\(\sqrt{x^2}.\left|x+2\right|=x\)
<=>\(\left|x\right|.\left|x+2\right|=x\)
Vì \(VT\ge0\)\(\Leftrightarrow x\ge0\Leftrightarrow\left|x\right|=x\Leftrightarrow\left|x+2\right|=x+2\)
<=>\(x\left(x+2\right)=x\)
<=>\(x\left(x+2\right)-x=0\)
<=>\(x\left(x+2-1\right)=0\)
<=>\(x\left(x+1\right)=0\)
<=>\(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(chọn\right)\\x=-1\left(lọai\right)\end{cases}}}\)
Vậy x=0