Đáp án:
\(\dfrac{{ - \left( {{x^2} + 4x - 4} \right)\left( {x - 2} \right)\left( {x + 1} \right)}}{{2x\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0\\
Q = \left[ {\dfrac{{ - {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} - \dfrac{{2{x^2}}}{{{x^3} + 4x + 2{x^2} + 8}}} \right].\left( {\dfrac{{{x^2} - x - 2}}{{{x^2}}}} \right)\\
= \left[ {\dfrac{{ - {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} - \dfrac{{2{x^2}}}{{x\left( {{x^2} + 4} \right) + 2\left( {{x^2} + 4} \right)}}} \right].\left( {\dfrac{{{x^2} - x - 2}}{{{x^2}}}} \right)\\
= \left[ {\dfrac{{ - {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} - \dfrac{{2{x^2}}}{{\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}} \right].\left( {\dfrac{{{x^2} - x - 2}}{{{x^2}}}} \right)\\
= \dfrac{{\left( {x + 2} \right)\left( { - {x^2} + 2x} \right) - 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\left( {\dfrac{{{x^2} - x - 2}}{{{x^2}}}} \right)\\
= \dfrac{{ - {x^3} + 2{x^2} - 2{x^2} + 4x - 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
= \dfrac{{ - {x^3} - 4{x^2} + 4x}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
= \dfrac{{ - x\left( {{x^2} + 4x - 4} \right)}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
= \dfrac{{ - \left( {{x^2} + 4x - 4} \right)\left( {x - 2} \right)\left( {x + 1} \right)}}{{2x\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}
\end{array}\)
( bạn xem lại đề bài có nhầm số hay dấu không nhé )
