\(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+13}=0\Leftrightarrow\left(x-5\right)^{x+1}.\left[1-\left(x-5\right)^{12}\right]=0\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{x+1}=0\\1-\left(x-5\right)^{12}=0\end{matrix}\right.\)
\(\left(x-5\right)^{x+1}=\Leftrightarrow0\left\{{}\begin{matrix}x-5=0\\x+1\ne0\end{matrix}\right.\Leftrightarrow x=5\)
\(1-\left(x-5\right)^{12}=0\Leftrightarrow\left(x-5\right)^{12}=1\Leftrightarrow\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)
Vậy \(x=4;x=5;x=6\)
Tham khảo:
Ta có:
( x - 5 )x + 1 - ( x - 5 )x + 13 = 0
\(\Leftrightarrow\) ( x - 5 )x + 1 = ( x - 5 )x + 13
\(\Leftrightarrow\) ( x - 5 )x + 1 = ( x - 5 )x + 1 . ( x - 5 )12
\(\Leftrightarrow\) ( x - 5 )12 = 1
\(\Leftrightarrow\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)
Vậy x = 4 hoặc x = 6.
