Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \frac{{\left( {2 - {x^2}} \right)\left( {3 - {x^3}} \right)}}{{1 - x + {x^2}}} = \frac{{6 - 2{x^3} - 3{x^2} + {x^5}}}{{1 - x + {x^2}}} = \frac{{{x^5} - 2{x^3} - 3{x^2} + 6}}{{{x^2} - x + 1}}\\
\Rightarrow y' = \frac{{\left( {{x^5} - 2{x^3} - 3{x^2} + 6} \right)'.\left( {{x^2} - x + 1} \right) - \left( {{x^2} - x + 1} \right)'.\left( {{x^5} - 2{x^3} - 3{x^2} + 6} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\
= \frac{{\left( {5{x^4} - 6{x^2} - 6x + 6} \right)\left( {{x^2} - x + 1} \right) - \left( {2x - 1} \right)\left( {{x^5} - 2{x^3} - 3{x^2} + 6} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\
= \frac{{\left( {5{x^6} - 5{x^5} - {x^4} + 6{x^2} - 12x + 6} \right) - \left( {2{x^6} - {x^5} - 2{x^4} - 4{x^3} + 3{x^2} + 12x - 6} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\
= \frac{{2{x^6} - 4{x^5} + {x^4} + 4{x^3} + 3{x^2} - 24x + 12}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}
\end{array}\)
