Bài 4:
a) \(4^{10}+4^7=4^7\left(4^3+1\right)=4^7.65⋮65\)
b) \(10^{10}-10^9-10^8=10^8\left(10^2-10-1\right)=10^8.89⋮89\)
Bài 5:
a) \(\Rightarrow n\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{1;2;4\right\}\)
b) \(\Rightarrow\left(n+2\right)+4⋮\left(n+2\right)\)
\(\Rightarrow\left(n+2\right)\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;2\right\}\)
c) \(\Rightarrow3\left(n-2\right)+7⋮\left(n-2\right)\)
\(\Rightarrow\left(n-2\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{1;3;9\right\}\)
d) \(\Rightarrow2\left(3n+9\right)⋮\left(2n-1\right)\)
\(\Rightarrow3\left(2n-1\right)+21⋮\left(2n-1\right)\)
\(\Rightarrow\left(2n-1\right)\inƯ\left(21\right)=\left\{-21;-7;-3;-1;1;3;7;21\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;1;2;4;11\right\}\)
Bài 4
Ta có: \(4^{10}+4^7=4^7.4^3+4^7=4^7\left(4^3+1\right)=4^7.65\)
Vì \(65⋮65\) nên \(4^{10}+4^7⋮65\)
b) Ta có:\(10^{10}-10^9-10^8=10^8.10^2-10^8.10-10^8=10^8\left(10^2-10-1\right)=10^8.89\)
Vì \(89⋮89\) nên \(10^{10}-10^9-10^8⋮65\)
