6. 23 Given the function $y=\frac{\sqrt{4-x^{2}}}{\lg (|x|+x)}$, then its domain is
(A) $[-2,0) \cup(0,2]$.
(B) $\left(0, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right]$.
(C) $\left[-2, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right]$.
(D) $(0,2]$.
(China Beijing High School Mathematics Competition, 1990)
[Solution] For the numerator to be meaningful, $4-x^{2} \geqslant 0$, then $|x| \leqslant 2$. For the denominator to be meaningful, $\lg (|x|+x) \neq 0$, and $|x|+x>0$, i.e.,
$$
x>0 \text {, and } \quad x \neq \frac{1}{2} \text {. }
$$

In summary, the domain of the function is $\left(0, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right]$. Therefore, the correct choice is $(B)$.