b) `|2x-1|-0,5=1,5`
`|2x-1|=2`
\(\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ \left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left|2x-1\right|-0,5=1,5\\ \left|2x-1\right|=2\\ \Rightarrow\left\{{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
a)\(\dfrac{2}{5}+\dfrac{3}{5}\cdot x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{3}{5}\cdot x=\dfrac{1}{2}-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{3}{5}\cdot x=\dfrac{1}{10}\)
\(\Rightarrow x=\dfrac{1}{10}:\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)
Vậy \(x=\dfrac{1}{6}\)
b) \(\left|2x-1\right|-0,5=1,5\)
\(\Rightarrow\left|2x-1\right|=1,5+0,5=2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{3}{2};-\dfrac{1}{2}\right\}\)
