\(b,\left|x+\dfrac{1}{2}\right|-\dfrac{3}{4}=\dfrac{5}{8}\)
\(\Rightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{11}{8}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{11}{8}\\x+\dfrac{1}{2}=-\dfrac{11}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{15}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{7}{8};-\dfrac{15}{8}\right\}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{11}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{11}{8}\\x+\dfrac{1}{2}=-\dfrac{11}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{15}{8}\end{matrix}\right.\)
