a: Ta có: \(\left|x-\dfrac{5}{2}\right|-\dfrac{3}{4}=0\)
\(\Leftrightarrow\left|x-\dfrac{5}{2}\right|=\dfrac{3}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{3}{4}\\x-\dfrac{5}{2}=-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{4}\\x=\dfrac{7}{4}\end{matrix}\right.\)
b: Ta có: \(\dfrac{1}{2}-\left|2x-\dfrac{5}{4}\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left|2x-\dfrac{5}{4}\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{5}{4}=\dfrac{1}{6}\\2x-\dfrac{5}{4}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{24}\\x=\dfrac{13}{24}\end{matrix}\right.\)
a) \(\Rightarrow\left|x-2,5\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x-2,5=\dfrac{3}{4}\\x-2,5=-\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{4}\\x=\dfrac{7}{4}\end{matrix}\right.\)
b) \(\Rightarrow\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{4}-2x=\dfrac{1}{6}\\\dfrac{5}{4}-2x=-\dfrac{1}{6}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{24}\\x=\dfrac{17}{24}\end{matrix}\right.\)
c) \(\Rightarrow\left|0,5x-2\right|=\left|x+\dfrac{2}{3}\right|\)
\(\Rightarrow\left[{}\begin{matrix}0,5x-2=x+\dfrac{2}{3}\left(x\ge4\right)\\0,5x-2=-x-\dfrac{2}{3}\left(x< 4\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{3}\left(ktm\right)\\x=\dfrac{8}{9}\left(tm\right)\end{matrix}\right.\)
d) \(đk:x\ge\dfrac{1}{4}\)
\(\Rightarrow\left|x+1\right|=2x-\dfrac{1}{2}\)
\(\Rightarrow x+1=2x-\dfrac{1}{2}\)( do \(x\ge\dfrac{1}{4}\))
\(\Rightarrow x=\dfrac{3}{2}\)
