Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 1;x \ne - 1\\
A = \left( {\dfrac{{1 - {x^3}}}{{1 - x}} - x} \right):\dfrac{{1 - {x^2}}}{{1 - x - {x^2} + {x^3}}}\\
= \left( {\dfrac{{\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}}{{1 - x}} - x} \right).\dfrac{{\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)}}{{1 - {x^2}}}\\
= \left( {1 + x + {x^2} - x} \right).\dfrac{{\left( {1 - x} \right)\left( {1 - {x^2}} \right)}}{{1 - {x^2}}}\\
= \left( {1 + {x^2}} \right).\left( {1 - x} \right)\\
b)x = - 1\dfrac{2}{3} = \dfrac{{ - 5}}{3}\left( {tmdk} \right)\\
\Rightarrow A = \left( {1 + {x^2}} \right).\left( {1 - x} \right)\\
= \left( {1 + \dfrac{{25}}{9}} \right).\left( {1 + \dfrac{5}{3}} \right)\\
= \dfrac{{34}}{9}.\dfrac{8}{3}\\
= \dfrac{{272}}{{27}}\\
c)A < 0\\
\Rightarrow \left( {1 + {x^2}} \right).\left( {1 - x} \right) < 0\\
\Rightarrow 1 - x < 0\left( {do:1 + {x^2} > 0} \right)\\
\Rightarrow x > 1\\
Vậy\,x > 1
\end{array}$
