Đáp án đúng:
Giải chi tiết:a) \(y' = \dfrac{{2.\left( { - 1} \right) - 3.5}}{{{{\left( {3x - 1} \right)}^2}}} = \dfrac{{ - 17}}{{{{\left( {3x - 1} \right)}^2}}}\)
b) \(y = \dfrac{{3x - 2}}{{5x + 1}}\) \( \Rightarrow y' = \dfrac{{3.1 - 5.\left( { - 2} \right)}}{{{{\left( {5x + 1} \right)}^2}}} = \dfrac{{13}}{{{{\left( {5x + 1} \right)}^2}}}\)
c) \(y = \dfrac{{{x^2} + 2x + 2}}{{x + 1}}\)
\(\begin{array}{l} \Rightarrow y' = \dfrac{{\left| {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right|{x^2} + 2\left| {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right|x + \left| {\begin{array}{*{20}{c}}2&2\\1&1\end{array}} \right|}}{{{{\left( {x + 1} \right)}^2}}}\\ \Rightarrow y' = \dfrac{{{x^2} + 2x}}{{{{\left( {x + 1} \right)}^2}}}\end{array}\)
d) \(y = \dfrac{{1 + x - {x^2}}}{{1 - x + {x^2}}} = \dfrac{{ - {x^2} + x + 1}}{{{x^2} - x + 1}}\)
\(\begin{array}{l} \Rightarrow y' = \dfrac{{\left| {\begin{array}{*{20}{c}}{ - 1}&1\\1&{ - 1}\end{array}} \right|{x^2} + 2\left| {\begin{array}{*{20}{c}}{ - 1}&1\\1&1\end{array}} \right|x + \left| {\begin{array}{*{20}{c}}1&1\\{ - 1}&1\end{array}} \right|}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\ \Rightarrow y' = \dfrac{{ - 4x + 2}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\end{array}\)
