1. Given $\sqrt{x}+\frac{1}{\sqrt{x}}=3$. Then the value of $\frac{x}{x^{2}+2018 x+1}$ is ( ).
(A)2020
(B) $\frac{1}{2020}$
(C) 2025
(D) $\frac{1}{2025}$
- 1. D.

Squaring both sides of the given equation and rearranging yields $x+\frac{1}{x}=7$.
Therefore, $\frac{x}{x^{2}+2018 x+1}=\frac{1}{7+2018}=\frac{1}{2025}$.