tìm các số nguyên n sao cho a , 3n + 2 : n – 1 b , 3n + 24 : n – 4 c, n^2 + 5 : n + 1

tìm các số nguyên n sao cho
a , 3n + 2 : n – 1
b , 3n + 24 : n – 4
c, n^2 + 5 : n + 1

0 bình luận về “tìm các số nguyên n sao cho a , 3n + 2 : n – 1 b , 3n + 24 : n – 4 c, n^2 + 5 : n + 1”

  1. `a)“ 3n+2 \vdots n-1`

    `⇔5 \vdots n-1`

    `⇔n-1∈{±5;±1}`

    `⇔n∈{6;-4;2;0}`

    `b) 3n+24 \vdots n-4`

    `⇔36 \vdots n-4`

    `⇔n-4∈{±36;±1;±2;±3;±4;±6;±12;±9;±18}`

    `⇔n∈{40;-32;6;2;5;3;7;0;10;-2;16;-8;13;4;22;14}`

    `c)n^2+5 \vdots n+1`

    `⇔(n+1)^2 -2n-2+2+4 \vdots n+1`

    `⇔6 \vdots n+1`

    `⇔n+1∈{±1;±2;±3;±6}`

    `⇔n∈{0;1;2;5;-2;-3;-4;-7}`

    $@FBoy24$

    Bình luận
  2. `a)(3n+2)`$\vdots$`n-1`

    `⇒[3(n-1)+5]`$\vdots$`n-1`

    `⇒5`$\vdots$`n-1`

    `⇒n-1∈Ư(5)={±1,±5}`

    `⇒n∈{2,0,6,-4}`

    `b)(3n+24)`$\vdots$`n-4`

    `⇒[3(n-4)+36]`$\vdots$`n-4`

    `⇒36`$\vdots$`n-4`

    `⇒n-4∈Ư(36)={±1,±2,±3,±4,±6,±9,±12,±18,±36}`

    `⇒n∈{5,3,6,2,7,1,8,0,10,-2,13,-5,16,-8,22,-14,40,-32}`

    `c)(n^2+5)`$\vdots$`n+1`

    `⇒ [(n^2-1)+6]` $\vdots$`n+1`

    `⇒[(n-1)(n+1)+6]` $\vdots$`n+1`

    `⇒6` $\vdots$`n+1`

    `⇒n+1∈Ư(6)={±1,±2,±3,±6}`

    `⇒n∈{0,-2,1,-3,2,-4,5,-7}`

     

     

    Bình luận

Viết một bình luận