Question

{x+2}^n+1={x+2}^n+11

Tìm X.Các ban giúp mik với 

Answer 1

#)Giải :

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

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

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

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

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

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

Answer 2

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