Đáp án:
`A=1/(x-1)`
Giải thích các bước giải:
`ĐKXĐ:x\ne 1;x\ne -1;x\ne 2`
Ta có:
`A=(\frac{1}{x+1}-\frac{1}{x^2-1}).\frac{x+1}{x-2}`
`=(\frac{1}{x+1}-\frac{1}{(x-1)(x+1)}).\frac{x+1}{x-2}`
`=(\frac{1.(x-1)}{(x+1)(x-1)}-\frac{1}{(x-1)(x+1)}).\frac{x+1}{x-2}`
`=\frac{x-1-1}{(x+1)(x-1)}.\frac{x+1}{x-2}`
`=\frac{x-2}{(x+1)(x-1)}.\frac{x+1}{x-2}`
`=\frac{(x-2)(x+1)}{(x+1)(x-1)(x-2)}`
`=\frac{1}{x-1}`
Vậy `A=1/(x-1)`
