Question

 Tìm x để A = \(\dfrac{2m}{m^{^2}+1}\) đạt min

Answer 1

\(A+1=\dfrac{2m}{m^2+1}+1=\dfrac{m^2+2m+1}{m^2+1}=\dfrac{(m+1)^2}{m^2+1} \ge0\)

\(\Rightarrow A\ge-1\)

Dấu '=' xảy ra \(\Leftrightarrow m+1=0\Leftrightarrow m=-1\)

Vậy \(A_{min}=-1 \Leftrightarrow m=-1\)