Ta có : (x - 4)2 = (x - 4)4
=> (x - 4)4 - (x - 4)2 = 0
=> (x - 4)2.[(x - 4)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-4=0\\\left(x-4\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x-4=0\\x-4=\pm1\end{cases}}\)
Nếu x - 4 = 0
=> x = 4
Nếu x - 4 = 1
=> x = 5
Nếu x - 1 = -1
=> x = 3
Vậy \(x\in\left\{3;4;5\right\}\)là giá trị cần tìm
( x - 4 )2 = ( x - 4 )4
<=> ( x - 4 )4 - ( x - 4 )2 = 0
<=> ( x - 4 )2[ ( x - 4 )2 - 1 ] = 0
<=> \(\orbr{\begin{cases}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{cases}}\)
+) ( x - 4 )2 = 0 => x - 4 = 0 => x = 4
+) ( x - 4 )2 - 1 = 0
=> ( x - 4 )2 = 1 = ( ±1 )2
=> \(\orbr{\begin{cases}x-4=1\\x-4=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)
=> x ∈ { 3 ; 4 ; 5 }
Ta có: \(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)^4=0\)
\(\Leftrightarrow\left(x-4\right)^2\left[1-\left(x-4\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{cases}}\Leftrightarrow x\in\left\{3;4;5\right\}\)