10. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ satisfying: $S_{n}+a_{n}=\frac{n-1}{n(n+1)}, n=1,2, \cdots$, then the general term $a_{n}=$ . $\qquad$
$$
S_{n}+a_{n}=2 S_{n}-S_{n-1}=\frac{n-1}{n(n+1)}=\frac{2}{n+1}-\frac{1}{n} \Rightarrow 2\left(S_{n}-\frac{1}{n+1}\right)=S_{n-1}-\frac{1}{n}
$$
$\Rightarrow$ The sequence $\left\{S_{n}-\frac{1}{n+1}\right\}$ is a geometric sequence with a common ratio of $\frac{1}{2}$, and $S_{1}-\frac{1}{2}=-\frac{1}{2}$,
thus $S_{n}-\frac{1}{n+1}=-\left(\frac{1}{2}\right)^{n} \Rightarrow S_{n}=\frac{1}{n+1}-\left(\frac{1}{2}\right)^{n}\left(n \in \mathbf{N}^{*}\right)$.
Therefore, $a_{n}=S_{n}-S_{n-1}=\frac{1}{n+1}-\left(\frac{1}{2}\right)^{n}-\frac{1}{n}+\left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^{n}-\frac{1}{n(n+1)}$.