\(A=\dfrac{5}{x-3}\\ \Rightarrow5⋮\left(x-3\right)\\ \Rightarrow x-3\in\text{Ư}\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau :
x-2 | 1 | -1 | 5 | -5 |
x | 3 | 1 | 7 | -3 |
~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(B=\dfrac{x+2}{x-5}=\dfrac{x-5-3}{x-5}\\ =1-\dfrac{3}{x-5}\)
Để `B in ZZ` thì `3/(x-5) in ZZ`
`=>3 vdots x-5`
`=>x-5 in Ư(3)={+-1;+-3}`
Ta có bảng :
x-5 | 1 | -1 | 3 | -3 |
x | 6 | 4 | 8 | 2 |