\(\left|2x+1\right|-2x=1\)
+) Xét \(x\ge\dfrac{-1}{2}\) có:
\(2x+1-2x=1\Rightarrow1=1\) ( loại )
+) Xét \(x< \dfrac{-1}{2}\) có:
\(-2x-1-2x=1\)
\(\Rightarrow-4x=2\Rightarrow x=\dfrac{-1}{2}\) ( không t/m )
Vậy không có giá trị x thỏa mãn
\(\left|2x+1\right|-2x=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1-2x=1\left(đk:2x+1\ge0\right)\\-\left(2x+1\right)-2x=1\left(đk:2x+1< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in R\left(đk:x\ge-\dfrac{1}{2}\right)\\x=-\dfrac{1}{2}\left(đk:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in R\left(đk:x\ge-\dfrac{1}{2}\right)\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\ge-\dfrac{1}{2}\forall x\in R\)
