If $f(x)=\frac{1}{2x+1}, x≠-\frac{1}{2}$ then $f[f(x)]$ is :

$\frac{2x+1}{2x+3}, $ provided $x≠-\frac{1}{2}$ and $x≠-\frac{3}{2}$

To find \( f[f(x)] \), we first need to find \( f(x) \), and then substitute it into \( f \) again.

Given that \( f(x) = \frac{1}{2x + 1} \), we can substitute this expression into \( f \) again:

\( f[f(x)] = f\left(\frac{1}{2x + 1}\right) \)

Now, replace \( x \) in \( f(x) \) with \( \frac{1}{2x + 1} \):

\( f[f(x)] = \frac{1}{2\left(\frac{1}{2x + 1}\right) + 1} \)

\( = \frac{1}{\frac{2}{2x + 1} + 1} \)

\( = \frac{1}{\frac{2}{2x + 1} + \frac{2x + 1}{2x + 1}} \)

\(= \frac{1}{\frac{2 + 2x + 1}{2x + 1}} \)

\( = \frac{1}{\frac{2x + 3}{2x + 1}} \)

\( = \frac{2x + 1}{2x + 3} \)

So, \( f[f(x)] = \frac{2x + 1}{2x + 3} \), where \( x \) cannot be \( -\frac{1}{2} \).