Để `Q in ZZ =>2 - 5 sqrt(x) vdots sqrtx + 3`
`<=> 17 - 15 - 5 sqrt(x) vdots sqrtx + 3`
`<=> 17 - 5(3 + sqrtx) vdots sqrtx + 3`
`<=> 17 vdots sqrt x + 3`
`<=> sqrt x + 3 in Ư(17)`
Do `sqrtx >= 0 => sqrt(x) + 3 >= 3` nên:
`sqrtx + 3 = 17`
`<=> sqrtx = 14`
`<=> x = 196.`
Vậy `x= 196.`