cj hc chx:)

Biểu cảm

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

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

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

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

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

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

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

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

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

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