Ta có:\(\frac{4x-5}{x-2}=\frac{4.\left(x-2\right)+3}{x-2}=4+\frac{3}{x-2}\)
Suy ra:\(3⋮\left(x-2\right)\)
Hoặc \(\left(x-2\right)\inƯ\left(3\right)\)
Vậy Ư(3) là:[1,-1,3,-3]
Do đó ta có bảng sau:
| x-2 | -3 | -1 | 1 | 3 |
| x | -1 | 1 | 3 | 5 |
Vậy x=-1;1;3;5
<=> 4x - 5 - 4(x - 2) chia hết cho x - 2
<=> 4x - 5 - 4x +2 chia hết cho x - 2
<=> - 3 chia hết cho x -2
\(\Leftrightarrow x-2\inƯ_3\)
\(\Leftrightarrow x-2\in\left\{1;3;-1;-3\right\}\)
\(\Leftrightarrow x\in\left\{3;5;1;-1\right\}\)
Vậy \(x\in\left\{3;5;1;-1\right\}\)
\(\frac{4x-3}{x-2}=\frac{4x-8+5}{x+2}=\frac{4\left(x-2\right)+5}{x-2}\)
mà \(4x-2⋮x-2\Leftrightarrow5⋮x-2\)
\(\Rightarrow x-2\inư\left(5\right)=\left\{\pm1;\pm5\right\}\)
nếu \(x-2=1\Rightarrow x=3\)
nếu \(x-2=-1\Rightarrow x=1\)
nếu \(x-2=5\Rightarrow x=7\)
nếu \(x-2=-5\Rightarrow x=-3\)
vậy \(x=\left\{-1;3;-3;7\right\}\)
4x - 5 chia hết cho x - 2
=> 4x - 8 + 2 chia hết cho x - 2
=> 4x - 2 chia hết cho x - 2
=> 2(x - 2) chia hết cho x - 2
<=> 2 chia hết cho x - 2
=> x - 2 thuộc Ư(2) = {1;-1;2;-2}
Vậy x thuộc {3;1;4;0}