Bài `5`

`a`

`A=2/(sqrtx-5) in ZZ`

`to 2 vdots sqrtx -5 `

`to sqrtx-5 in Ư_((2))={-2;-1;1;2}`

`to sqrtx in {3 ; 4 ; 6 ;  7}`

`to x in {9 ; 16 ; 36  ; 49 } (tm)`

Vậy `x in {9 ; 16 ; 36  ; 49 }`

`b`

`B=(sqrtx+2)/(sqrtx-1)=(sqrtx-1+3)/(sqrtx-1) =1+3/(sqrtx-1)`

Để `B in ZZ` thì `3/(sqrtx-1) in ZZ`

`to 3 vdots sqrtx-1`

`to sqrtx-1 in Ư_((3))={-3;-1;1;3}`

Vì `sqrtx-1 >= -1`

`to sqrtx-1 in {-1;1;3}`

`to sqrtx in {0 ;2 ;4}`

`to x in {0;4;16 } (tm)`

Vậy `x in {0;4;16 }`