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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: Tìm n biết: $n^{2}$+10 chia hết n+1

Toán Lớp 6: Tìm n biết:
$n^{2}$+10 chia hết n+1

Comments ( 1 )

  1. Thêm điều kiện n là số tự nhiên (n∈N)
    Ta có: n + 1 vdots n+1 
    => n^2 + n vdots n+1
    => n^2 + n – (n^2 + 10) vdots n+1
    => n^2 + n – n^2 – 10 vdots n+1
    => n – 10 vdots n+1
    => n + 1 – (n – 10) vdots n+1 
    => n + 1 – n + 10 vdots n+1
    => 11 vdots n+1
    mà n ∈ N => n + 1 ∈ N*
    => n + 1 ∈ Ư(11) = {1;11}
    => n ∈ {0;10}
    Vậy n = 0;10 thì n^2 + 10 vdots n+1
     

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222-9+11+12:2*14+14 = ? ( )

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