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