Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 0;x \ne - 1\\
a)D = 1 + \left( {\dfrac{{x + 1}}{{{x^3} + 1}} - \dfrac{1}{{x - {x^2} - 1}} - \dfrac{2}{{x + 1}}} \right)\\
:\dfrac{{{x^3} - 2{x^2}}}{{{x^3} - {x^2} + x}}\\
= 1 + \left( {\dfrac{{x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} + \dfrac{1}{{{x^2} - x + 1}} - \dfrac{2}{{x + 1}}} \right)\\
.\dfrac{{x\left( {{x^2} - x + 1} \right)}}{{{x^2}\left( {x - 2} \right)}}\\
= 1 + \dfrac{{x + 1 + x + 1 - 2\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{x\left( {{x^2} - x + 1} \right)}}{{{x^2}\left( {x - 2} \right)}}\\
= 1 + \dfrac{{2x + 2 - 2{x^2} + 2x - 2}}{{x + 1}}.\dfrac{1}{{x\left( {x - 2} \right)}}\\
= 1 + \dfrac{{4x - 2{x^2}}}{{x + 1}}.\dfrac{1}{{x\left( {x - 2} \right)}}\\
= 1 + \dfrac{{ - 2x\left( {x - 2} \right)}}{{x + 1}}.\dfrac{1}{{x\left( {x - 2} \right)}}\\
= 1 - \dfrac{2}{{x + 1}}\\
= \dfrac{{x + 1 - 2}}{{x + 1}}\\
= \dfrac{{x - 1}}{{x + 1}}\\
b)\left| {x - \dfrac{3}{4}} \right| = \dfrac{5}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{3}{4} = \dfrac{5}{4} \Leftrightarrow x = 2\left( {ktm} \right)\\
x - \dfrac{3}{4} = - \dfrac{5}{4} \Leftrightarrow x = - \dfrac{2}{4} = \dfrac{{ - 1}}{2}\left( {tm} \right)
\end{array} \right.\\
+ Khi:x = - \dfrac{1}{2}\\
\Leftrightarrow D = \dfrac{{x - 1}}{{x + 1}} = \dfrac{{ - \dfrac{1}{2} - 1}}{{ - \dfrac{1}{2} + 1}} = \dfrac{{\dfrac{{ - 3}}{2}}}{{\dfrac{1}{2}}} = - 3\\
c)D = \dfrac{{x - 1}}{{x + 1}} = \dfrac{{x + 1 - 2}}{{x + 1}} = 1 - \dfrac{2}{{x + 1}}\\
D \in Z\\
\Leftrightarrow \dfrac{2}{{x + 1}} \in Z\\
\Leftrightarrow \left( {x + 1} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Leftrightarrow x \in \left\{ { - 3; - 2;0;1} \right\}\\
Do:x \ne - 1;x \ne 0;x \ne 2\\
\Leftrightarrow x \in \left\{ { - 3; - 2;1} \right\}\\
Vậy\,x \in \left\{ { - 3; - 2;1} \right\}\\
d)D < 0\\
\Leftrightarrow \dfrac{{x - 1}}{{x + 1}} < 0\\
\Leftrightarrow - 1 < x < 1\\
Vậy\, - 1 < x < 1;x \ne 0\\
d)D > 2\\
\Leftrightarrow \dfrac{{x - 1}}{{x + 1}} > 2\\
\Leftrightarrow \dfrac{{x - 1}}{{x + 1}} - 2 > 0\\
\Leftrightarrow \dfrac{{x - 1 - 2x - 2}}{{x + 1}} > 0\\
\Leftrightarrow \dfrac{{ - x - 3}}{{x + 1}} > 0\\
\Leftrightarrow \dfrac{{x + 3}}{{x + 1}} < 0\\
\Leftrightarrow - 3 < x < - 1\\
Vậy\, - 3 < x < - 1
\end{array}$
