a/ Ta có :
\(\left|2x-1\right|=\left|x+1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=1+1\\2x+x=-1+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy ..........
b/ \(\left|\dfrac{1}{3}-4x\right|=\left|2x+\dfrac{2}{3}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}-4x=2x+\dfrac{2}{3}\\\dfrac{1}{3}-4x=-2x-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}-\dfrac{2}{3}=2x+4x\\\dfrac{1}{3}+\dfrac{2}{3}=-2x+4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{3}=6x\\1=2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{18}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ..
