\(f'\left(x\right)=\dfrac{1-x}{\sqrt{2x-x^2}}\)
\(f'\left(x\right)\ge1\Leftrightarrow\dfrac{1-x}{\sqrt{2x-x^2}}\ge1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-x^2>0\\1-x>0\\\left(1-x\right)^2\ge2x-x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0< x< 2\\x< 1\\2x^2-4x+1\ge0\end{matrix}\right.\) \(\Rightarrow0< x\le\dfrac{2-\sqrt{2}}{2}\)
f'(x)=\(\dfrac{2-2x}{2\sqrt{2x-x^2}}\) = \(\dfrac{1-x}{\sqrt{2x-x^2}}\)
để f'(x) \(\ge\) 1 \(\Leftrightarrow\) \(\dfrac{1-x}{\sqrt{2x-x^2}}\) \(\ge\) 1 \(\Leftrightarrow\) 1-x \(\ge\) \(\sqrt{2x-x^2}\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x-x^2>0\\1-2x+x^2\ge2x-x^2\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}0< x< 2\\\left\{{}\begin{matrix}x< \dfrac{2-\sqrt{2}}{2}\\x>\dfrac{2+\sqrt{2}}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\) 0<x\(\le\) \(\dfrac{2-\sqrt{2}}{2}\)
