Đáp án:
$\begin{array}{l}
\dfrac{1}{{x + 1}} < \dfrac{1}{{{{\left( {x - 1} \right)}^2}}}\\
\Leftrightarrow \dfrac{1}{{x + 1}} - \dfrac{1}{{{{\left( {x - 1} \right)}^2}}} < 0\\
\Leftrightarrow \dfrac{{{{\left( {x - 1} \right)}^2} - \left( {x + 1} \right)}}{{\left( {x + 1} \right){{\left( {x - 1} \right)}^2}}} < 0\\
\Leftrightarrow \dfrac{{{x^2} - 2x + 1 - x - 1}}{{\left( {x + 1} \right){{\left( {x - 1} \right)}^2}}} < 0\\
\Leftrightarrow \dfrac{{{x^2} - 3x}}{{x + 1}} < 0\left( {x \ne 1} \right)\\
\Leftrightarrow \dfrac{{x\left( {x - 3} \right)}}{{x + 1}} < 0\left( {x \ne 1} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x\left( {x - 3} \right) < 0\\
x + 1 > 0\\
x \ne 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x\left( {x - 3} \right) > 0\\
x + 1 < 0\\
x \ne 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
0 < x < 3\\
x > - 1\\
x \ne 1
\end{array} \right.\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 3\\
x < 0
\end{array} \right.\\
x < - 1\\
x \ne 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
0 < x < 3;x \ne 1\\
x < - 1
\end{array} \right.\\
Vậy\,x < - 1\,hoặc\,0 < x < 3;x \ne 1
\end{array}$
