Condition :\(\left|x+2\right|\ge0\Rightarrow2x-1\ge0\Rightarrow x\ge\dfrac{1}{2}\)

1^st case : \(x+2\ge0\Leftrightarrow x\ge-2\),then :

\(x+2=2x-1\Leftrightarrow2x-x=2+1\Leftrightarrow x=3\) (satisfied)

2^nd case : \(x+2< 0\Leftrightarrow x< -2\),then :

\(-x-2=2x-1\Leftrightarrow2x+x=-2+1\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\)

                                                                                                (absurd)

Tran Kim Anh,you copied my style,right ?

We have |x+2| = 2x -1 \(\Rightarrow\) 2x - 1 \(\ge\) 0 \(\Rightarrow\) x \(\ge\)\(\dfrac{1}{2}\)

1\(^{st}\)case: x \(\ge\) 0 \(\Rightarrow\)|x+2| = x+2 \(\Rightarrow\)x + 2 = 2x -1 \(\Rightarrow\)  x = 3 (satisfied)

2\(^{nd}\)case: x < 0 \(\Rightarrow\) |x+2| = - (x+2) \(\Rightarrow\)-x -2 = 2x - 1\(\Rightarrow\)x = \(\dfrac{-1}{3}\)(absurd)

Hence, x = 3.