Question

\(\dfrac{5}{2}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{3}\)

Answer 1

`5/2-|x+1/2|=1/3`

`|x+1/2|=5/2-1/3=13/6`

`@TH1:x+1/2=13/6=>x=13/6-1/2=5/3`

`@TH2:x+1/2=-13/6=>x=-13/6-1/2=-8/3`

Vậy `x in {5/3;-8/3}`

Answer 2

=>|x+1/2|=5/2-1/3=15/6-2/6=13/6

=>x+1/2=13/6 hoặc x+1/2=-13/6

=>x=5/3 hoặc x=-8/3

Answer 3

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{13}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{13}{6}\\x+\dfrac{1}{2}=-\dfrac{13}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\)

Answer 4

\(\dfrac{5}{2}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{3}\)

        \(\left|x+\dfrac{1}{2}\right|=\dfrac{5}{2}-\dfrac{1}{3}\)

        \(\left|x+\dfrac{1}{2}\right|=\dfrac{15}{6}-\dfrac{2}{6}\)

        \(\left|x+\dfrac{1}{2}\right|=\dfrac{13}{6}\)

\(\Rightarrow x+\dfrac{1}{2}=\pm\dfrac{13}{6}\)

\(TH1:x+\dfrac{1}{2}=\dfrac{13}{6}\)

          \(x=\dfrac{13}{6}-\dfrac{1}{2}=\dfrac{13}{6}-\dfrac{3}{6}=\dfrac{10}{6}=\dfrac{5}{3}\)

\(TH2:x+\dfrac{1}{2}=-\dfrac{13}{6}\)

           \(x=-\dfrac{13}{6}-\dfrac{1}{2}=-\dfrac{13}{6}-\dfrac{3}{6}=-\dfrac{16}{6}=-\dfrac{8}{3}\)

Vậy \(x\in\left\{\dfrac{5}{3};-\dfrac{8}{3}\right\}\)