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

Toán Lớp 8: (9n-2)^2 -(2n-9)^2 chia hết cho 11 với mọi n thuộc Z

Toán Lớp 8: (9n-2)^2 -(2n-9)^2 chia hết cho 11 với mọi n thuộc Z

Comments ( 2 )

  1. $(9n-2)^2 – (2n-9)^{2}$
    $=(9n-2-2n+9) (9n-n+2n-9)^{}$ 
    $=(7n+7) (11n-11)^{}$ 
    $=7 . (n+1) . 11 . (n-1)^{}$ 
    $=11 . [ 7 . (n-1) (n+1) ]^{}$ 
    $=11 . (7n^2-7)^{}$  $\vdots$ $11$ $∀$ $n$ $\in$ Z
     Vậy $(9n-2)^2 – (2n-9)^{2}$ $\vdots$ $11$ $∀$ $n$ $\in$ Z
     

  2. ~ Bạn tham khảo ~
    Ta có :
    (9n-2)^2 – (2n-9)^2
    = (9n-2-2n+9)(9n-2+2n-9)
    = (7n+7)(11n-11)
    = 11(n-1)(7n+7)\vdots11
    Vậy (9n-2)^2 -(2n-9)^2\vdots11∀n\inZZ

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