Đáp án:Điều kiện:`x>=0`
Đặt `B=(2\sqrtx)/(x-sqrtx+1)`
Ta có:`2\sqrtx>=0AAx>=0`.
`x-sqrtx+1`
`=x-2*sqrtx*1/2+1/4+3/4`
`=(sqrtx-1/2)^2+3/4>=3/4>0`
`=>B>=0`
`**B=0<=>x=0`
`**x>0` chia cả tử và mẫu cho `sqrtx`
`=>B=2/(sqrtx-1+1/sqrtx)`
Áp dụng bđt cosi ta có:
`sqrtx+1/sqrtx>=2`
`=>sqrtx-1+1/sqrtx>=1>0`
`=>B<=2`
Mặt khác `B>0AAx>0`
`=>0<B<=2`
Mà `B in ZZ`
`=>B in {1,2}`
`**B=1`
`<=>x-sqrtx+1=2sqrtx`
`<=>x-3sqrtx+1=0`
Đặt `sqrtx=a(a>0)`
`pt<=>a^2-3a+1=0`
Ta có:`Delta=9-4=5`
`<=>` \(\left[ \begin{array}{l}a=\dfrac{-b+\sqrt{\Delta}}{2a}\\a=\dfrac{-b-\sqrt{\Delta}}{2a}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}a=\dfrac{3-\sqrt5}{2}\\a=\dfrac{3+\sqrt5}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\Big(\dfrac{3-\sqrt5}{2}\Big)\\x=\Big(\dfrac{3+\sqrt5}{2}\Big)^2\end{array} \right.\)
`**B=2`
`<=>2x-2sqrtx+2=2sqrt2`
`<=>2x-4sqrtx+2=0`
`<=>x-2sqrt+1=0`
`<=>(sqrtx-1)^2=0`
`<=>x=1`
Vậy `x in {1,0,(3+sqrt5)/2,(3-sqrt5)/2}`.
