Câu trả lời:
Ta có:\(\dfrac{2-7n}{n+1}\)
=\(\dfrac{-7n+2}{n+1}\)
=\(\dfrac{-7n+\left(-7\right)+7+2}{n+1}\)
=\(\dfrac{-7\left(n+1\right)+9}{n+1}\)
=\(-7+\dfrac{9}{n+1}\)
Để \(\dfrac{2-7n}{n+1}\)\(\) \(\in\) Z suy ra \(-7+\dfrac{9}{n+1}\)\(\in\) Z suy ra\(\dfrac{9}{n+1}thu\text{ộc}\) Z
suy ra n+1 thuộc Ư(9)={ \(\pm1\);\(\pm3\);\(\pm9\)}
suy ra n\(\in\){0;-2;2;-4;8;-10}