Chứng minh:

A

ailatrieuphu

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V

vipboycodon

a) $n^5-n$
= $n(n^4-1)$
= $n(n^2-1)(n^2+1)$
= $n(n-1)(n+1)(n^2-4+5)$
= $(n-2)(n-1)n(n+1)(n+2)+5n(n-1)(n+1)$ chia hết cho 30

b) $n^3+6n^2+8n = n(n^2+6n+8) = n(n+2)(n+4)$ chia hết cho 48 với n chẵn
 
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