\(\left|x+1\right|=x\left(đk:x\ge0\right)\)
\(\Leftrightarrow x+1=x\)
\(\Leftrightarrow1=0\left(VLý\right)\)
Vậy \(S=\varnothing\)
\(|x+1|=x\) (đk: \(x\ge-1\))
<=> \(\left[{}\begin{matrix}x+1=x\\x+1=-x\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}0=-1\left(VLí\right)\\x=\dfrac{-1}{2}\left(TM\right)\end{matrix}\right.\)
Vậy nghiệm PT là \(S=\left\{\dfrac{-1}{2}\right\}\)