[Toán đại 7] Tìm n nguyên

K

kieu6a4nbk123

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N

nguyenbahiep1

[TEX]\frac{n}{2n +1} = \frac{1}{2}.( 1 - \frac{1}{2n +1})\\ 2n +1 = 1\Rightarrow n = 0, 2n +1 = -1 \Rightarrow n =-1[/TEX]
 
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H

harrypham

Ta có $A= \dfrac{n}{2n+1} \in \mathbb{Z} \iff 2A \in \mathbb{Z} \iff \dfrac{2n+1-1}{2n+1} \in \mathbb{Z} \iff 1- \dfrac{1}{2n+1} \in \mathbb{Z} \iff 2n+1 = \pm 1$
Do đó $n= -1,0$. Thử lại với $A$ ta thấy thỏa mãn.
 
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