Tìm số nguyên n: a)n+7 chia hết cho n+2 b)9-n chia hết cho n-3 c)n^2+n+17 chia hết cho n+1 d)n^2 +25 chia hết cho n+2

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1. a,

   $n+7⋮n+2$

   ⇔ $n+2+5⋮n+2$

   ⇒ $n+2∈Ư(5)= {-5;-1;1;5}$

   ⇔ $n∈{-7;-3;-1;3}$

   b,

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   c,

   $n²+n+17⋮n+1$

   ⇔ $n.(n+1)+17⋮n+1$

   ⇒ $n+1∈Ư(17)={-17;-1;1;17}$

   ⇔ $n∈{-18;-2;0;16}$

   d,

   $n²+25⋮n+2$

   ⇔ $n²-4+29⋮n+2$

   ⇔ $(n-2).(n+2)+29⋮n+2$

   ⇒ $n+2∈Ư(29)={-29;-1;1;29}$

   ⇔ $n∈{-31;-3;-1;27}$

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2. a, Ta có: n+7$\vdots$n+2

   ⇒n+2+5$\vdots$n+2

   ⇒n+2∈Ư(5)={±1;±5}

   n+2=1⇒n=-1

   n+2=-1⇒n=-3

   n+2=5⇒n=3

   n+2=-5⇒n=-7

   Vậy n∈{-1;-3;3;-7}

   b, Ta có: 9-n$\vdots$n-3

   ⇒-(n-3)+6$\vdots$n-3

   ⇒n-3∈Ư(6)={±1;±2;±3;±6}

   n-3=-6⇒n=-3

   n-3=-3⇒n=0

   n-3=-2⇒n=1

   n-3=-1⇒n=2

   n-3=1⇒n=4

   n-3=2⇒n=5

   n-3=3⇒n=6

   n-3=6⇒n=9

   Vậy n={0,1,2,-3,4,5,6,9}

   c, Ta có: n²+n+17$\vdots$n+1

   ⇒n(n+1)+17$\vdots$n+1

   ⇒n+1∈Ư(17)={±1;±17}

   n+1=-17⇒n=-18

   n+1=-1⇒n=-2

   n+1=1⇒n=0

   n+1=17⇒n=16

   Vậy n={-18,-2,0,16}

   d, Ta có: n²+25$\vdots$n+2

   ⇒n²-4-21$\vdots$n+2

   ⇒(n-2)(n+2)-21$\vdots$n+2

   ⇒n+2∈Ư(21)={±1;±3;±7;±21}

   n+2=-21⇒n=-23

   n+2=-7⇒n=-9

   n+2=-3⇒n=-5

   n+2=-1⇒n=-3

   n+2=1⇒n=-1

   n+2=3⇒n=1

   n+2=7⇒n=5

   n+2=21⇒n=19

   Vậy n={-23,-9,-5,-3,-1,1,5,19}

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