10. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ satisfying
$$
S_{n}+a_{n}=\frac{n-1}{n(n+1)}(n=1,2, \cdots) \text {. }
$$

Then the general term $a_{n}=$
10. $\frac{1}{2^{n}}-\frac{1}{n(n+1)}$.

Notice
$$
\begin{array}{l}
a_{n+1}=S_{n+1}-S_{n} \\
=\frac{n}{(n+1)(n+2)}-a_{n+1}-\frac{n-1}{n(n+1)}+a_{n},
\end{array}
$$

which means
$$
\begin{array}{l}
2 a_{n+1}=\frac{n+2-2}{(n+1)(n+2)}-\frac{1}{n+1}+\frac{1}{n(n+1)}+a_{n} \\
=\frac{-2}{(n+1)(n+2)}+a_{n}+\frac{1}{n(n+1)} .
\end{array}
$$

From this, we get
$$
\begin{array}{l}
2\left[a_{n+1}+\frac{1}{(n+1)(n+2)}\right]=a_{n}+\frac{1}{n(n+1)} . \\
\text { Let } b_{n}=a_{n}+\frac{1}{n(n+1)}, b_{1}=a_{1}+\frac{1}{2}=\frac{1}{2}
\end{array}
$$
$\left(a_{1}=0\right)$. Then $b_{n+1}=\frac{1}{2} b_{n}$. Therefore, $b_{n}=\frac{1}{2^{n}}$.
So, $a_{n}=\frac{1}{2^{n}}-\frac{1}{n(n+1)}$.