Đáp án:
\( - 1 < x < \dfrac{{ - 5 + \sqrt {13} }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + 5x + 3 < 0\\
\dfrac{{{x^2} - 4 - 2{x^2} - 2x - x - 1}}{{x + 1}} \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( {\dfrac{{ - 5 - \sqrt {13} }}{2};\dfrac{{ - 5 + \sqrt {13} }}{2}} \right)\\
\dfrac{{ - {x^2} - 3x - 5}}{{x + 1}} \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( {\dfrac{{ - 5 - \sqrt {13} }}{2};\dfrac{{ - 5 + \sqrt {13} }}{2}} \right)\\
x + 1 > 0\left( {do: - {x^2} - 3x - 5 \le 0\forall x \in R} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( {\dfrac{{ - 5 - \sqrt {13} }}{2};\dfrac{{ - 5 + \sqrt {13} }}{2}} \right)\\
x > - 1
\end{array} \right.\\
KL: - 1 < x < \dfrac{{ - 5 + \sqrt {13} }}{2}
\end{array}\)
