đkxđ: x≥-1
\(\sqrt{x^2+2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)^2}=\sqrt{x+1}\)
\(\Leftrightarrow\left(x+1\right)^2=x+1\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\Leftrightarrow x=-1\end{matrix}\right.\)(t/m)
Vậy pt có 2 nghiệm.......
\(\sqrt{x^2+2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2+2x+1=x+1\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x^2+2x+1=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-x=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
