2.209. $\left(\frac{x-9}{x+3 x^{1 / 2}+9}: \frac{x^{0.5}+3}{x^{1.5}-27}\right)^{0.5}-x^{0.5}$.
## Solution.

Domain of definition: $0 \leq x \neq 9$.

$$
\begin{aligned}
& \left(\frac{x-9}{x+3 x^{1 / 2}+9}: \frac{x^{0.5}+3}{x^{1.5}-27}\right)^{0.5}-x^{0.5}=\left(\frac{x-9}{(\sqrt{x})^{2}+3 \sqrt{x}+9}: \frac{\sqrt{x}+3}{(\sqrt{x})^{3}-3^{3}}\right)^{0.5}- \\
& -\sqrt{x}=\sqrt{\frac{(\sqrt{x})^{2}-3^{2}}{(\sqrt{x})^{2}+3 \sqrt{x}+9} \cdot \frac{(\sqrt{x})^{3}-3^{3}}{\sqrt{x}+3}-\sqrt{x}=} \\
& =\sqrt{\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(\sqrt{x})^{2}+3 \sqrt{x}+9} \cdot \frac{(\sqrt{x}-3)\left((\sqrt{x})^{2}+3 \sqrt{x}+9\right)}{\sqrt{x}+3}}-\sqrt{x}=
\end{aligned}
$$

$=\sqrt{(\sqrt{x}-3)^{2}}-\sqrt{x}=|\sqrt{x}-3|-\sqrt{x}=$

$=\left\{\begin{array}{l}-\sqrt{x}+3-\sqrt{x}=3-2 \sqrt{x} \text { for } x \in[0 ; 9) \\ \sqrt{x}-3-\sqrt{x}=-3 \text { for } x \in(9 ; \infty)\end{array}\right.$

Answer: $3-2 \sqrt{x}$, if $x \in[0 ; 9) ;-3$, if $x \in(9 ; \infty)$.