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

Toán Lớp 6: Câu 3. Tìm số nguyên n sao cho (n+ 3) chia hết cho (n-1)

Toán Lớp 6: Câu 3. Tìm số nguyên n sao cho (n+ 3) chia hết cho (n-1)

Comments ( 2 )

  1. (n+3)\vdots(n-1)
     <=>n-1+4\vdotsn-1
    <=>4\vdotsn-1
    <=>n-\inƯ(4)={1;-1;2;-2;4;-4}
    <=>[([(n-1=1),(n-1=-1):}),([(n-1=2),(n-1=-2):}),([(n-1=4),(n-1=-4):}):}
    <=>[([(n=2),(n=0):}),([(n=3),(n=-1):}),([(n=5),(n=-3):}):}
    Vậy n\in{(2;0),(3;-1),(5;-3)}

  2. n + 3 ⋮ n – 1
    Ta có : $\left \{ {{n+3⋮n-1} \atop {n-1⋮n-1}} \right.$ 
    ⇒ ( n + 3 ) – ( n – 1 ) ⋮ n – 1 ⇒ 4 ⋮ n – 1 ⇔ n – 1 ∈ Ư ( 4 ) ⇔ n – 1 ∈ { -1 ; 1 ; -2 ; 2 ; -4 ; 4 }
    ⇔ n ∈ {0 ; 2 ; -1 ; 3 ; -3 ; 5 }

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