\(\Rightarrow\left(x-3\right)^3\left[\left(x-3\right)^2-1\right]=0\)
\(\Rightarrow\left(x-3\right)^3\left(x-2\right)\left(x-4\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=3\\x=4\end{array}\right.\)
Vậy x = 2 ; x = 3 ; x = 4
\(\left(x-3\right)^5=\left(x-3\right)^3\)
\(\Leftrightarrow x-3=1\)
\(\Leftrightarrow x=4\)
\(\left(x-3\right)^5=\left(x-3\right)^3\)
\(\Rightarrow\left(x-3\right)^5-\left(x-3\right)^3=0\\ \)
\(\left(x-3\right)^3\left[\left(x-3\right)^2-1\right]=0\)
\(\Rightarrow\begin{cases}\left(x-3\right)^3=0\\\left(x-3\right)^2-1=0\end{cases}\)
\(\Rightarrow\begin{cases}x=3\\\left(x-3\right)^2=1\end{cases}\)
\(\Rightarrow\begin{cases}x=3\\x=4;2\end{cases}\)
Vậy x= {3;4;2}
