`1.``A=(x+1/(sqrtx)).((sqrtx-1)/(x-sqrtx+1)-1/(sqrtx+1))`

`ĐK: {(x>0),(x-sqrtx+1≠0),(sqrtx+1≠0):}`

`<=> x>0`

`A=(xsqrtx+1)/(sqrtx) . [((sqrtx-1)(sqrtx+1))/((sqrtx+1)(x-sqrtx+1))-(x-sqrtx+1)/((sqrtx+1)(x-sqrtx+1))]`

`A=(xsqrtx+1)/(sqrtx) . [(x-1)/((sqrtx+1)(x-sqrtx+1))-(x-sqrtx+1)/((sqrtx+1)(x-sqrtx+1))]`

`A=(sqrtx^3+1)/(sqrtx) . (sqrtx-2)/(sqrtx^3+1)`

`A=(sqrtx-2)/(sqrtx)`

`b)``A=(sqrtx-2)/(sqrtx)=1-2/(sqrtx)`

Để `A in ZZ` thì `2/(sqrtx) in ZZ`

`=>sqrtx in Ư(2)`

`=> sqrtx in {1;2}`

`<=> x in {1;4}`

Vậy `x in {1;4}` thì `A` nguyên

`- \text{Lamtoanbangcatinhmang} -`