\(\frac{x+8}{x-1}=\frac{x-1+9}{x-1}=1+\frac{9}{x-1}\)
\(\frac{x+8}{x-1}\in Z\Leftrightarrow\frac{9}{x-1}\in Z\Leftrightarrow x-1\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
ta có bảng:
x-1 | 1 | 3 | 9 | -1 | -3 | -9 |
x | 2 | 4 | 10 | 0 | -2 | -8 |
TL | TM | TM | TM | TM | TM | TM |
vậy.............