One, (Full marks 14 points) Solve the equation
$$
1+\frac{1}{1+\frac{1}{1+\frac{1}{x}}}=\left|\frac{3 x+2}{2 x+1}\right|
$$
First, find the allowed values of $x$.
$$
\left\{\begin{array} { l } 
{ x \neq 0 , } \\
{ 1 + \frac { 1 } { x } \neq 0 , } \\
{ 1 + \frac { 1 } { 1 + \frac { 1 } { x } } \neq 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x \neq 0, \\
x \neq-1, \\
x \neq-\frac{1}{2} .
\end{array}\right.\right.
$$

From the left $=\frac{3 x+2}{2 x+1}$, from the original equation we know $\frac{3 x+2}{2 x+1} \geqslant 0$.
Solving this, we get $x \leqslant-\frac{2}{3}$ or $x>-\frac{1}{2}$.
Combining (1) and (2), the solution to the original equation is
$x \leqslant-\frac{2}{3}$ and $x \neq-1$ or $x>-\frac{1}{2}$ and $x \neq 0$.