đkxđ: `x≥0; x\ne 4`
`A=(\sqrtx-2)/[x\sqrtx-8]`
`A=(\sqrtx-2)/[(\sqrtx-2)(x+2\sqrtx+4)]`
`A=1/(x+2\sqrtx+4)`
`A=1/[(\sqrtx+1)^2+3]`
Vì `\sqrtx≥0`
`⇒(\sqrtx+1)^2+3≥1+3=4`
`⇒A=1/[(\sqrtx+1)^2+1]≤1/4`
Dấu `=` xảy ra `⇔\sqrtx=0⇒x=0`
Vậy $Max_A=\dfrac{1}{4}⇔=0$