Tìm số tự nhiên n để: a, n ² + 5 chia hết n – 2 b, 5n chia hết n – 1

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1. Đáp án:

   a) n²+5⋮n−2n+5⋮n−2

   =n(n−2)+(2n+5)⋮n−2=n(n−2)+(2n+5)⋮n−2

   Vì n(n−1)⋮n−2⇒2n+5⋮n−2n(n−1)⋮n−2⇒2n+5⋮n−2

   ⇒n−2+7⋮n−2⇒n−2+7⋮n−2

   Vì n−2⋮n−2⇒7⋮n−2n−2⋮n−2⇒7⋮n−2

   ⇒n−2∈Ư(7)={±1;±7}⇒n−2∈Ư(7)={±1;±7}

   ⋅ n−2=−1⇒n=1· n−2=−1⇒n=1

   ⋅ n−2=1⇒n=3· n−2=1⇒n=3

   ⋅ n−2=−7⇒n=−5· n−2=−7⇒n=−5

   ⋅ n−2=7⇒n=9· n−2=7⇒n=9

   b) 5n⋮n−15n⋮n−1

   =5n−5+5⋮n−1=5n−5+5⋮n−1

   =5(n−1)+5⋮n−1=5(n−1)+5⋮n−1

   Vì 5(n−1)⋮n−1⇒5⋮n−15(n−1)⋮n−1⇒5⋮n−1

   ⇒n−1∈Ư(5)={±1;±5}⇒n−1∈Ư(5)={±1;±5}

   ⋅ n−1=−1⇒n=0· n−1=−1⇒n=0

   ⋅ n−1=1⇒n=2· n−1=1⇒n=2

   ⋅ n−1=−5⇒n=−4· n−1=−5⇒n=−4

   ⋅ n−1=5⇒n=6

    

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2. Đáp án:

   a) $n^2 + 5\;\vdots\; n – 2$

   $= n(n-2)+(2n+5)\;\vdots\; n – 2$

   Vì $n(n-1)\;\vdots\; n – 2 ⇒ 2n+5\;\vdots\; n – 2$

   $⇒n-2+7\;\vdots\; n – 2$

   Vì $n-2\;\vdots\; n – 2 ⇒ 7\;\vdots\; n – 2$

   $⇒n-2\in Ư(7)=\{±1;±7\}$

   $·\ n-2=-1 ⇒ n=1$

   $·\ n-2=1 ⇒ n=3$

   $·\ n-2=-7 ⇒n=-5$

   $·\ n-2=7⇒n=9$

   b) $5n\;\vdots\; n-1$

   $= 5n-5+5\;\vdots\; n-1$

   $= 5(n-1)+5\;\vdots\; n-1$

   Vì $5(n-1)\;\vdots\; n-1 ⇒ 5\;\vdots\; n-1$

   $⇒n-1\in Ư(5)=\{±1;±5\}$

   $·\ n-1=-1 ⇒ n=0$

   $·\ n-1=1 ⇒ n=2$

   $·\ n-1=-5⇒n=-4$

   $·\ n-1=5⇒n=6$

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