Đáp án: $m > 1$
Giải thích các bước giải:
$\begin{array}{l}
0 < \dfrac{{ - {m^3} + 1}}{{ - {m^3} - 1}} < 1\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{ - {m^3} + 1}}{{ - {m^3} - 1}} > 0\\
\dfrac{{ - {m^3} + 1}}{{ - {m^3} - 1}} - 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{{m^3} - 1}}{{{m^3} + 1}} > 0\\
\dfrac{{ - {m^3} + 1 + {m^3} + 1}}{{ - {m^3} - 1}} < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
{m^3} > 1\\
{m^3} < - 1
\end{array} \right.\\
\dfrac{2}{{ - {m^3} - 1}} < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
- {m^3} - 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
{m^3} > - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
m > - 1
\end{array} \right.\\
\Leftrightarrow m > 1\\
Vậy\,m > 1
\end{array}$
