Đáp án:
$\begin{array}{l}
y = \dfrac{{x + 1}}{{{x^2} - 2x}}\\
\Rightarrow y' = \dfrac{{{x^2} - 2x - \left( {x + 1} \right).\left( {2x - 2} \right)}}{{{{\left( {{x^2} - 2x} \right)}^2}}}\\
= \dfrac{{{x^2} - 2x - \left( {2{x^2} - 2} \right)}}{{{{\left( {{x^2} - 2x} \right)}^2}}}\\
= \dfrac{{ - {x^2} - 2x + 2}}{{{{\left( {{x^2} - 2x} \right)}^2}}} = 0\\
\Rightarrow {x^2} + 2x - 2 = 0\\
\Rightarrow {\left( {x + 1} \right)^2} = 3\\
\Rightarrow \left[ \begin{array}{l}
x = - 1 + \sqrt 3 \\
x = - 1 - \sqrt 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
CD:x = - 1 + \sqrt 3 \\
CT:x = - 1 - \sqrt 3
\end{array} \right.
\end{array}$
