Ta có \(A=\frac{x^2+2x+1}{x^2-1}\left(x\ne\pm1\right)\)
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)
Để A nguyên => \(\frac{2}{x-1}\)nguyên => 2 chia hết cho x-1
x nguyên => x-1 nguyên => x-1 \(\in\)Ư(2)={-2;-1;1;2}
ta có bảng
x-1 | -2 | -1 | 1 | 2 |
x | -1 | 0 | 2 | 3 |