Đáp án đúng:
Giải chi tiết:a) \(f\left( x \right) = \dfrac{{{x^7}}}{7} - \dfrac{9}{4}{x^4} + 8x - 3\) \( \Rightarrow f'\left( x \right) = {x^6} - 9{x^3} + 8\)
\(\begin{array}{l} \Rightarrow f'\left( x \right) > 0 \Leftrightarrow {x^6} - 9{x^3} + 8 > 0\\ \Leftrightarrow \left( {{x^3} - 8} \right)\left( {{x^3} - 1} \right) > 0\\ \Leftrightarrow \left[ \begin{array}{l}{x^3} > 8\\{x^3} < 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x > 2\\x < 1\end{array} \right.\end{array}\)
Vậy \(x \in \left( { - \infty ;1} \right) \cup \left( {2; + \infty } \right)\).
b) \(g\left( x \right) = \dfrac{{{x^2} - 5x + 4}}{{x - 2}}\)
\(\begin{array}{l} \Rightarrow g'\left( x \right) = \dfrac{{\left( {2x - 5} \right)\left( {x - 2} \right) - {x^2} + 5x - 4}}{{{{\left( {x - 2} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2{x^2} - 9x + 10 - {x^2} + 5x - 4}}{{{{\left( {x - 2} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{{x^2} - 4x + 6}}{{{{\left( {x - 2} \right)}^2}}}\\ \Rightarrow g'\left( x \right) \le 0 \Leftrightarrow {x^2} - 4x + 6 \le 0 \Leftrightarrow {\left( {x - 2} \right)^2} + 2 \le 0\end{array}\)
\( \Rightarrow \) vô nghiệm.
Vậy \(S = \emptyset \).
