ĐKXĐ: \(x\ge1\)
\(\sqrt{x+1}=x-1\\ \Rightarrow x+1=x^2-2x+1\\ \Leftrightarrow x^2-2x+1-x-1=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\left(khôngthỏamãnĐKXĐ\right)\\x=3\left(thỏamãnĐKXĐ\right)\end{matrix}\right.\\ Vậy...\)