(x - 5)4 = (x - 5)6
=> (x - 5)6 - (x - 5)4 = 0
=> (x - 5)4.[(x - 5)2 - 1] = 0
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x-5=0\\\left(x-5\right)^2=1\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x-5=0\\x-5=1\\x-5=-1\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=5\\x=6\\x=4\end{array}\right.\)
Vậy \(x\in\left\{4;5;6\right\}\)
