Question

tính GTNN của biểu thức

\(A=\dfrac{x^2+2x+3}{\left(x+2\right)^2}\)

Answer 1

\(A=\dfrac{x^2+2x+3}{\left(x+2\right)^2}=\dfrac{x^2+2x+3}{x^2+4x+4}\)

\(=\dfrac{\left(\dfrac{2}{3}x^2+\dfrac{8}{3}x+\dfrac{8}{3}\right)+\left(\dfrac{1}{3}x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)}{x^2+4x+4}\)

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

\(=\dfrac{2}{3}+\dfrac{\dfrac{1}{3}\left(x-1\right)^2}{x^2+4x+4}\ge\dfrac{2}{3}\forall x\in R\)

Vậy: \(Min_A=\dfrac{2}{3}\Leftrightarrow x=1\)