ĐKXĐ: x>=-1
\(2x+1=\sqrt{x+1}\)
=>\(\left\{{}\begin{matrix}x+1=\left(2x+1\right)^2\\x>=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\4x^2+4x+1-x-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\4x^2+3x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(4x+3\right)=0\\x>=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\\left[{}\begin{matrix}x=0\\4x+3=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x=0\)
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