6.005. $\frac{1}{x(x+2)}-\frac{1}{(x+1)^{2}}=\frac{1}{12}$.
Solution.

Domain of definition: $x \neq 0, x \neq-1, x \neq-2$.

$\frac{1}{x(x+2)}-\frac{1}{(x+1)^{2}}=\frac{1}{12} \Leftrightarrow \frac{1}{x^{2}+2 x}-\frac{1}{x^{2}+2 x+1}=\frac{1}{12}$.

Let $x^{2}+2 x=z$, then

$\frac{z^{2}+z+12}{z(z+1)}=0 \Rightarrow z^{2}+z+12=0 \Rightarrow z_{1}=-4, z_{2}=3$.

To find $x$, solve the two equations:

$x^{2}+2 x=-4$ or $x^{2}+2 x=3$.

Solving them, we find: $x_{1,2} \in \varnothing (D<0), x_{3}=-3, x_{4}=1$.

Answer: $x_{1,2} \in \varnothing, x_{3}=-3, x_{4}=1$.