Question

Tìm GTNN help cần gấp

Answer 1

\(A=\dfrac{x^2+x+5}{\left(x+1\right)^2}\)

Đặt \(x+1=t\Rightarrow x=t-1\)

\(\Rightarrow A=\dfrac{\left(t-1\right)^2+t-1+5}{t^2}=\dfrac{t^2-t+5}{t^2}=\dfrac{5}{t^2}-\dfrac{1}{t}+1=5\left(\dfrac{1}{t}-\dfrac{1}{10}\right)^2+\dfrac{19}{20}\ge\dfrac{19}{20}\)

\(A_{min}=\dfrac{19}{20}\) khi \(t=10\) hay \(x=9\)