\(\left(x-2\right)^{x+3}=\left(x-2\right)^{x+1}\)
\(\Leftrightarrow\left(x-2\right)^{x+3}-\left(x-2\right)^{x+1}=0\)
\(\Leftrightarrow\left(x-2\right)^{x+1}\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^{x+1}=0\\\left(x-2\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x-2=1;x-2=-1\end{cases}}\)
Bt trên đúng \(\Leftrightarrow x=2;x=3;x=1\)
\(\left(x-2\right)^{x+3}=\left(x-2\right)^{x+1}\)
\(< =>\frac{\left(x-2\right)^x.\left(x-2\right)^3}{\left(x-2\right)^x.x-2}=1\)
\(< =>\frac{\left(x-2\right)^3}{x-2}=1< = >\left(x-2\right)^2=1\)
\(< =>\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}< =>\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
(x-2)x+3=(x-2)x+1
(x-2)x+3-(x-2)x+1=0
(x-2)x+1.[(x-2)2-1]=0
\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{x+1}=0\\\left(x-2\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3;x=1\end{cases}}\)
Vậy \(x\in\left\{1;2;3\right\}\)
Ta có:
\(\left(x-2\right)^{x+3}=\left(x-2\right)^{x+1}\)
\(\Rightarrow\left(x-2\right)^{x+3}-\left(x-2\right)^{x+1}=0\)
\(\Rightarrow\left(x-2\right)^{x+1}.\left[\left(x-2\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{x+1}=0\\\left(x-2\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3;x=1\end{cases}}\)
Vậy \(x\in\left\{1;2;3\right\}\)
\(\left(x-2\right)^{x+3}=\left(x-2\right)^{x+1}\)
\(\Leftrightarrow\left(x-2\right)^{x+3}-\left(x-2\right)^{x+1}=0\)
\(\Leftrightarrow\left(x-2\right)^{x+1}\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^{x+1}=0\\\left(x-2\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x-2=\pm1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3;1\end{cases}}}\)
