Question

\(\dfrac{-10}{|2x-1|+10}\)

Giá trị nhỏ nhất

Answer 1

-10 nha bạn

Answer 2

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 = \(\dfrac{1}{2}\)

Answer 3

ý bạn Vân là bạn giải hẳn ra