Question

Tìm GTLC của \(A=\dfrac{x}{x^2+2x+1}\)

Answer 1

\(A=\dfrac{x}{x^2+2x+1}=\dfrac{x}{\left(x+1\right)^2}\)

Đặt \(x+1=a\) , ta có :

\(A=\dfrac{x+1-1}{\left(x+1\right)^2}=\dfrac{a-1}{a^2}=\dfrac{1}{a}-\dfrac{1}{a^2}\)

Đặt \(\dfrac{1}{a}=t\) , ta có : \(A=t-t^2=-\left(t^2-t+\dfrac{1}{4}\right)+\dfrac{1}{4}=\dfrac{1}{4}-\left(t-\dfrac{1}{2}\right)^2\le\dfrac{1}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow t=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{a}=\dfrac{1}{2}\Leftrightarrow a=2\Leftrightarrow x+1=2\Leftrightarrow x=1\)

Vậy Max A là : \(\dfrac{1}{4}\Leftrightarrow x=1\)

Answer 2