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khanhvy.hoduong

Bài 1:

a) Ta có:

$A= \dfrac{8n+ 193}{4n+ 3}= \dfrac{8n+ 6+ 187}{4n+ 3}= \dfrac{2.(4n+3)+ 187}{4n+ 3}= 2+ \dfrac{187}{4n+ 3}$

Để $A$ $\in$ $N$ thì $187 \vdots 4n+ 3$

\Rightarrow $4n+ 3$ $\in$ $Ư(187)$

\Rightarrow $4n+ 3$ $\in$ {$17; 11; 187$}

+ $4n+ 3= 11$ \Rightarrow $n= 2$

+ $4n+ 3= 17$ \Rightarrow $n \notin N$ (loại)

+ $4n+ 3= 187$ \Rightarrow $n= 46$

Vậy: Để $A \in N$ thì $n= 2$ hoặc $n= 46$
 
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