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

Toán Lớp 7: Bài 7: Tìm n nguyên dương sao cho 2n – 3 ⋮ n + 1

Toán Lớp 7: Bài 7: Tìm n nguyên dương sao cho 2n – 3 ⋮ n + 1

Comments ( 2 )

  1. Giải đáp:
    (2n -3) \vdots ( n +1)
    = (2n + 2) – 5 \vdots (n+1)
    = 2(n +1) -5 \vdots (n+1)
    => (n +1) in Ư(5) ={ ±1 ; ±5}
    Ta có bảng sau:
    \begin{array}{|c|c|c|}\hline \text{n +1}&\text{1}&\text{-1}&\text{5}&\text{-5}\\\hline \text{n}&\text{0(KTM)}&\text{-2(KTM)}&\text{4(TM)}&\text{-6(KTM)}\\\hline\end{array}
    Vậy n =4

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

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