a
xét x +19-(x+1)=18
để x+19 chia hết cho x+1 thì 18 chia hết cho x+1
suy ra x+1 thuộc ư(18)={1;2;3;6;9;18}
x+1=1thì x =0
x+1=2 thì x=1
x+1=3thì x=2
x+1=6thì x =5
x+1=9 thì x=8
x+1=18thì x=17
xậy ......................
a) (x + 19) ⋮ (x + 1)
=> \(\left(x+19\right)-\left(x+1\right)⋮\left(x+1\right)\)
=> \(\left(x+19-x-1\right)⋮\left(x+1\right)\)
=> \(18⋮\left(x-1\right)\)
=> \(\left(n+1\right)\inƯ\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
ta có bảng sau
n+1 | 1 | 2 | 3 | 6 | 9 | 18 |
n | 0 | 1 | 2 | 5 | 8 | 17 |
vậy x\(\in\left\{0;1;2;5;8;17\right\}\)