Đáp án:

$\begin{array}{l}
C = \left( {\frac{{{x^2} + x}}{{{x^3} + {x^2} + x + 1}} + \frac{1}{{{x^2} + 1}}} \right):\left( {\frac{1}{{x - 1}} - \frac{{2x}}{{{x^3} - {x^2} + x - 1}}} \right)\\
a)Đkxđ:\\
\left\{ \begin{array}{l}
{x^3} + {x^2} + x + 1 \ne 0\\
{x^2} + 1 \ne 0\left( {ld} \right)\\
x - 1 \ne 0\\
{x^3} - {x^2} + x - 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {x + 1} \right)\left( {{x^2} + 1} \right) \ne 0\\
\left( {x - 1} \right)\left( {{x^2} + 1} \right) \ne 0\\
x \ne 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x \ne  - 1\\
x \ne 1
\end{array} \right.\\
b)Đkxđ:x \ne 1;x \ne  - 1\\
C = \left( {\frac{{{x^2} + x}}{{{x^3} + {x^2} + x + 1}} + \frac{1}{{{x^2} + 1}}} \right):\left( {\frac{1}{{x - 1}} - \frac{{2x}}{{{x^3} - {x^2} + x - 1}}} \right)\\
 = \left( {\frac{{x\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{1}{{{x^2} + 1}}} \right):\left( {\frac{1}{{x - 1}} - \frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}} \right)\\
 = \left( {\frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right):\frac{{{x^2} + 1 - 2x}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}\\
 = \frac{{x + 1}}{{{x^2} + 1}}.\frac{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \frac{{x + 1}}{{x - 1}}\\
c)Đkxđ:x \ne 1;x \ne  - 1\\
C = \frac{{x + 1}}{{x - 1}} = \frac{{x - 1 + 2}}{{x - 1}} = 1 + \frac{2}{{x - 1}}\\
C \in Z \Rightarrow \frac{2}{{x - 1}} \in Z\\
 \Rightarrow 2 \vdots \left( {x - 1} \right)\\
 \Rightarrow \left( {x - 1} \right) \in {\rm{\{ }} - 2; - 1;1;2\} \\
 \Rightarrow x \in {\rm{\{ }} - 1;0;2;3\} \\
Mà:x \ne 1;x \ne  - 1\\
 \Rightarrow x \in {\rm{\{ }}0;2;3\} 
\end{array}$

a, Để C xác định

$\Leftrightarrow x\neq\pm1$

b, $C=(\dfrac{x^2+x}{x^3+x^2+x+1}+\dfrac{1}{x^2+1}):(\dfrac{1}{x-1}-\dfrac{2x}{x^3-x^2+x-1})$ $(x\neq\pm1)$

$C=(\dfrac{x^2+x}{x^2(x+1)+(x+1)}+\dfrac{1}{x^2+1}):(\dfrac{1}{x-1}-\dfrac{2x}{x^2(x-1)+(x-1)})$

$C=\dfrac{x(x+1)}{(x+1)(x^2+1)}+\dfrac{1}{x^2+1}):(\dfrac{1}{x-1}-\dfrac{2x}{(x^2+1)(x-1)})$

$C=\dfrac{x+1}{x^2+1}:\dfrac{x^2+1-2x}{(x-1)(x^2+1)}$

$C=\dfrac{x+1}{x^2+1}\cdot\dfrac{(x-1)(x^2+1)}{(x-1)^2}$

$C=\dfrac{x+1}{x-1}$

Vậy $C=\dfrac{x+1}{x-1}$ với $x\neq\pm1$

c, $C=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}$

Để C nguyên

$\dfrac{2}{x-1}$ nguyên

`⇒x-1∈Ư(2)={±1;±2}`.

· $x-1=1⇒x=2(tm)$

· $x-1=-1⇒x=0(ktm)$

· $x-1=2⇒x=3(tm)$

· $x-1=-2⇒x=-1(ktm)$

Vậy để C nguyên thì `x\in{2;3}`.