Question

6-\(\left|\dfrac{1}{2}-x\right|\)=\(\dfrac{2}{5}\)

Answer 1

\(6-\left|\dfrac{1}{2}-x\right|=\dfrac{2}{5}\)

\(\left|\dfrac{1}{2}-x\right|=6-\dfrac{2}{5}\)

\(\left|\dfrac{1}{2}-x\right|=\dfrac{28}{5}\)

=> \(\dfrac{1}{2}-x=\dfrac{28}{5}\) hoặc \(\dfrac{1}{2}-x=-\dfrac{28}{5}\)

TH1 :

\(\dfrac{1}{2}-x=\dfrac{28}{5}\)

\(x=\dfrac{1}{2}-\dfrac{28}{5}\)

\(x=-\dfrac{51}{10}\)

TH2 :

\(\dfrac{1}{2}-x=-\dfrac{28}{5}\)

\(x=\dfrac{1}{2}--\dfrac{28}{5}\)

\(x=\dfrac{28}{5}\)

Vậy x \(\in\left\{\dfrac{28}{5};-\dfrac{51}{10}\right\}\)

Answer 2

x = -5,1

va

x = 6,1

Answer 3

\(6-\left|\dfrac{1}{2}-x\right|=\dfrac{2}{5}\)

xét: th1 \(x\le\dfrac{1}{2}\) thì \(6-\left|\dfrac{1}{2}-x\right|=\dfrac{2}{5}\Leftrightarrow6-\left(\dfrac{1}{2}-x\right)=\dfrac{2}{5}\Leftrightarrow6-\dfrac{1}{2}+x=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}+\dfrac{1}{2}-6=\dfrac{-51}{10}\)

th2: \(x>\dfrac{1}{2}\) thì \(6-\left|\dfrac{1}{2}-x\right|=\dfrac{2}{5}\Leftrightarrow6-\left(x-\dfrac{1}{2}\right)=\dfrac{2}{5}\Leftrightarrow6-x+\dfrac{1}{2}=\dfrac{2}{5}\)

\(\Leftrightarrow x=6+\dfrac{1}{2}-\dfrac{2}{5}=\dfrac{61}{10}\)

vậy \(x=\dfrac{-51}{10};x=\dfrac{61}{10}\)