1.)
\(\left|x+2\right|+\left|2x+1\right|=4x\)
TH1 : \(x< -2\)
\(\Rightarrow\left|x+2\right|+\left|2x+1\right|=\left(-x-2\right)-\left(2x+1\right)=4x\\ \Leftrightarrow-3x-3=4x\\ \Rightarrow7x=-3\\ \Rightarrow x=-\dfrac{-3}{7}\left(loai\right)\)
TH2 : \(-2\le x\le-\dfrac{1}{2}\)
\(\Rightarrow\left|x+2\right|+\left|2x+1\right|=\left(x+2\right)+\left(2x+1\right)=4x\\ \Leftrightarrow3x+3=4x\\ \Leftrightarrow x=3\left(loai\right)\)
TH3 : x>-1/2
\(\left|x+2\right|+\left|2x+1\right|=\left(x+2\right)+\left(2x+1\right)=4x\\ \Leftrightarrow x=3\left(TM\right)\)
Vậy x=3