Bg
Ta có: \(M=\left\{k\in N\left|0< \frac{3k+1}{2}< 10\right|\right\}\)
Xét 0 < \(\frac{3k+1}{2}\)< 10:
Vì \(\frac{3k+1}{2}\)< 10 nên \(\frac{3k+1}{2}\)< 10
=> 3k + 1 < 10 x 2
=> 3k + 1 < 20
=> 3k < 20 - 1
=> 3k < 19
=> k < 19 : 3
=> k <\(\frac{19}{3}\)
=> 3k < 18 (vì 18 \(⋮\)3) (đổi thành bé hơn hoặc bằng)
=> k < 18 : 3
=> k < 6
=> M = {0; 1; 2; 3; 4; 5; 6}