Đáp án:
b) \( - \dfrac{{3\left( {x - 3} \right)}}{{x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left( {2x - 1} \right)\left( {2x + 1} \right)\left[ {\dfrac{{2x + 1 - 2x + 1 - 4{x^2} + 1}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}} \right]\\
= - 4{x^2} + 3\\
b)\left[ {\dfrac{{3x + 9 - 9}}{{{{\left( {x + 3} \right)}^2}}}} \right]:\left[ {\dfrac{{3 - x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right]\\
= \dfrac{{3x}}{{{{\left( {x + 3} \right)}^2}}}.\dfrac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{ - x}}\\
= - \dfrac{{3\left( {x - 3} \right)}}{{x + 3}}\\
c)\dfrac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{{x^2} + 1}}.\left[ {\dfrac{{4x + 1}}{{x\left( {x - 4} \right)}}.\dfrac{{4x - 1}}{{x\left( {x + 4} \right)}}} \right]\\
= \dfrac{{16{x^2} - 1}}{{{x^2}\left( {{x^2} + 1} \right)}}
\end{array}\)
