Question

Tìm GTLN của biểu thức :

\(\text{A}=\frac{5}{\left(2x-1\right)^2+3}\).

Answer 1

\(\left(2x-1\right)^2+3\ge3\Rightarrow A=\frac{5}{\left(2x-1\right)^2+3}\le\frac{5}{3}\)

\(\text{Dấu = xảy ra khi }2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

\(\text{Vậy Max}A=\frac{5}{3}\Leftrightarrow x=\frac{1}{2}\)

Answer 2

GIẢI :

Ta có : \(\left(2x-1\right)^2\ge0\)

\(\Rightarrow(2x-1)^2+3\ge3\)

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

\(\Rightarrow\frac{5}{\left(2x-1\right)^2+3}\le\frac{5}{3}\)

\(\Rightarrow\text{A}_{max}=\frac{5}{3}\).

Dấu "=" xảy ra khi : \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

Vậy \(\text{A}_{max}=\frac{5}{3}\) khi \(x=\frac{1}{2}\).