Đáp án:
$\begin{array}{l}
a)Dkxd:x\# 1;x\# - 3\\
C = \left( {\dfrac{{2{x^2} + 1}}{{{x^3} - 1}} - \dfrac{1}{{x - 1}}} \right):\left( {1 - \dfrac{{{x^2} - 2}}{{{x^2} + x + 1}}} \right)\\
= \dfrac{{2{x^2} + 1 - \left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}:\dfrac{{{x^2} + x + 1 - {x^2} + 2}}{{{x^2} + x + 1}}\\
= \dfrac{{2{x^2} + 1 - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 3}}\\
= \dfrac{{ - {x^2} + x}}{{x - 1}}.\dfrac{1}{{x + 3}}\\
= \dfrac{{ - x\left( {x - 1} \right)}}{{x - 1}}.\dfrac{1}{{x + 3}}\\
= \dfrac{{ - x}}{{x + 3}}\\
b)\left| {1 - x} \right| + 2 = 3\left( {x + 1} \right)\\
\Leftrightarrow \left| {1 - x} \right| = 3x + 3 - 2\\
\Leftrightarrow \left| {x - 1} \right| = 3x + 1\\
Dkxd:x \ge - \dfrac{1}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 3x + 1\\
x - 1 = - 3x - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = - 2\\
4x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\left( {tm} \right)\\
x = 0\left( {tm} \right)
\end{array} \right.\\
+ Khi:x = - 1 \Leftrightarrow C = \dfrac{{ - x}}{{x + 3}} = \dfrac{1}{2}\\
+ Khi:x = 0 \Leftrightarrow C = \dfrac{{ - x}}{{x + 3}} = 0\\
c)C = \dfrac{{ - x}}{{x + 3}} = \dfrac{{ - x - 3 + 3}}{{x + 3}} = - 1 + \dfrac{3}{{x + 3}}\\
C \in {Z^ + } \Leftrightarrow C\# 0 \Leftrightarrow x\# 0\\
\Leftrightarrow - 1 + \dfrac{3}{{x + 3}} \in Z\\
\Leftrightarrow 3 \vdots \left( {x + 3} \right)\\
\Leftrightarrow \left( {x + 3} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
\Leftrightarrow x \in \left\{ { - 6; - 4; - 2;0} \right\}\\
Do:x\# - 1;x\# - 3;x\# 0\\
\Leftrightarrow x \in \left\{ { - 6; - 4; - 2} \right\}\\
d)\left| C \right| > C\\
\Leftrightarrow C < 0\\
\Leftrightarrow \dfrac{{ - x}}{{x + 3}} < 0\\
\Leftrightarrow \dfrac{x}{{x + 3}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 0\\
x < - 3
\end{array} \right.\\
Vậy\,x < - 3\,hoac\,x > 0;x\# 1\\
e){C^2} - C + 1\\
= {C^2} - 2.C.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {C - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\Leftrightarrow GTNN = \dfrac{3}{4}\,khi:C = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{ - x}}{{x + 3}} = \dfrac{1}{2}\\
\Leftrightarrow - 2x = x + 3\\
\Leftrightarrow 3x = - 3\\
\Leftrightarrow x = - 1\left( {tmdk} \right)\\
Vậy\,x = - 1
\end{array}$
