a, Có\(\frac{3n+2}{n}=3+\frac{2}{n}\)
Vì \(3\inℤ\)=> Để \(a\inℤ\)thì \(\frac{2}{n}\inℤ\)<=> \(n\in U\left(2\right)=\left\{\pm1;\pm2\right\}\)
b, Có
\(\frac{a+n}{b+n}=1-\frac{b-a}{b+n}\)
\(\frac{a}{b}=1-\frac{b-a}{b}\)
Vì\(b+n\ge b\)=> \(\frac{b-a}{b+n}\le\frac{b-a}{b}\)=> \(1-\frac{b-a}{b+n}\ge1-\frac{b-a}{b}\)=> \(\frac{a+n}{b+n}\ge\frac{a}{b}\)