2.214. $\frac{x^{2}+4}{x \sqrt{4+\left(\frac{x^{2}-4}{2 x}\right)^{2}}}$.

2.214. $\frac{x^{2}+4}{x \sqrt{4+\left(\frac{x^{2}-4}{2 x}\right)^{2}}}$.

(Note: The mathematical expression is the same in both languages, so the translation is identical to the original text.)
Solution.

Domain of definition: $x \neq 0$.

$$
\begin{aligned}
& \frac{x^{2}+4}{x \sqrt{4+\left(\frac{x^{2}-4}{2 x}\right)^{2}}}=\frac{x^{2}+4}{x \sqrt{4+\frac{x^{4}-8 x^{2}+16}{4 x^{2}}}}=\frac{x^{2}+4}{x \sqrt{\frac{16 x^{2}+x^{4}-8 x^{2}+16}{4 x^{2}}}}= \\
& =\frac{x^{2}+4}{x \sqrt{\frac{x^{4}+8 x^{2}+16}{4 x^{2}}}}=\frac{x^{2}+4}{x \sqrt{\frac{\left(x^{2}+4\right)^{2}}{4 x^{2}}}}=\frac{x^{2}+4}{\frac{x\left(x^{2}+4\right)}{2 \cdot|x|}}=\frac{2 \cdot|x|}{x}= \\
& =\left\{\begin{array}{l}
-\frac{2 x}{x}=-2, \text { if } x \in(-\infty ; 0) ; \\
\frac{2 x}{x}=2, \text { if } x \in(0 ; \infty) .
\end{array}\right.
\end{aligned}
$$

Answer: -2, if $x \in(-\infty ; 0) ; 2$, if $x \in(0 ; \infty)$.