`\sqrt{x+8}-\sqrt{x}=\sqrt{x+3}` $(1)$
$ĐK: \begin{cases}x+8\ge 0\\x\ge 0\\x+3\ge 0\end{cases}$ $⇔\begin{cases}x\ge -8\\x\ge 0\\x\ge -3\end{cases}$ `=>x\ge 0`
`(1)<=>\sqrt{x+3}+\sqrt{x}=\sqrt{x+8}`
`<=>(\sqrt{x+3}+\sqrt{x})^2=(\sqrt{x+8})^2`
`<=>x+3+2\sqrt{x(x+3)}+x=x+8`
`<=>2\sqrt{x(x+3)}=-x+5`
$⇔\begin{cases}-x+5\ge 0\\4x(x+3)=(-x+5)^2\end{cases}$
$⇔\begin{cases}x\le 5\\4x^2+12x=x^2-10x+25\end{cases}$
$⇔\begin{cases}x\le 5\\3x^2+22x-25=0\end{cases}$
$⇔\left\{\begin{matrix}x\le 5\\\left[\begin{array}{l}x=1\ (nhận)\\x=\dfrac{-25}{3}\ (loại\ vì \ đk\ x\ge 0)\end{array}\right.\end{matrix}\right.$
Vậy `S={1}`
