1. Given $a+\frac{1}{a+1}=b+\frac{1}{b-1}-2$, and $a-$ $b+2 \neq 0$. Then the value of $a b-a+b$ is $\qquad$ .
$=、 1.2$.
Let $a+1=x$,
$$
b-1=y \text {. }
$$

Then $x-y$
$$
\begin{array}{l}
=a-b+2 \\
\neq 0 .
\end{array}
$$

Substitute into the given equation,
we get $x+\frac{1}{x}=y+\frac{1}{y}$,
which means $(x-y)(x y-1)=0$.
Since $x-y \neq 0$, it follows that $x y=1$.
Therefore, $(a+1)(b-1)=1$, which means $a b-a+b=2$.