Ta có:
\(\dfrac{-10}{\left|2x-1\right|+10}\)
Ta có:
|2x - 1| \(\ge\) 0 \(\forall\) x \(\in\) R
\(\Leftrightarrow\) |2x - 1| + 10 \(\ge\) 10
\(\Leftrightarrow\dfrac{1}{\left|2x-1\right|+10}\le\dfrac{1}{10}\)
\(\Leftrightarrow\) \(\dfrac{-10}{\left|2x-1\right|+10}\) \(\ge\) \(\dfrac{-10}{10}\)
\(\Leftrightarrow\) \(\dfrac{-10}{\left|2x-1\right|+10}\ge-1\)
Do đó \(\left(\dfrac{-10}{\left|2x-1\right|+10}\right)_{min}\) = -1 khi |2x - 1| = 0
\(\Leftrightarrow\) 2x - 1 = 0
\(\Leftrightarrow\) 2x = 1
\(\Leftrightarrow\) x = \(\dfrac{1}{2}\)
Vậy \(\left(\dfrac{-10}{\left|2x-1\right|+10}\right)_{min}\) = -1 khi x = \(\dfrac{1}{2}\)
