1. For any integer $x$, the function $f(x)$ satisfies $f(x+1)=\frac{1+f(x)}{1-f(x)}$ If $f(1)=2$, then the value of $f(1993)$ is $\qquad$
1. 2

Since $f(x+2)=\frac{1+f(x+1)}{1-f'(x+1)}=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}}=-\frac{1}{f(x)}$,
thus $f(x+4)=-\frac{1}{f(x+2)}=f(x)$.
Therefore, $f(x)$ is a periodic function with a period of 4, so $f(1993)=f(1)=2$.