2. Given the function $f(x)$ satisfies $f\left(\frac{2}{x+|x|}\right)=$ $\log _{2} \sqrt{x|x|}$, then the analytical expression of $f(x)$ is ( ).
(A) $\log _{2} x(\mathrm{~B})-\log _{2} x(\mathrm{C}) 2^{-x}(\mathrm{D}) x^{-2}$
2. B.

From $x|x|>0$, we get $x>0$. Therefore, $f\left(\frac{1}{x}\right)=\log _{2} x$. Thus, $f(x)=\log _{2} \frac{1}{x}=-\log _{2} x$.