Question

Tìm min :

N = \(\dfrac{3x^2+2x+5}{4x^2+4x+1}\)

Answer 1

\(N=\dfrac{57x^2+38x+95}{19\left(4x^2+4x+1\right)}=\dfrac{14\left(4x^2+4x+1\right)+\left(x^2-18x+81\right)}{19\left(4x^2+4x+1\right)}=\dfrac{14}{19}+\left(\dfrac{x-9}{2x+1}\right)^2\ge\dfrac{14}{19}\)

\(N_{min}=\dfrac{14}{19}\) khi \(x=9\)

Answer 2

Nếu đặt ẩn: \(N=\dfrac{3x^2+2x+5}{\left(2x+1\right)^2}\)

Đặt \(2x+1=t\Leftrightarrow x=\dfrac{t-1}{2}\)

\(\Rightarrow N=\dfrac{3\left(\dfrac{t-1}{2}\right)^2+2\left(\dfrac{t-1}{2}\right)+5}{t^2}=\dfrac{3t^2-2t+19}{4t^2}=\dfrac{19}{4t^2}-\dfrac{1}{2t}+\dfrac{3}{4}\)

\(N=\dfrac{19}{4}\left(\dfrac{1}{t}-\dfrac{1}{19}\right)^2+\dfrac{14}{19}\ge\dfrac{14}{19}\)