\(A=x^2+\frac{1}{x}+\frac{1}{x}\ge3\sqrt[3]{x^2.\frac{1}{x}.\frac{1}{x}}=3\)
\(\Rightarrow A_{min}=3\) khi \(x=1\)
\(B=\frac{x^3}{2}+\frac{x^3}{2}+\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{x^2}\ge5\sqrt[5]{\frac{x^6}{4x^6}}=\frac{5}{\sqrt[5]{4}}\)
\(\Rightarrow B_{min}=\frac{5}{\sqrt[5]{4}}\) khi \(x=\sqrt[5]{2}\)