2.273. $\frac{\sqrt{(3 x+2)^{2}-24 x}}{3 \sqrt{x}-\frac{2}{\sqrt{x}}}$.

2.273. $\frac{\sqrt{(3 x+2)^{2}-24 x}}{3 \sqrt{x}-\frac{2}{\sqrt{x}}}$.

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## Solution.

Domain of definition: $0<x \neq \frac{2}{3}$.

$$
\begin{aligned}
& \frac{\sqrt{(3 x+2)^{2}-24 x}}{3 \sqrt{x}-\frac{2}{\sqrt{x}}}=\frac{\sqrt{9 x^{2}+12 x+4-24 x}}{\frac{3 x-2}{\sqrt{x}}}=\frac{\sqrt{9 x^{2}-12 x+4} \cdot \sqrt{x}}{3 x-2}= \\
& =\frac{\sqrt{(3 x-2)^{2}} \cdot \sqrt{x}}{3 x-2}=\frac{|3 x-2| \cdot \sqrt{x}}{3 x-2}=\left\{\begin{array}{l}
\frac{-(3 x-2) \cdot \sqrt{x}}{3 x-2}=-\sqrt{x}, \text { if } x \in\left(0 ; \frac{2}{3}\right) \\
\frac{(3 x-2) \cdot \sqrt{x}}{3 x-2}=\sqrt{x}, \text { if } x \in\left(\frac{2}{3} ; \infty\right)
\end{array}\right. \\
& \text { Answer: }-\sqrt{x}, \text { if } x \in\left(0 ; \frac{2}{3}\right) ; \sqrt{x}, \text { if } x \in\left(\frac{2}{3} ; \infty\right) .
\end{aligned}
$$