Bài `1:`
`a)` $\dfrac{13}{15}$ và $\dfrac{45}{47}$
Ta có:
`1-` $\dfrac{13}{15}$ `=` $\dfrac{2}{15}$
`1-` $\dfrac{45}{47}$ `=` $\dfrac{2}{47}$
Vì `15<47` nên $\dfrac{2}{15}$ `>` $\dfrac{2}{47}$
Hay $\dfrac{13}{15}$ `<` $\dfrac{45}{47}$
Vậy $\dfrac{13}{15}$ `<` $\dfrac{45}{47}$
`b)` $\dfrac{35}{65}$ và $\dfrac{17}{47}$
Ta có:
`1-` $\dfrac{35}{65}$ `=` $\dfrac{30}{65}$
`1-` $\dfrac{17}{47}$ `=` $\dfrac{30}{47}$
Vì `65>47` nên $\dfrac{30}{65}$ `<` $\dfrac{30}{47}$
Hay $\dfrac{35}{65}$ `>` $\dfrac{17}{47}$
Vậy $\dfrac{35}{65}$ `>` $\dfrac{17}{47}$
Bài `2:`
`a)` $\dfrac{2}{3}$ `+` $\dfrac{2}{15}$ `+` $\dfrac{2}{35}$ `+` $\dfrac{2}{63}$ `+...+` $\dfrac{2}{101×103}$
`=` $\dfrac{2}{1×3}$ `+` $\dfrac{2}{3×5}$ `+` $\dfrac{2}{5×7}$ `+` $\dfrac{2}{7×9}$ `+...+` $\dfrac{2}{101×103}$
`=` `1-` $\dfrac{1}{3}$ `+` $\dfrac{1}{3}$ `-` $\dfrac{1}{5}$ `+...+` $\dfrac{1}{101}$ `-` $\dfrac{1}{103}$
`=` `1-` $\dfrac{1}{103}$
`=` $\dfrac{102}{103}$
`b)` $\dfrac{4}{1×5}$ `+` $\dfrac{4}{5×9}$ `+...+` $\dfrac{4}{101×105}$
`=` `1-` $\dfrac{1}{5}$ `+` $\dfrac{1}{5}$ `-` $\dfrac{1}{9}$ `+...+` $\dfrac{1}{101}$ `-` $\dfrac{1}{105}$
`=` `1-` $\dfrac{1}{105}$
`=` $\dfrac{104}{105}$
