\(\dfrac{5}{3}=\dfrac{35}{21}\)
\(\dfrac{8}{7}=\dfrac{24}{21}\)
do 35>24 nên \(\dfrac{5}{3}>\dfrac{8}{7}\)→a>b
\(ta.c\text{ó}:BCNN=21\\ \dfrac{5}{3}=\dfrac{5\cdot7}{3\cdot7}=\dfrac{35}{21};\dfrac{8}{7}=\dfrac{8\cdot3}{7\cdot3}=\dfrac{24}{21}\\ 35>24\\ \Rightarrow\dfrac{5}{3}>\dfrac{8}{7}hay.a>b\)