Đáp án:
2) \(x \in \left( { - \infty ;1} \right) \cup \left( {3; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:\left\{ \begin{array}{l}
\frac{{x + 1}}{{{{\left( {x - 2} \right)}^2}}} < x + 1\\
x - 2 \ne 0
\end{array} \right.\\
2)\left\{ \begin{array}{l}
\frac{{x + 1}}{{{{\left( {x - 2} \right)}^2}}} < \frac{{x + 1}}{1}\\
x - 2 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 2\\
{\left( {x - 2} \right)^2} > 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 2\\
{x^2} - 4x + 4 - 1 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 2\\
{x^2} - 4x + 3 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 2\\
\left( {x - 3} \right)\left( {x - 1} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 2\\
x \in \left( { - \infty ;1} \right) \cup \left( {3; + \infty } \right)
\end{array} \right.\\
KL:x \in \left( { - \infty ;1} \right) \cup \left( {3; + \infty } \right)
\end{array}\)
