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

Toán Lớp 6: Tìm n thuộc N để: a) n+8 chia hết cho n+3 b) n+6 chia hết cho n-1

Toán Lớp 6: Tìm n thuộc N để:
a) n+8 chia hết cho n+3
b) n+6 chia hết cho n-1

Comments ( 2 )

  1. a) n+8 \vdots n+3
    ⇒(n+3)+5 \vdots n+3
    ⇒5 \vdots n+3
    Hay n+3 \in Ư(5)={-5;-1;1;5}
    ⇒n={-8;-4;-2;2} 
    ⇒n=2 ( Vì n\in N )
    b) n+6 \vdots n-1
    ⇒(n-1)+7 \vdots n-1
    ⇒7 \vdots n-1
    Hay n-1 \in Ư(7)={-7;-1;1;7}
    ⇒n={-6;0,2,8} 
    ⇒n={0,2,8} (Vì n\in N)
     

  2. Answer
    $\text{Ta có:}$ n + 8 \vdots n + 3
    $\text{Mà:}$ n + 3 \vdots n + 3
    $\text{Lập hiệu:}$
    (n + 8) – (n + 3)
    = n + 8 – n – 3
    = (n- n) + (8 – 3)
    = 5 \vdots n + 3
    $\text{Vì}$ n \in NN $\text{nên}$ n + 3 \in NN
    => n + 3 \in Ư(5) = {-1 ; 1 ; ;-5 ; 5}
    => n + 3 \in {-1 ; 1 ; ;-5 ; 5}
    => n \in {-4 ; -2 ; -8 ; 2}
    $\text{Vậy}$ n = 2
    _________________________
    $\text{Ta có:}$ n + 6 \vdots n – 1
    $\text{Mà:}$ n – 1 \vdots n – 1
    $\text{Lập hiệu:}$
    (n + 6) – (n – 1)
    = n + 6 – n + 1
    = (n – n) + (6 + 1)
    = 7 \vdots n – 1
    $\text{Vì}$ n \in NN $\text{nên}$ n – 1 \in NN
    => n – 1 \in Ư(7) = {-1 ; 1 ; ;-7 ; 7}
    => n – 1 \in {-1 ; 1 ; ;-7 ; 7}
    => n \in {0 ; 2 ; -6 ; 8}
    $\text{Vậy}$ n \in {0 ; 2 ; 8}

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

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