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

Toán Lớp 6: Tìm n ∈ N để : 3n + 14 chia hết cho n + 1

Toán Lớp 6: Tìm n ∈ N để :
3n + 14 chia hết cho n + 1

Comments ( 2 )

  1. 3n + 14 ⋮ n + 1
    Ta có : $\left \{ {{3n + 14⋮ n+1} \atop {n+1⋮n+1}} \right.$ ⇒ $\left \{ {{3n+14⋮n+1} \atop {3(n+1)⋮n+1}} \right.$ ⇔ $\left \{ {{3n+14⋮n+1} \atop {3n+3⋮n+1}} \right.$ 
    ⇒ ( 3n + 14 ) – ( 3n + 3 ) ⋮ n + 1 ⇒ 3n + 14 – 3n – 3 ⋮ n + 1 ⇔ 11 ⋮ n + 1 ⇔ n + 1 ∈ Ư ( 11 )
    ⇔ n + 1 ∈ { 1 ; 11 } ⇔ n ∈ { 0 ; 10 }
    Vậy với n ∈ { 0 ; 10 } thì 3n + 14 ⋮ n + 1

  2. (3n+14)\vdots(n+1)
    =>(3n+3)+11\vdots(n+1)
    =>3(n+1)+11\vdots(n+1)
    Vì 3(n+1)\vdots(n+1)
    =>11\vdots(n+1)
    =>(n+1)\in\text{Ư}(11)={1;11}
    =>n\in{0;10}

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