Question

[Nguyễn Đức Anh]

 Tìm x biết :

\(x+1=\left(x+1\right)^2\)

Answer 1

[Võ Đông Anh Tuấn]

\(x+1=\left(x+1\right)^{^2}\)

\(\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0\)

\(=\left(x+1\right)x=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)

Answer 2

[Tài Nguyễn Tuấn]

Ta có : 

\(x+1=\left(x+1\right)^2\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\x+1=1\end{array}\right.\)

\(=>\left[\begin{array}{nghiempt}x=0-1=-1\\x=1-1=0\end{array}\right.\)

Vậy \(x\in\left\{-1;0\right\}\)