a) \(\dfrac{x^2-3x-4}{x^2-2x+1}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-3x-4\ge0\\x^2-2x+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-3x-4\le0\\x^2-2x+1\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le-1\\x\ge4\end{matrix}\right.\\x=1\end{matrix}\right.\)