b)x+30%=1,3
x+0,3=1,3
x=1,3-0,3
x=1
Ta có:
A=\(\dfrac{3.n+2}{n-1}\)=\(\dfrac{3.\left(n-1\right)+5}{n-1}\)=\(\dfrac{3.\left(n-1\right)}{n-1}\)+\(\dfrac{5}{n-1}\)=3+\(\dfrac{5}{n-1}\)
Để A có giá trị nguyên thì 5\(⋮\)n-1 hay n-1\(\in\)Ư(5)
| n-1 | 1 | -1 | 5 | -5 |
| n | 2 | 0 | 6 | -4 |
Vậy n\(\in\){2;0;6;-4}