a, đk : x khác 0 ; - 5
\(A=\dfrac{1}{x}+\dfrac{1}{x+5}+\dfrac{x-5}{x\left(x+5\right)}=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)
Vậy ta có đpcm
b, Thay x = -4 ta được : \(\dfrac{3}{-4+5}=3\)
c, Ta có : \(\dfrac{3}{x+5}\Rightarrow x+5\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x + 5 | 1 | -1 | 3 | -3 |
| x | -4 | -6 | -2 | -8 |


