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

Toán Lớp 6: Tìm n thuộc N để : a) n + 8 : n – 3 b) 2n : n – 1 c) 2n + 5 : n + 1 d ) 3n +4 : n – 1

Toán Lớp 6: Tìm n thuộc N để :
a) n + 8 : n – 3
b) 2n : n – 1
c) 2n + 5 : n + 1
d ) 3n +4 : n – 1

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:

    a, n+8 vdots n-3

    <=> (n-3) + 8 + 3 vdots n-3

    <=> 11 vdots n-3 ( Vì n-3 vdots n-3 )

    => n-3 in Ư(11) = { -1; 1; -11; 11 }

    => n in { 2; 4; -8; 14 }

    -Mà n in N => n in {2;4;14}

    b, 2n vdots n-1

    <=> 2 . (n-1) + 2 vdots n-1

    <=> 2 vdots n-1 ( Vì 2.(n-1) vdots n-1 )

    => n-1 in Ư(2) = { -1; 1; -2; 2 }

    => n in { 0; 2; -1; 3 }

    -Mà n in N => n in { 0; 2; 3 }

    c, 2n+5 vdots n+1

    <=> 2.(n+1) – 2 + 5 vdots n+1

    <=> 3 vdots n+1 ( Vì 2.(n+1) vdots n+1 )

    => n+1 in Ư(3) = { -1; 1; -3; 3 }

    => n in { -2; 0; -4; 2 }

    -Mà n in N => n in { 0; 2 }

    d, 3n+4 vdots n-1

    <=> 3.(n-1) + 3 + 4 vdots n-1

    <=> 7 vdots n-1 ( Vì 3.(n-1) vdots n-1 )

    => n-1 in Ư(7) = { -1; 1; -7; 7 }

    => n in { 0; 2; -6; 8 }

    -Mà n in N => n in { 0; 2; 8 }

  2. Tham khảo:

    a)

    n + 8 \vdots n – 3

    => n – 3 + 11 \vdots n – 3

    Mà n – 3 \vdots n – 3

    => 11 \vdots n – 3

    => n – 3 \in Ư(11) = { -11; -1; 1; 11 }

    => n \in { -8; 2; 4; 14 }

    Mà n \in NN => n \in { 2; 4; 14 }

    b)

    2n \vdots n – 1

    => 2n – 1 + 1 \vdots n – 1

    Mà 2n – 1 \vdots n – 1

    => 1 \vdots n – 1

    => n – 1 \in Ư(1) = {-1; 1}

    => n \in { 0; 2 }

    c)

    2n + 5 \vdots n + 1

    => 2n + 1 + 4 \vdots n + 1

    Mà 2n + 1 \vdots n + 1

    => 4 \vdots n + 1

    => n + 1 \in Ư(4) = {-4; -2; -1; 1; 2; 4}

    => n \in { -5; -3; -2; 0; 1; 3 }

    Mà n \in NN => n \in { 0; 1; 3 }

    d)

    3n + 4 \vdots n – 1

    => 3n – 1 + 5 \vdots n – 1

    Mà 3n – 1 \vdots n – 1

    => 5 \vdots n – 1

    => n – 1 \in Ư(5) = { -5; -1; 1; 5 }

    => n \in { -4; 0; 2; 6 }

    Mà n \in NN => n \in { 0; 2; 6 }

    #Sad

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