2.253. $(4 x-1) \cdot\left(\frac{1}{8 x} \cdot(\sqrt{8 x-1}+4 x)^{-1}-(\sqrt{8 x-1}-4 x)^{-1}\right)^{1 / 2}$.
## Solution.

Domain of definition: $x \geq \frac{1}{8}, x \neq \frac{1}{4}$.

$$
\begin{aligned}
& (4 x-1) \cdot\left(\frac{1}{8 x} \cdot(\sqrt{8 x-1}+4 x)^{-1}-(\sqrt{8 x-1}-4 x)^{-1}\right)^{1 / 2}= \\
& =(4 x-1) \cdot \sqrt{\frac{1}{8 x} \cdot\left(\frac{1}{\sqrt{8 x-1}+4 x}-\frac{1}{\sqrt{8 x-1}-4 x}\right)}= \\
& =(4 x-1) \cdot \sqrt{\frac{\sqrt{8 x-1}-4 x-\sqrt{8 x-1}-4 x}{8 x(\sqrt{8 x-1}+4 x)(\sqrt{8 x-1}-4 x)}}=
\end{aligned}
$$

$$
\begin{aligned}
& =(4 x-1) \cdot \sqrt{\frac{-8 x}{8 x\left((\sqrt{8 x-1})^{2}-(4 x)^{2}\right)}}=(4 x-1) \cdot \sqrt{\frac{-1}{8 x-1-16 x^{2}}}= \\
& =(4 x-1) \cdot \sqrt{\frac{1}{16 x^{2}-8 x+1}}=(4 x-1) \cdot \sqrt{\frac{1}{(4 x-1)^{2}}}=(4 x-1) \cdot \frac{1}{|4 x-1|}= \\
& =\frac{4 x-1}{|4 x-1|}=\left\{\begin{array}{l}
\frac{4 x-1}{-(4 x-1)}=-1, \text { if } x \in\left[\frac{1}{8} ; \frac{1}{4}\right) \\
\frac{4 x-1}{4 x-1}=1, \text { if } x \in\left(\frac{1}{4} ; \infty\right)
\end{array}\right.
\end{aligned}
$$

Answer: -1, if $x \in\left[\frac{1}{8} ; \frac{1}{4}\right) ; 1$, if $x \in\left(\frac{1}{4} ; \infty\right)$