Đáp án:

 c) x=0

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ { - 1;\dfrac{1}{2};1} \right\}\\
b)M = \left[ {\dfrac{{2{x^3} + {x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{x\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right].\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {2x - 1} \right)\left( {x + 1} \right)}} + \dfrac{x}{{2x - 1}}\\
 = \left[ {\dfrac{{2{x^3} + {x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{x}{{x - 1}}} \right].\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {2x - 1} \right)\left( {x + 1} \right)}} + \dfrac{x}{{2x - 1}}\\
 = \dfrac{{2{x^3} + {x^2} - x - {x^3} - {x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {2x - 1} \right)\left( {x + 1} \right)}} + \dfrac{x}{{2x - 1}}\\
 = \dfrac{{{x^3} - 2x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {2x - 1} \right)\left( {x + 1} \right)}} + \dfrac{x}{{2x - 1}}\\
 = \dfrac{{{x^3} - 2x}}{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{x}{{2x - 1}}\\
 = \dfrac{{{x^3} - 2x + {x^3} + {x^2} + x}}{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{2{x^3} + {x^2} - x}}{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{x\left( {2x - 1} \right)\left( {x + 1} \right)}}{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{x\left( {x + 1} \right)}}{{{x^2} + x + 1}} = \dfrac{{{x^2} + x}}{{{x^2} + x + 1}}\\
c)M = \dfrac{{{x^2} + x}}{{{x^2} + x + 1}}\\
 = \dfrac{{{x^2} + x + 1 - 1}}{{{x^2} + x + 1}} = 1 - \dfrac{1}{{{x^2} + x + 1}}\\
M \in Z \Leftrightarrow \dfrac{1}{{{x^2} + x + 1}} \in Z\\
 \Leftrightarrow {x^2} + x + 1 \in U\left( 1 \right)\\
 \to \left[ \begin{array}{l}
{x^2} + x + 1 = 1\\
{x^2} + x + 1 =  - 1\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x =  - 1\left( l \right)
\end{array} \right.
\end{array}\)