ĐKXĐ: $x>0$
$A.\sqrt{x} = 2(x-1)^2 < 8$
$\Leftrightarrow (x-1)^2 < 4$
$\Leftrightarrow x^2 - 2x+1-4<0$
$\Leftrightarrow x^2-2x-3<0$
$\Leftrightarrow (x-3)(x+1)<0$
- TH1: $x-3<0; x+1>0$
$\Leftrightarrow x<3; x>-1$
$\Rightarrow -1<x<3$
- TH2: $x-3>0; x+1<0$
$\Leftrightarrow x>3; x<-1$ (loại)
Vậy $S= \{x\in R | -1<x<3\}$