Question

 

 Tìm x biết :

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

Answer 1

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

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

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

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

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

Answer 2

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 \(x\in\left\{-1;0\right\}\)

Answer 3

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

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

<=>\(\left(x+1\right).\left(-x\right)=0\)

<=>x=-1 hoặc x=0(nhận cả 2)

Vậy S={-1;0}