=>(x-5)6/(x-5)4=1
=> (x-5)2=1
=> x-5 =1 hoac -1
=>x=6 hoac 4
K NHA
Ta có : \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Rightarrow\left(x-5\right)^4\left(1-\left(x-5\right)^2\right)=0\)
\(\Rightarrow\left(x-5\right)^4\left(1+x-5\right)\left(1-x+5\right)=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(6-x\right)=0\)
\(\hept{\begin{cases}\left(x-5\right)^4=0\\x-4=0\\6-x=0\end{cases}}\Rightarrow\hept{\begin{cases}x-5=0\\x=4\\x=6\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\x=4\\x=6\end{cases}\left(TM\right)}\)
Vậy \(x=4;x=5;x=6\)
