n^2 + 2n – 3 chia hết cho n + 1 3n + 1 chia hết cho 2n -1

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1. Đáp án:

   1.$n\in\{0,-2,1,-3,3,-5\}$

   2.$n\in\{1,3,0,-2\}$ 

   Giải thích các bước giải:

   1.$n^2+2n-3\quad\vdots\quad n+1$

   $\leftrightarrow (n+1)^2-4\quad\vdots\quad n+1$

   $\leftrightarrow 4\quad\vdots\quad n+1$

   $\leftrightarrow n+1\in\{1,-1,2,-2,4,-4\}$

   $\leftrightarrow n\in\{0,-2,1,-3,3,-5\}$

   2.$3n+1\quad\vdots\quad 2n-1$

   $\leftrightarrow 2(3n+1)\quad\vdots\quad 2n-1$

   $\leftrightarrow 6n+2\quad\vdots\quad 2n-1$

   $\leftrightarrow 6n-3+5\quad\vdots\quad 2n-1$

   $\leftrightarrow 3(2n-1)+5\quad\vdots\quad 2n-1$

   $\leftrightarrow 5\quad\vdots\quad 2n-1$ 

   $\leftrightarrow 2n-1\in\{1,5,-1,-5\}$ 

   $\leftrightarrow n\in\{1,3,0,-2\}$ 

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