ĐKXĐ: \(x\ge1\)
\(\sqrt{x+1}=x-1\)
\(\Rightarrow x+1=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1-x-1=0\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
Vậy \(S=\left\{3\right\}\)
đk x >= -1
\(x+1=\left(x-1\right)^2\Leftrightarrow x+1=x^2-2x+1\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow x=0\left(loai\right);x=3\)
ĐKXĐ: \(x\ge-1\)
\(\sqrt{x+1}=x-1\)
Bình phương hai vế ta được:
\(x+1=x^2-2x+1\)
\(x^2-3x=0\)
x(x-3)=0
=> x=0(l) hoặc x-3=0
x=3
