Question

Tìm x, biết

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

Answer 1

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

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

\(\Rightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\)

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

Answer 2

Tìm x, biết:

x + 1 = (x + 1)²

(x + 1) - (x + 1)² = 0

(x + 1)[1 - (x + 1)] = 0

(x + 1)(1 - x - 1) = 0

(x + 1)(- x) = 0

Suy ra: \(\left\{{}\begin{matrix}x+1=0\\-x=0\end{matrix}\right.\left\{{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{-1;0\right\}\).