\(\left|\dfrac{3}{4}x-\dfrac{1}{2}\right|-1=\dfrac{1}{4}\)
\(\Rightarrow\)\(\left|\dfrac{3}{4}x-\dfrac{1}{2}\right|=\dfrac{5}{4}\)
\(\Rightarrow\)\(\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{5}{4}\\\dfrac{3}{4}x-\dfrac{1}{2}=-\dfrac{5}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{4}\\\dfrac{3}{4}x=-\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-1\end{matrix}\right.\)
Vậy \(x=\dfrac{7}{3}\) hoặc \(x=-1\)
\(pt\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{1}{2}\right|-1=\dfrac{1}{4}\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{1}{2}\right|=\dfrac{5}{4}\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{5}{4}\\\dfrac{3}{4}x-\dfrac{1}{2}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-1\end{matrix}\right.\)
Vậy ...........
Ta có: \(\left|\dfrac{3}{4}x-\dfrac{1}{2}\right|-1=\dfrac{1}{4}\)
\(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{1}{2}\right|=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{-5}{4}\\\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{-3}{4}\\\dfrac{3}{4}x=\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{3}\end{matrix}\right.\)