Ta có:2n-1 là bội của n+3
=>2n-1 chia hết cho n+3
Ta có 2n-1=n+n-1
=n+n+3+3-1-6
=2(n+3)-(1+6)
=2(n+3)-7
Vì 2(n+3) chia hết cho n+3 nên để 2n-1 chia hết cho n+3 thì 7 phải chia hết cho n+3
=>n+3\(\in\)Ư(7)={-1;-7;1;7}
n+3 | -1 | -7 | 1 | 7 |
n | -4 | -10 | -2 | 4 |
=> x={-4;-10;-2;4}
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tìm số nguyên n sao cho 2n-1 là bội của n+2