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

Toán Lớp 6: Tìm n ∈ N để : 8n + 11 ⋮ 2n + 1

Toán Lớp 6: Tìm n ∈ N để :
8n + 11 ⋮ 2n + 1

Comments ( 2 )

  1. Ta có : 2n + 1 vdots 2n + 1
    ⇔ 4 . ( 2n + 1 ) vdots 2n + 1
    ⇔ 8n + 4 vdots 2n + 1
    ⇒ ( 8n + 11 ) – ( 8n + 4 ) vdots 2n + 1
    ⇔ 7 vdots 2n + 1
    ⇔ 2n + 1 ∈ Ư( 7 ) = { 1 ; 7 ; – 1 ; – 7 }
    ⇔ n ∈ { 0 ; 3 ; – 1 ; – 4 }
    Mà n ∈ N ⇒ n ∈ { 0 ; 3 }
    Vậy , n ∈ { 0 ; 3 } .

  2. 8n + 11 ⋮ 2n + 11
    Ta có : $\left \{ {{8n+11⋮2n+1} \atop {2n+1⋮2n+1}} \right.$ ⇒ $\left \{ {{8n+11⋮2n+1} \atop {4(2n+1)⋮n+1}} \right.$ ⇒ $\left \{ {{8n+11⋮2n+1} \atop {8n+4⋮2n+1}} \right.$ 
    ⇒ ( 8n + 11 ) – ( 8n + 4 ) ⋮ 2n + 1 ⇒ 7 ⋮ 2n + 1 ⇔ 2n + 1 ∈ Ư ( 7 ) ⇔ 2n + 1 ∈ { 1 ; 7 } ⇔ 2n ∈ { 0 ; 6 }
    ⇒ n ∈ { 0 ; 3 }
    Vậy với n ∈ { 0 , 3 } thì 8n + 11 ⋮ 2n + 1

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

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