Đáp án:
\(\dfrac{{\left( { - {x^2} + 4x - 8} \right)\left( {x + 1} \right)}}{{2x\left( {x - 2} \right)\left( {{x^2} + 4} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ {0;1;2} \right\}\\
C = \left[ {\dfrac{{\left( {2x - {x^2}} \right)\left( {x - 2} \right) - 2.2x}}{{2\left( {x - 2} \right)\left( {{x^2} + 4} \right)}}} \right].\left( {\dfrac{{2x - x + 1}}{{{x^2}}}} \right)\\
= \left[ {\dfrac{{2{x^2} - {x^3} - 4x + 2{x^2} - 4x}}{{2\left( {x - 2} \right)\left( {{x^2} + 4} \right)}}} \right].\dfrac{{x + 1}}{{{x^2}}}\\
= \dfrac{{ - {x^3} + 4{x^2} - 8x}}{{2\left( {x - 2} \right)\left( {{x^2} + 4} \right)}}.\dfrac{{x + 1}}{{{x^2}}}\\
= \dfrac{{x\left( { - {x^2} + 4x - 8} \right)}}{{2\left( {x - 2} \right)\left( {{x^2} + 4} \right)}}.\dfrac{{x + 1}}{{{x^2}}}\\
= \dfrac{{\left( { - {x^2} + 4x - 8} \right)\left( {x + 1} \right)}}{{2x\left( {x - 2} \right)\left( {{x^2} + 4} \right)}}
\end{array}\)
