Question

\(\sqrt{\left(x+2\right)^2}=2x+1\)

Answer 1

ĐK: \(x\ge\frac{-1}{2}\)

\(\sqrt{\left(x+2\right)^2}=2x+1\)

\(\Leftrightarrow\left|x+2\right|=2x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x+1\\x+2=-2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(chon\right)\\x=-1\left(loai\right)\end{matrix}\right.\)

Vậy...

Answer 2

\(\sqrt{\left(x+2\right)^2}=2x+1\\ x+2=2x+1\\ -1+2=2x-x\\ x=1\)

Vậy \(x=1\)