a)ta có:\(\frac{a}{b}=\frac{a.\left(b+m\right)}{b.\left(b+m\right)}=\frac{ab+am}{b^2+bm}\)
\(\frac{a+m}{b+m}=\frac{\left(a+m\right)b}{\left(b+m\right)b}=\frac{ab+bm}{bm+b^2}\)
vì a<b =>am<bm=>ab+am<ab+bm
hay\(\frac{a}{b}< \frac{a+m}{b+m}\)
b)tương tự như phần a