Ta có: A=\(\frac{\frac{\left(2m+2\right)\left[\frac{\left(2m-2\right)}{2}+1\right]}{2}}{m}\)=\(\frac{\frac{2\left(m+1\right)m}{2}}{m}=\frac{\left(m+1\right)m}{m}=m+1\)
B=\(\frac{\frac{\left(2n+2\right)\left[\frac{\left(2n-2\right)}{2}+1\right]}{2}}{m}=\frac{\frac{2\left(n+1\right)n}{2}}{n}=\frac{\left(n+1\right)n}{n}=n+1\)
Mà A>B
=>m+1>n+1
=>m>n