2. Given a positive number $x$ satisfies
$$
x^{10}+x^{5}+\frac{1}{x^{5}}+\frac{1}{x^{10}}=15250 \text {. }
$$

Then the value of $x+\frac{1}{x}$ is . $\qquad$
2.3.

Let $a=x^{5}+\frac{1}{x^{5}}$. Then $x^{10}+\frac{1}{x^{10}}=a^{2}-2$.
Thus, the original equation becomes
$$
\begin{array}{l}
a^{2}+a-15252=0 \\
\Rightarrow(a-123)(a+124)=0
\end{array}
$$
$\Rightarrow a=123$ or $a=-124$ (discard).
Therefore, $x^{5}+\frac{1}{x^{5}}=123$.
Let $x+\frac{1}{x}=b>0$. Then
$$
\begin{array}{l}
x^{2}+\frac{1}{x^{2}}=b^{2}-2, \\
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right) \\
=b\left(b^{2}-3\right)=b^{3}-3 b .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x^{5}+\frac{1}{x^{5}}\right)+\left(x+\frac{1}{x}\right) . \\
\text { Hence }\left(b^{2}-2\right)\left(b^{3}-3 b\right)=123+b \\
\Rightarrow b^{5}-5 b^{3}+5 b-123=0 .
\end{array}
$$

Since $123=3 \times 41$, when $b=3$, both sides of equation (1) are equal. Therefore, $b=3$ is a solution to equation (1).
Dividing the left side of equation (1) by $b-3$, we get the quotient $b^{4}+3 b^{3}+16 b+41$.
Since this quotient is always greater than 0, we know $b=3$, i.e., $x+\frac{1}{x}=3$.