Quảng cáo
2 câu trả lời 4991
\[\begin{array}{l}
f\left( x \right) = \frac{{{x^2} - 2x + 5}}{{x - 1}}\\
dk:x \ne 1\\
f'\left( x \right) = \frac{{\left( {{x^2} - 2x + 5} \right)'\left( {x - 1} \right) - \left( {{x^2} - 2x + 5} \right)\left( {x - 1} \right)'}}{{{{\left( {x - 1} \right)}^2}}}\\
= \frac{{\left( {2x - 2} \right)\left( {x - 1} \right) - \left( {{x^2} - 2x + 5} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \frac{{2{x^2} - 2x - 2x + 2 - {x^2} + 2x - 5}}{{{{\left( {x - 1} \right)}^2}}}\\
= \frac{{{x^2} - 2x - 3}}{{{{\left( {x - 1} \right)}^2}}}\\
vay:f'\left( x \right) = \frac{{{x^2} - 2x - 3}}{{{{\left( {x - 1} \right)}^2}}}
\end{array}\]
Quảng cáo
Bạn muốn hỏi bài tập?
