ĐKXĐ: ...
Đặt \(\sqrt{x+1}=a\ge0\Rightarrow a^2-x=1\)
\(\Rightarrow x^2+a=a^2-x\)
\(\Leftrightarrow x^2-a^2+x+a=0\)
\(\Leftrightarrow\left(x+a\right)\left(x-a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-x\\a=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=-x\left(x\le0\right)\\\sqrt{x+1}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2\\x+1=\left(x+1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x=0\end{matrix}\right.\)
