17:
a: \(n+8⋮n+3\)
=>\(n+3+5⋮n+3\)
=>\(n+3\inƯ\left(5\right)\)
=>\(n+3\in\left\{1;-1;5;-5\right\}\)
=>\(n\in\left\{-2;-4;2;-8\right\}\)
mà n>=0
nên n=2
b: \(A=5+5^2+5^3+...+5^8\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)\)
\(=30\left(1+5^2+5^4+5^6\right)⋮30\)
a. ta có :
\(\text{n + 8 ⋮ n + 3}\)
\(\Rightarrow\left(n+3\right)+5⋮\left(n+3\right)\)
\(\Rightarrow5⋮n+3\)
\(\Rightarrow n+3\in\left\{1;5\right\}\)
\(+\text{)}n+3=1\rightarrow n=-2\left(KTM\right)\)
\(+\text{)}n+3=5\rightarrow n=2\left(TM\right)\)
Vậy \(n=2\)
b. \(\text{A = 5 + 5^2 + 5^3 + ... + 5^8}⋮30\)
\(\text{A = ( 5 + 5^2 ) + 5^2 . ( 5 + 5^2 ) + ... + 5^6 . ( 5^7 + 5^8 )}\)
\(\text{A = 30 + 5^2 . 30 + ... + 5^6 . 30}\)
\(\text{A = 30 . ( 1 + 5^2 + ... + 5^6 )}⋮30\left(đpcm\right)\)
