\(\left|2x-6\right|-1=x\\ < =>\left|2x-6\right|=x+1\\ < =>\left[{}\begin{matrix}2x-6=x+1\\2x-6=-x-1\end{matrix}\right.\\< =>\left[{}\begin{matrix}2x-x=1+6\\2x+x=-1+6\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\3x=5\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\x=\dfrac{5}{3}\end{matrix}\right. \)
`|2x - 6| - 1 = x`
`<=> |2x - 6|=x+1`
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+1\\2x-6=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=1+6\\2x+x=-1+6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{5}{3}\end{matrix}\right.\)
\(\left|2x-6\right|-1=x\)
\(\Rightarrow\left|2x-6\right|=\pm\left(x+1\right)\)
TH1: \(x+1>0\Rightarrow x>-1\)
\(2x-6=x+1\Rightarrow2x-x=1+6\Rightarrow x=7\)
TH2: \(x+1< 0\Rightarrow x< -1\)
\(2x-6=-x-1\Rightarrow2x+x=-1+6\Rightarrow x=\dfrac{5}{3}\)
Vậy \(x\in\left\{7;\dfrac{5}{3}\right\}\).
