Đáp án:
\( - \dfrac{{4x}}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \pm 2\\
A = \dfrac{{2 + x}}{{2 - x}} - \dfrac{{4{x^2}}}{{{x^2} - 4}} - \dfrac{{2 - x}}{{x + 2}}\\
= \dfrac{{ - \left( {x + 2} \right)\left( {x + 2} \right) - 4{x^2} - \left( {2 - x} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ - {x^2} - 4x - 4 - 4{x^2} + {x^2} - 4x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ - 4{x^2} - 8x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{ - 4x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= - \dfrac{{4x}}{{x - 2}}
\end{array}\)
