Question

Tìm min hoặc max:

A = 2x - \(\sqrt{x}\)

Answer 1

ĐK: \(x\ge0\)

\(A=2x-\sqrt{x}\)

\(A=2\left(x-\frac{\sqrt{x}}{2}\right)\)

\(A=2\left(x-2\cdot\sqrt{x}\cdot\frac{1}{4}+\frac{1}{16}-\frac{1}{16}\right)\)

\(A=2\left[\left(\sqrt{x}-\frac{1}{4}\right)^2-\frac{1}{16}\right]\)

\(A=2\left(\sqrt{x}-\frac{1}{4}\right)^2-\frac{1}{8}\ge\frac{-1}{8}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\frac{1}{4}\Leftrightarrow x=\frac{1}{16}\)