不等式 $\frac{x+5}{(x-1)^{2}} \geqslant 2$ 的解是( )
解析 $\frac{x+5}{(x-1)^{2}} \geqslant 2 \Leftrightarrow\left\{\begin{array}{l}x+5 \geqslant 2(x-1)^{2} \\ x-1 \neq 0\end{array}\right.$

$\Leftrightarrow\left\{\begin{array}{l}-\frac{1}{2} \leqslant x \leqslant 3, \quad \therefore x \in\left[-\frac{1}{2}, 1\right) \cup(1,3] . \\ x \neq 1,\end{array}\right.$