Đáp án:
$\begin{array}{l}
y = \dfrac{{{{\left( {x + 1} \right)}^2}.\left( {x - 4} \right)}}{{x - 2}}\\
= \dfrac{{\left( {{x^2} + 2x + 1} \right)\left( {x - 4} \right)}}{{x - 2}}\\
= \dfrac{{{x^3} - 2{x^2} - 7x - 4}}{{x - 2}}\\
y' = \dfrac{{\left( {3{x^2} - 4x - 7} \right)\left( {x - 2} \right) - {x^3} + 2{x^2} + 7x + 4}}{{{{\left( {x - 2} \right)}^2}}}\\
= \dfrac{{3{x^3} - 10{x^2} + x + 14 - {x^3} + 2{x^2} + 7x + 4}}{{{{\left( {x - 2} \right)}^2}}}\\
= \dfrac{{2{x^3} - 8{x^2} + 8x + 18}}{{{{\left( {x - 2} \right)}^2}}}
\end{array}$
