\(\sqrt{x^2+2x+1}-\sqrt{x+1}=0\left(đk:x\ge-1\right)\)
\(\Leftrightarrow\sqrt{\left(x+1\right)^2}-\sqrt{x+1}=0\)
\(\Leftrightarrow\left|x+1\right|-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1-\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(ĐKXĐ:x\ge-1\)
\(\sqrt{x^2+2x+1}-\sqrt{x+1}=0\\ \Rightarrow\sqrt{x^2+2x+1}=\sqrt{x+1}\\ \Rightarrow x^2+2x+1=x+1\\ \Rightarrow x^2+x=0\\ \Rightarrow x\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=0\left(Tm\right)\end{matrix}\right.\)
Vậy \(x\in\left\{-1;0\right\}\)
