Đáp án:

 a) \(\dfrac{{x + 1}}{{2x}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ {0;2} \right\}\\
Q = \left( {\dfrac{{2x - {x^2}}}{{2{x^2} + 8}} - \dfrac{{2{x^2}}}{{{x^3} - 2{x^2} + 4x - 8}}} \right).\left( {\dfrac{2}{{{x^2}}} - \dfrac{{x - 1}}{x}} \right)\\
 = \left[ {\dfrac{{2x - {x^2}}}{{2{x^2} + 8}} - \dfrac{{2{x^2}}}{{{x^2}\left( {x - 2} \right) + 4\left( {x - 2} \right)}}} \right].\dfrac{{2 - {x^2} + x}}{{{x^2}}}\\
 = \left[ {\dfrac{{2x - {x^2}}}{{2\left( {{x^2} + 4} \right)}} - \dfrac{{2{x^2}}}{{\left( {{x^2} + 4} \right)\left( {x - 2} \right)}}} \right].\dfrac{{\left( {2 - x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \left[ {\dfrac{{\left( {2x - {x^2}} \right)\left( {x - 2} \right) - 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x - 2} \right)}}} \right].\dfrac{{\left( {2 - x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{ - {x^3} + 2{x^2} - 4x + 2{x^2} - 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x - 2} \right)}}.\dfrac{{\left( {2 - x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{ - {x^3} - 4x}}{{2\left( {{x^2} + 4} \right)\left( {x - 2} \right)}}.\dfrac{{\left( {2 - x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{ - x\left( {{x^2} + 4} \right)}}{{2\left( {{x^2} + 4} \right)\left( {x - 2} \right)}}.\dfrac{{\left( {2 - x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{x + 1}}{{2x}}\\
b)Q > \dfrac{1}{2}\\
 \to \dfrac{{x + 1}}{{2x}} > \dfrac{1}{2}\\
 \to \dfrac{{2x + 2 - 2x}}{{4x}} > 0\\
 \to \dfrac{2}{{4x}} > 0\\
 \to x > 0;x \ne 2\\
c)Q = \dfrac{{x + 1}}{{2x}}\\
 \to 2Q = \dfrac{{x + 1}}{x} = 1 + \dfrac{1}{x}\\
Q \in Z \Leftrightarrow \dfrac{1}{x} \in Z\\
 \Leftrightarrow x \in U\left( 1 \right)\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  - 1
\end{array} \right.
\end{array}\)