có:
a<b
c<d
m<n
=> a+c+m<b+d+n
=> a+c+m+a+c+m<a+c+m+b+d+n
=>2(a+c+m) < a+b+c+d+m+n
C1 : có : $\frac{a+c+m}{a+b+c+d+m+n}$ < $\frac{a+c+m}{2(a+c+m)}$ = $\frac{1}{2}$
C2 : có : $\frac{a+c+m}{a+b+c+d+m+n}$ = $\frac{2(a+c+m)}{2(a+b+c+d+m+n)}$ < $\frac{a+b+c+d+m+n}{2(a+b+c+d+m+n)}$ = $\frac{1}{2}$
chúc bn học tốt !