1. Given $\sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a}$. Then the value of $\sqrt{4 x+x^{2}}$ is ( ).
(A) $a-\frac{1}{a}$
(B) $\frac{1}{a}-a$
(C) $a+\frac{1}{a}$
(D) cannot be determined
\begin{array}{l}-1 .(\mathrm{B}) . \\ \sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a} \geqslant 0, x=\frac{1}{a}+a-2, \frac{1}{a} \geqslant a, \\ 4 x+x^{2}=(x+2)^{2}-4 \\ =\left(\frac{1}{a}+a\right)^{2}-4=\left(\frac{1}{a}-a\right)^{2}, \\ \therefore \sqrt{4 x+x^{2}}=\frac{1}{a}-a .\end{array}