求下列函数的定义域:

(1) $y=\frac{1}{\log _{2}(x+1)-1}$;

(2) $y=\ln (x+1)+\frac{3 x^{2}}{\sqrt{1-x}}$
(1)要使函数有意义, 需满足 $\left\{\begin{array}{l}x+1>0, \\ \log _{2}(x+1)-1 \neq 0,\end{array} \quad\right.$ 即 $\left\{\begin{array}{l}x>-1, \\ x+1 \neq 2,\end{array}\right.$

$\therefore x>-1$ 且 $x \neq 1$,

$\therefore$ 函数的定义域为 $\{x \mid x>-1$ 且 $x \neq 1\}$.

(2)要使函数有意义，需满足 $\left\{\begin{array}{l}x+1>0 ， \\ 1-x>0 ，\end{array}\right.$

$\therefore-1<x<1$,

$\therefore$ 函数的定义域为 $\{x \mid-1<x<1\}$.