toan hinh

H

harrypham

Ta có [TEX]\frac{n^3-8n^2+2n}{n^2+1}= \frac{n+8}{n^2+1}+n-8[/TEX].
Khi đó [TEX]\frac{n+8}{n^2+1}[/TEX] nguyên.
Ta tìm được [TEX]n \in \{ 2,0 \}[/TEX]
 
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