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

Toán Lớp 8: Tìm tất cả số nguyên dương n sao cho n^2 +9n−2 chia hết cho 11.

Toán Lớp 8: Tìm tất cả số nguyên dương n sao cho n^2 +9n−2 chia hết cho 11.

Comments ( 2 )

  1. $n^2+9n-2$
    $= (n^2+9n+9)-11$
    $= (n+3)^2-11$
    Để $(n+3)^2-11 \;\vdots\; 11$
    $\to \begin{cases}(n+3)^2\;\vdots\; 11\\11\;\vdots\; 11\end{cases}$
    $\to n+3 \;\vdots\; 11$
    $\to n+3 \in B(11)=\{0;11;22;33;44;….\}$
    $\to n\in \{8;19;30;41;….\}$

  2. n^2 +9n -2 = (n^2+9n+9) -11 = (n+3)^2 – 11 \vdots 11
    Mà 11 \vdots 11 -> (n+3)^2 \vdots 11
    -> n +3 \in B(11) = {0 ; 11 ; 22 ; 33 ; 44 ; 55 ; … }
    -> n \in {-3 ; 8 ; 19 ; 30 ; 41 ; 52 ; … }
    Do n \in Z+
    -> n \in {8 ; 19 ; 30 ; 41 ; 52 ; … }

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