Đáp án:
\[\left[ \begin{array}{l}
x > - 1\\
- 5 < x < - 2
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^3} + 8{x^2} + 17x + 10 > 0\\
\Leftrightarrow \left( {{x^3} + {x^2}} \right) + \left( {7{x^2} + 7x} \right) + \left( {10x + 10} \right) > 0\\
\Leftrightarrow {x^2}\left( {x + 1} \right) + 7x\left( {x + 1} \right) + 10\left( {x + 1} \right) > 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 7x + 10} \right) > 0\\
\Leftrightarrow \left( {x + 1} \right)\left[ {\left( {{x^2} + 2x} \right) + \left( {5x + 10} \right)} \right] > 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 5} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > - 1\\
- 5 < x < - 2
\end{array} \right.
\end{array}\)
