Question

Tìm x, biết : \(\left(x+2\right)^{n+1}=\left(x+2\right)^{n+11}\) với n là số tự nhiên

Answer 1

\(\left(x+2\right)^{n+1}=\left(x+2\right)^{n+11}\)

\(\Rightarrow\left(x+2\right)^{n+11}-\left(x+2\right)^{n+1}=0\)

\(\Rightarrow\left(x+2\right)^{n+1}\left[\left(x+2\right)^{10}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)^{n+1}=0\\\left(x+2\right)^{10}-1=0\end{matrix}\right.\)

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

+) \(\left(x+2\right)^{10}-1=0\Rightarrow\left(x+2\right)^{10}=1\)

\(\Rightarrow\left[{}\begin{matrix}x+2=1\\x+2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{-2;-1;-3\right\}\)