Question

Tìm giá trị nhỏ nhất của hàm số

a, y= \(x+\dfrac{1}{x}\) với \(x\ge2\)

b, y= \(x^3+\dfrac{3}{x^2}\) với \(x>0\)

Answer 1

a/ \(y=\dfrac{3x}{4}+\dfrac{x}{4}+\dfrac{1}{x}\ge\dfrac{3x}{4}+2\sqrt{\dfrac{x}{4}.\dfrac{1}{x}}\ge\dfrac{3.2}{4}+1=\dfrac{5}{2}\)

\(\Rightarrow y_{min}=\dfrac{5}{2}\) khi \(x=2\)

b/ \(y=\dfrac{x^3}{2}+\dfrac{x^3}{2}+\dfrac{1}{x^2}+\dfrac{1}{x^2}+\dfrac{1}{x^2}\ge5\sqrt[5]{\dfrac{x^3}{2}.\dfrac{x^3}{2}.\dfrac{1}{x^2}.\dfrac{1}{x^2}.\dfrac{1}{x^2}}=\dfrac{5}{\sqrt[5]{4}}\)

\(\Rightarrow y_{min}=\dfrac{5}{\sqrt[5]{4}}\) khi \(x=\sqrt[5]{2}\)