Đáp án:
$\begin{array}{l}
a)f\left( x \right) = 2x + \dfrac{1}{{{x^2}}}\left( {x > 0} \right)\\
= x + x + \dfrac{1}{{{x^2}}}\\
Theo\,Cô - si:\\
x + x + \dfrac{1}{{{x^2}}} \ge 3.\sqrt[3]{{x.x.\dfrac{1}{{{x^2}}}}} = 3\\
\Rightarrow f\left( x \right) \ge 3\\
\Rightarrow GTNN:f\left( x \right) = 3\\
Khi:x = x = \dfrac{1}{{{x^2}}}\\
\Rightarrow {x^3} = 1\\
\Rightarrow x = 1\\
b)f\left( x \right) = 3x + \dfrac{1}{{{x^3}}}\\
= x + x + x + \dfrac{1}{{{x^3}}}\\
Theo\,Cô - si:x > 0\\
x + x + x + \dfrac{1}{{{x^3}}} \ge 4.\sqrt[4]{{x.x.x.\dfrac{1}{{{x^3}}}}} = 4\\
\Rightarrow f\left( x \right) \ge 4\\
\Rightarrow GTNN:f\left( x \right) = 4\\
Khi:x = x = x = \dfrac{1}{{{x^3}}}\\
\Rightarrow {x^4} = 1\\
\Rightarrow x = 1\left( {do:x > 0} \right)\\
c)f\left( x \right) = x + \dfrac{1}{{x - 1}}\left( {x > 1} \right)\\
= x - 1 + \dfrac{1}{{x - 1}} + 1\\
Theo\,Cô - si:\\
\left( {x - 1} \right) + \dfrac{1}{{x - 1}} \ge 2\sqrt {\left( {x - 1} \right).\dfrac{1}{{x - 1}}} = 2\\
\Rightarrow f\left( x \right) \ge 2 + 1 = 3\\
\Rightarrow f\left( x \right) \ge 3\\
\Rightarrow GTNN:f\left( x \right) = 3\\
Khi:\left( {x - 1} \right) = \dfrac{1}{{x - 1}}\\
\Rightarrow {\left( {x - 1} \right)^2} = 1\\
\Rightarrow x - 1 = 1\\
\Rightarrow x = 2\left( {do;x > 1} \right)\\
d)f\left( x \right) = 2{x^2} + \dfrac{1}{{{x^3}}}\\
= \dfrac{2}{3}{x^2} + \dfrac{2}{3}{x^2} + \dfrac{2}{3}{x^2} + \dfrac{1}{{2{x^3}}} + \dfrac{1}{{2{x^3}}}\\
Theo\,Cô - si:x > 0\\
\Rightarrow \dfrac{2}{3}{x^2} + \dfrac{2}{3}{x^2} + \dfrac{2}{3}{x^2} + \dfrac{1}{{2{x^3}}} + \dfrac{1}{{2{x^3}}}\\
\ge 5\sqrt[5]{{\dfrac{2}{3}{x^2}.\dfrac{2}{3}{x^2}.\dfrac{2}{3}{x^2}.\dfrac{1}{{2{x^3}}}.\dfrac{1}{{2{x^3}}}}}\\
\Rightarrow f\left( x \right) \ge 5.\sqrt[5]{{\dfrac{2}{{27}}}}\\
\Rightarrow GTNN:f\left( x \right) = 5.\sqrt[5]{{\dfrac{2}{{27}}}}\\
Khi:\dfrac{2}{3}{x^2} = \dfrac{1}{{2{x^3}}}\\
\Rightarrow {x^5} = \dfrac{3}{4}\\
\Rightarrow x = \sqrt[5]{{\dfrac{3}{4}}}
\end{array}$
