\(\left(x+3\right)⋮\left(x+2\right)\)
=> \(\left(x+3\right)-\left(x+2\right)⋮\left(x+2\right)\)
=> \(\left(x+3-x-2\right)⋮\left(x+2\right)\)
=> \(1⋮\left(x+2\right)\)
=> \(x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
ta có bảng sau
x+2 | -1 | 1 |
x | -3 | -1 |
vậy x=-3 hoặc x=-1