Đáp án:
Điều kiện:`x>=0`
`=>x-sqrtx+1`
`=(sqrtx-1/2)^2+3/4>=3/4>0`
Mà `2sqrtx>=0`
`=>A>=0`
`**A=0<=>x=0`
`**x>0=>A>0(1)`
Chia cả tử và mẫu cho `sqrtx` ta có:
`A=2/(sqrtx-1+1/sqrtx)`
Áp dụng bđt cosi ta có:
`sqrtx+1/sqrtx>=2`
`=>sqrtx-1+1/sqrtx>=2-1=1`
`=>A<=2(2)`
`(1)(2)=>0<A<=2`
Mà `A in ZZ`
`=>A in {1,2}`
`+)A=1`
`<=>x-sqrtx+1=2sqrtx`
`<=>x-3sqrtx+1=0`
Đặt `sqrtx=a(a>0)`
`pt<=>a^2-3a+1=0`
`<=>a=(3+-sqrt5)/2`
`<=>x=(3+-sqrt5)^2/4 cancelin ZZ(l)`
`+)A=2`
`<=>2x-2sqrtx+2=2sqrtx`
`<=>x-sqrtx+1=sqrtx`
`<=>x-2sqrtx+1=0`
`<=>(sqrtx-1)^2=0`
`<=>x=1`
Vậy `x in {0,1}` thì `Q in ZZ`.
