b: Đặt \(A=9\cdot10^n+18\)
\(=9\left(10^n+2\right)\)
\(=9\cdot\left(10000...2\right)\)
Vì \(1000...2\) chia hết cho 3
nên \(A=9\cdot\left(1000...2\right)⋮3\cdot9\)
=>\(A⋮27\)
a: \(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+13}=0\)
=>\(\left(x-5\right)^{x+13}-\left(x-5\right)^{x+1}=0\)
=>\(\left(x-5\right)^{x+1}\cdot\left[\left(x-5\right)^{12}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{12}-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=-1+5=4\end{matrix}\right.\)
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