Question

Tìm X Biết:

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

Answer 1

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

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

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

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

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

vậy..........

Answer 2

\(x+1=\left(x+1\right)^2\) \(\Leftrightarrow x+1=x^2+2x+1\)
\(\Leftrightarrow x^2+2x+1-x-1=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x+1=0\)
\(\Leftrightarrow x\in\left\{0;-1\right\}\)
Vậy \(x\in\left\{0;-1\right\}\)