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

Toán Lớp 6: Tìm số tự nhiên n: a, 20 chia hết cho 2n + 1 b, 12 chia hết cho n – 1 c, n . ( n + 1 ) = 6

Toán Lớp 6: Tìm số tự nhiên n:
a, 20 chia hết cho 2n + 1
b, 12 chia hết cho n – 1
c, n . ( n + 1 ) = 6

Comments ( 2 )

  1. a. $20\;\vdots\; 2n+1$
    $⇒ 2n+1\in Ư(20)=\{1;2;4;5;10;20\}$
    $⇒ 2n \in \{0;1;3;4;9;19\}$
    $⇒ n\in \{0;2\}$
    b. $12\;\vdots\;n-1$
    $⇒ n-1\in Ư(12)=\{1;2;3;4;6;12\}$
    $⇒ n\in \{2;3;4;5;7;13\}$
    c. $n(n+1)=6$
    $⇒ n^2+n-6=0$
    $⇒ n^2-2n+3n-6=0$
    $⇒ n(n-2)+3(n-2)=0$
    $⇒ (n+3)(n-2)=0$
    $⇒ n=2$

  2.  a, Để 20 chia hết cho 2n+1
    => 2n+1∈ Ư(20)={1; 2; 4; 5; 10; 20} (Vì n∈ N)
    => 2n∈ {0; 1; 3; 4; 9; 19}
    => n∈ {0; 1/2; 3/2; 2; 9/2; 19/2}
    Mà n∈ N
    => n∈ {0; 2}
    Vậy x∈ {0;2} thì 20 chia hết cho 2n+1
    b, Để 12 chia hết cho n-1
    =>  n-1∈ Ư(12)={1; 2; 3; 4; 6; 12}  (Vì n∈ N)
    => n∈ {2; 3; 4; 5; 7; 13}
    Vậy n∈ {2; 3; 4; 5; 7; 12} thì 12 chia hết cho n-1
    c, n(n+1)=6
    => n^2+n-6=0
    => n^2-2n+3n-6=0
    => n(n-2)+3(n-2)=0
    => (n+3)(n-2)=0
    => $\left[\begin{matrix} n+3=0\\ n-2=0\end{matrix}\right.$
    => $\left[\begin{matrix} n=-3\\n=2 \end{matrix}\right.$
    Vậy N∈ {-3;2}
    XIN CTLHN

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