\(A=x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)
\(\Rightarrow A_{min}=2\) khi \(x=1\)
b/ \(x\le\frac{1}{2}\Rightarrow\frac{1}{x}\ge2\)
\(B=x^2+\frac{1}{x}=x^2+\frac{1}{8x}+\frac{1}{8x}+\frac{3}{4x}\ge3\sqrt[3]{\frac{x^2}{64x^2}}+\frac{3}{4}.2=\frac{9}{4}\)
\(B_{min}=\frac{9}{4}\) khi \(x=\frac{1}{2}\)
c/
\(C=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge3\sqrt[3]{\frac{x}{2}.\frac{x}{2}.\frac{1}{x^2}}=\frac{3}{\sqrt[3]{4}}\)
\(C_{min}=\frac{3}{\sqrt[3]{4}}\) khi \(\frac{x}{2}=\frac{1}{x^2}\Leftrightarrow x=\sqrt[3]{2}\)
d/
\(x\le\frac{1}{4}\Rightarrow\frac{1}{x}\ge4\Rightarrow\frac{1}{x^2}\ge16\)
\(D=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{128x^2}+\frac{127}{128x^2}\ge3\sqrt[3]{\frac{x^2}{2.2.128x^2}}+\frac{127}{128}.16=\frac{65}{4}\)
\(D_{min}=\frac{65}{4}\) khi \(x=\frac{1}{4}\)
