Ta có: |x+2|-2x=1
=>|x+2|=2x+1
=>\(\begin{cases}2x+1\ge0\\ \left(2x+1\right)^2=\left(x+2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-\frac12\\ \left(2x+1-x-2\right)\left(2x+1+x+2\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-\frac12\\ \left(x-1\right)\left(3x+3\right)=0\end{cases}\Rightarrow x=1\)
