1.
a) \(\dfrac{5}{4}\) và \(\dfrac{5}{3}\)
Có : Cùng tử 5 mà 4 > 3
=> \(\dfrac{5}{4}< \dfrac{5}{3}\)
b) \(\dfrac{97}{96}\) và \(\dfrac{98}{99}\)
Có : \(\dfrac{97}{96}>1\) (tử lớn hơn mẫu)
\(\dfrac{98}{99}< 1\) ( tử bé hơn mẫu)
\(\Rightarrow\dfrac{97}{96}>\dfrac{98}{99}\)
c)
\(\dfrac{6}{7}=\dfrac{6\times8}{7\times8}=\dfrac{48}{56}\) ; \(\dfrac{7}{8}=\dfrac{7\times7}{8\times7}=\dfrac{49}{56}\)
\(\Rightarrow\dfrac{48}{56}< \dfrac{49}{56}\) HAY \(\dfrac{6}{7}< \dfrac{7}{8}\)
2.
a)\(\dfrac{19}{42};\dfrac{19}{35};\dfrac{19}{21};\dfrac{19}{19};\dfrac{19}{17}\)
b)\(\dfrac{8}{10}=\dfrac{4}{5}=\dfrac{4\times7}{5\times7}=\dfrac{28}{35}\)
\(\dfrac{16}{14}=\dfrac{8}{7}=\dfrac{8\times5}{7\times5}=\dfrac{40}{35}\)
\(\dfrac{6}{7}=\dfrac{6\times5}{7\times5}=\dfrac{30}{35}\)
\(\Rightarrow\dfrac{28}{35};\dfrac{30}{35};\dfrac{37}{35};\dfrac{40}{35}\) hay \(\dfrac{8}{10};\dfrac{6}{7};\dfrac{37}{35};\dfrac{16}{14}\)
