Đáp án:
$\begin{array}{l}
đkxđ:x \ne - 1\\
y = \frac{{{x^2} - 3x + 5}}{{x + 1}} = x - 4 + \frac{9}{{x + 1}}\\
y' = 1 - \frac{9}{{{{\left( {x + 1} \right)}^2}}}\\
hs\,nghịch\,biến \Leftrightarrow y' < 0\\
\Rightarrow 1 - \frac{9}{{{{\left( {x + 1} \right)}^2}}} < 0\\
\Rightarrow 1 < \frac{9}{{{{\left( {x + 1} \right)}^2}}} \Rightarrow {\left( {x + 1} \right)^2} < 9\\
\Rightarrow - 3 < x + 1 < 3\\
\Rightarrow - 4 < x < 2\\
Kết\,hợp\,dkxd: \Rightarrow \left( { - 4; - 1} \right)va\,\left( { - 1;2} \right)
\end{array}$
