Question

[antrang tram]

Tìm GTLN của A=\(\frac{2}{x-\sqrt{x}+1}\)

><  >< huhuhuhu

Answer 1

[Trí Tiên亗]

Ta có : \(x-\sqrt{x}+1=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow\frac{2}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{2}{\frac{3}{4}}=\frac{8}{3}\)

hay : \(A\le\frac{8}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

Vậy : Max \(A=\frac{8}{3}\) tại \(x=\frac{1}{4}\)