`#tnvt`
`a,`
`A=(\frac{x}{x+2}-\frac{x}{2-x}+\frac{8}{x^2-4}).\frac{x^2-2x}{2x^2+8}(x\ne+-2)`
`=(\frac{x(x-2)+x(2+x)+8}{(x-2)(x+2)}).\frac{x(x-2)}{2(x^2+4)}`
`=\frac{x^2-2x+2x+x^2+8}{(x-2)(x+2)}.\frac{x(x-2)}{2x^2+8}`
`=\frac{2x^2+8}{(x-2)(x+2)}.\frac{x(x-2)}{2x^2+8}`
`=\frac{x}{x+2}`
`b,`
`|x-1|=1`
`=>[(x-1=1),(x-1=-1):}`
`<=>[(x=2(\text{Loại})),(x=0(\text{Nhận})):}`
Vậy `x=0`
`->A=\frac{0}{0+2}=0`
Vậy tại `|x-1|=1` thì `A=0`
`c,`
`A=\frac{x}{x+2}`
`=\frac{x+2-2}{x+2}`
`=1-\frac{2}{x+2}`
`\text{Để A}\inZZ` `\text{thì}` `2\vdots x+2`
`->x+2\inƯ(2)\in{+-1;+-2}`
Ta lập bảng giá trị như sau.
\begin{array}{|c|c|c|}\hline x+2 &-1&1&-2&2\\\hline x&-3(tm)&-1(tm)&-4(tm)&0(tm)\\\hline\end{array}
Vậy `x\in{-4;-3;-1;0}` thì `A\inZZ`
