đó ta có bảng sau:
| x-3 | -3 | -1 | 1 | 3 |
| x | 0 | 2 | 4 | 6 |
Vậy x=0;2;4;6
\(3x-4=3x-9+5=3\left(x-3\right)+5\) (*)
\(x-3⋮x-3\Rightarrow3\left(x-3\right)⋮x-3\) (**)
Từ (*) (**) => 5 chia hết cho x - 3
=> x - 3 thuộc Ư(5) = {1;-1;5;-5}
\(\Rightarrow x-3=1\Rightarrow x=4\)
\(\Rightarrow x-3=-1\Rightarrow x=2\)
\(\Rightarrow x-3=5\Rightarrow x=8\)
\(\Rightarrow x-3=-5\Rightarrow x=-2\)
\(\frac{3x-12}{x-3}=\frac{3\left(x-3\right)-2}{x-3}=\frac{3\left(x-3\right)}{x-3}-\frac{2}{x-3}=3-\frac{3}{x-3}\in Z\)
\(\Rightarrow3⋮x-3\)
\(\Rightarrow x-3\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{4;2;6;0\right\}\)
Ta có
\(3x-12=3x-9-3=3\left(x-3\right)-3\)
=> Sai đề vì đề không xác định 3 có chia hết cho x - 3 hay 0
