(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 $a_{n}=$ $\qquad$
10 Because
$$
\begin{aligned}
a_{n+1} & =S_{n+1}-S_{n} \\
& =\frac{n}{(n+1)(n+2)}-a_{n+1}-\frac{n-1}{n(n+1)}+a_{n},
\end{aligned}
$$

that is
$$
\begin{aligned}
2 a_{n+1} & =\frac{n+2-2}{(n+1)(n+2)}-\frac{1}{n+1}+\frac{1}{n(n+1)}+a_{n} \\
& =\frac{-2}{(n+1)(n+2)}+a_{n}+\frac{1}{n(n+1)},
\end{aligned}
$$

from which we get
$$
2\left(a_{n+1}+\frac{1}{(n+1)(n+2)}\right)=a_{n}+\frac{1}{n(n+1)} .
$$

Let
$$
b_{n}=a_{n}+\frac{1}{n(n+1)},
$$

thus
$$
b_{1}=a_{1}+\frac{1}{2}=\frac{1}{2}\left(a_{1}=0\right), b_{n+1}=\frac{1}{2} b_{n},
$$

hence $b_{n}=\frac{1}{2^{n}}$, we get $a_{n}=\frac{1}{2^{n}}-\frac{1}{n(n+1)}$.