Question

Tìm x
\(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\)

 

Answer 1

\(\left(x+1\right)^{x+2}=\left(x+1\right)^{x+3}\)

\(\Rightarrow\left(x+1\right)^{x+1}-\left(x+1\right)^{x+3}=0\)

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

+) \(\left(x+1\right)^{x+1}=0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

+) \(1-\left(x+1\right)^2=0\)

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

\(\Rightarrow x+1=\pm1\)

+ \(x+1=1\Rightarrow x=0\)

+ \(x+1=-1\Rightarrow x=-2\)

Vậy \(x\in\left\{-2;-1;0\right\}\)