heo my pờ lít''''''

`( sqrtx + 1 )/( sqrtx - 1 ) = ( sqrtx - 1 + 2 )/( sqrtx - 1 ) = 1 + 2/( sqrtx - 1 ) ( x \ge 0 ; x \ne 1 )`

`-> 2 \vdots sqrtx - 1`

`-> sqrtx - 1 in { +-1 ; +-2 ; 0 }`

`-> sqrtx - 1 in { 1 ; 2 ; 0 ; - 1}`

`o. sqrtx - 1 = - 1 => sqrtx = 0 => x = 0`

`o. sqrtx - 1 = 0 => sqrtx = 1 => x = 1`

`o. sqrtx - 1 = 1 => sqrtx = 2 => x = 4`

`o. sqrtx - 1 = 2 => sqrtx = 3 => x = 9`

Vậy `x in { 4 ; 9 ; 1 ; 0 }`