Question

[NGUYỄN ĐỖ BẢO VY]

Tìm x

x - căn bậc hai(x+1) -1 = 0

Answer 1

[Dang Tung]

\(x-\sqrt{x+1}-1=0\\ =>\sqrt{x+1}=x-1\\ < =>\left\{{}\begin{matrix}x+1=\left(x-1\right)^2\\x-1\ge0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x+1=x^2-2x+1\\x\ge1\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x^2-3x=0\\x\ge1\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\left(x-3\right)=0\\x\ge1\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\x\ge1\end{matrix}\right.< =>x=3\)

Vậy \(S=\left\{3\right\}\)