(2) (2007 - Ningxia Hainan) Let the function $f(x)=\frac{(x+1)(x+a)}{x}$ be an odd function, then $a=$
From the given, we have $f(-x)=-f(x)$, which means $\frac{(-x+1)(-x+a)}{-x}=-\frac{(x+1)(x+a)}{x}$, or equivalently, $-(a+1) x=(a+1) x$ for any real number $x \neq 0$. Therefore, $a=-1$.

