Ta có :\(|x-2|-|1-2x|=0\)
\(\Rightarrow|x-2|=|1-2x|\)
\(\Rightarrow\orbr{\begin{cases}x-2=1-2x\\x-2=2x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Vậy \(x=1\)hoặc \(x=-1\)
\(\left|x-2\right|-\left|1-2x\right|=0\)
\(\Leftrightarrow\left|x-2\right|=\left|1-2x\right|\)Xet tung TH :
TH1 : \(x-2=1-2x\Leftrightarrow3x=3\Leftrightarrow x=1\)
TH2 : \(x-2=-1+2x\Leftrightarrow-x=1\Leftrightarrow x=-1\)
Vay \(x=\left\{\pm1\right\}\)