Question

Tìm GTNN: \(M=\dfrac{x^2-2x+\sqrt{2015}}{x^2}\) (với \(x\ne0\))

Answer 1

\(\Leftrightarrow Mx^2=x^2-2x+\sqrt{2015}\\ \Leftrightarrow x^2\left(M-1\right)+2x-\sqrt{2015}=0\)

Ta có \(\Delta'\ge0\Leftrightarrow1+\sqrt{2015}\left(M-1\right)\ge0\)

\(\Leftrightarrow1+\sqrt{2015}M-\sqrt{2015}\ge0\\ \Leftrightarrow M\ge\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\)

Vậy \(M_{min}=\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\Leftrightarrow x=-\dfrac{b'}{a}=-\dfrac{1}{M-1}=\dfrac{-\sqrt{2015}}{\sqrt{2015}-1}\)