Đáp án:
\[y' = \frac{{ - 5{x^2} + 6x + 8}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \frac{{5x - 3}}{{{x^2} + x + 1}}\\
\Rightarrow y' = \frac{{\left( {5x + 3} \right)'.\left( {{x^2} + x + 1} \right) - \left( {{x^2} + x + 1} \right)'.\left( {5x - 3} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}\\
= \frac{{5.\left( {{x^2} + x + 1} \right) - \left( {2x + 1} \right).\left( {5x - 3} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}\\
= \frac{{5{x^2} + 5x + 5 - 10{x^2} + x + 3}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}\\
= \frac{{ - 5{x^2} + 6x + 8}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}
\end{array}\)
