Đáp án:
\(d)\dfrac{1}{{x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{4{y^2}}}{{11{x^4}}}.\left( { - \dfrac{{3{x^2}}}{{8y}}} \right)\\
= - \dfrac{{3y}}{{11.2.{x^2}}}\\
b)\dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{3\left( {x + 4} \right)}}.\dfrac{{x + 4}}{{2\left( {x - 2} \right)}}\\
= \dfrac{{x + 2}}{6}\\
c)\dfrac{{4 - 2 + x}}{{x - 1}} = \dfrac{{x + 2}}{{x - 1}}\\
d)\dfrac{{ - \left( {x + 1} \right) + 2x}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \dfrac{{x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{1}{{x + 1}}
\end{array}\)
