i: \(\left|2x-\dfrac{1}{3}\right|-\dfrac{2}{5}=0\)
=>\(\left|2x-\dfrac{1}{3}\right|=\dfrac{2}{5}\)
=>\(\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{2}{5}\\2x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{11}{15}\\2x=-\dfrac{2}{5}+\dfrac{1}{3}=-\dfrac{1}{15}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{11}{30}\\x=-\dfrac{1}{30}\end{matrix}\right.\)
d: |2x-1|=|x-2|
=>\(\left[{}\begin{matrix}2x-1=x-2\\2x-1=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-2+1\\3x=3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
e: \(\left(x-\dfrac{2}{3}\right)+1=2x-\dfrac{5}{4}\)
=>\(x+\dfrac{1}{3}=2x-\dfrac{5}{4}\)
=>\(x-2x=-\dfrac{5}{4}-\dfrac{1}{3}\)
=>\(-x=-\dfrac{15}{12}-\dfrac{4}{12}=-\dfrac{19}{12}\)
=>\(x=\dfrac{19}{12}\)
f: |2x-1|-2=3
=>|2x-1|=3+2=5
=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
h: \(\left(2x+1\right)^3=-\dfrac{1}{7}\)
=>\(2x+1=-\dfrac{1}{\sqrt[3]{7}}\)
=>\(2x=-\dfrac{1}{\sqrt[3]{7}}-1=\dfrac{-1-\sqrt[3]{7}}{\sqrt[3]{7}}\)
=>\(x=\dfrac{-1-\sqrt[3]{7}}{2\cdot\sqrt[3]{7}}\)
