\(\left|x+1\right|+x=2\)
\(\Rightarrow\left|x+1\right|=2-x\)
\(\Rightarrow\left[{}\begin{matrix}x+1=2-x\left(ĐK:x\ge-1\right)\\-x-1=2-x\left(ĐK:x< 1\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2-x-1\Rightarrow x=1-x\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\left(TM\right)\\-x=3-x\Rightarrow0=3\left(KTM\right)\end{matrix}\right.\)
| x + 1 | + x = 2
\(\Leftrightarrow\)x+1+x=2
\(\Leftrightarrow\)x+x+1=2
\(\Leftrightarrow\)2x+1=2
\(\Rightarrow2x=2-1\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=1:2\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy x=\(\dfrac{1}{2}\)
