Question

 

Tìm x biết : (x+2)^n+1=(x+2)^n+11(với n là số tự nhiên)

Các bn giúp mk nha

Answer 1

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

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

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

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

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

Với \(1-\left(x+2\right)^{10}=0\Rightarrow\left(x+2\right)^{10}=1\Rightarrow\orbr{\begin{cases}x+2=1\\x+2=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)