\(\left(x-3\right)^{x+2017}-\left(x-3\right)^{x+2019}=0\)
\(\Leftrightarrow\left(x-3\right)^{x+2017}=\left(x-3\right)^{x+2019}\).
Nếu \(x-3=0\Leftrightarrow x=3\). Khi đó:
\(0^{3+2017}=0^{3+2019}\)\(\Leftrightarrow0=0\) (luôn đúng).
Nếu \(x-3\ne0\Leftrightarrow x\ne3\). Khi đó:
\(\left(x-3\right)^{x+2017}=\left(x-3\right)^{x+2019}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)^{x+2019}}{\left(x-3\right)^{x+2017}}=1\)\(\Leftrightarrow\left(x-3\right)^2=1\)\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\).
Vậy \(x\in\left\{-1;1;3\right\}\).
