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Câu hỏi: Cho \(f\left( x \right) = {2 \over x}, g\left(x \right) = {{{x^2}} \over 2} - {{{x^3}} \over 3}.\)
Giải bất phương trình \(f\left( x \right) \le g'\left(x \right).\)
Giải bất phương trình \(f\left( x \right) \le g'\left(x \right).\)
Lời giải chi tiết
\(\begin{array}{l}g'\left( x \right) = \dfrac{{2x}}{2} - \dfrac{{3{x^2}}}{3} = x - {x^2}\\f\left(x \right) \le g'\left(x \right) \Leftrightarrow \dfrac{2}{x} \le x - {x^2}\\ \Leftrightarrow {x^2} - x + \dfrac{2}{x} \le 0\\ \Leftrightarrow \dfrac{{{x^3} - {x^2} + 2}}{x} \le 0\\ \Leftrightarrow \dfrac{{\left({x + 1} \right)\left({{x^2} - 2x + 2} \right)}}{x} \le 0\\ \Leftrightarrow \dfrac{{x + 1}}{x} \le 0\\ \Leftrightarrow - 1 \le x < 0\end{array}\)
Do \({x^2} - 2x + 2 = {\left( {x - 1} \right)^2} + 1 > 0,\) \(\forall x \in \mathbb{R}\).
Vậy \(x \in \left[ { - 1; 0} \right)\).
\(\begin{array}{l}g'\left( x \right) = \dfrac{{2x}}{2} - \dfrac{{3{x^2}}}{3} = x - {x^2}\\f\left(x \right) \le g'\left(x \right) \Leftrightarrow \dfrac{2}{x} \le x - {x^2}\\ \Leftrightarrow {x^2} - x + \dfrac{2}{x} \le 0\\ \Leftrightarrow \dfrac{{{x^3} - {x^2} + 2}}{x} \le 0\\ \Leftrightarrow \dfrac{{\left({x + 1} \right)\left({{x^2} - 2x + 2} \right)}}{x} \le 0\\ \Leftrightarrow \dfrac{{x + 1}}{x} \le 0\\ \Leftrightarrow - 1 \le x < 0\end{array}\)
Do \({x^2} - 2x + 2 = {\left( {x - 1} \right)^2} + 1 > 0,\) \(\forall x \in \mathbb{R}\).
Vậy \(x \in \left[ { - 1; 0} \right)\).