\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
<=> \(\left[{}\begin{matrix}x+\dfrac{3}{4}-\dfrac{1}{3}=0\left(x\ge\dfrac{3}{4}\right)\\-x-\dfrac{3}{4}-\dfrac{1}{3}=0\left(x< \dfrac{3}{4}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{1}{3}-\dfrac{3}{4}\\x=-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-5}{12}\left(loại\right)\\x=-\dfrac{13}{12}\left(TM\right)\end{matrix}\right.\)
Vậy nghiệm của PT là: \(S=\left\{\dfrac{-13}{12}\right\}\)
\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
