Ta có S = \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}=\left(\right.\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\left.\right)+\left(\right.\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\left.\right)+\left(\right.\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\left.\right)\)⇒ S < \(\frac{1}{30} \cdot 10 + \frac{1}{40} \cdot 10 + \frac{1}{50} \cdot 10 = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{47}{60} < \frac{48}{60} = \frac{4}{5}\)

Vậy S < \(\frac{4}{5}\)

Ta có S = \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}=\left(\right.\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\left.\right)+\left(\right.\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\left.\right)+\left(\right.\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\left.\right)\)⇒ S < \(\frac{1}{30} \cdot 10 + \frac{1}{40} \cdot 10 + \frac{1}{50} \cdot 10 = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{47}{60} < \frac{48}{60} = \frac{4}{5}\)

Vậy S < \(\frac{4}{5}\)

Ta có S = \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}=\left(\right.\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\left.\right)+\left(\right.\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\left.\right)+\left(\right.\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\left.\right)\)⇒ S < \(\frac{1}{30} \cdot 10 + \frac{1}{40} \cdot 10 + \frac{1}{50} \cdot 10 = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{47}{60} < \frac{48}{60} = \frac{4}{5}\)

Vậy S < \(\frac{4}{5}\)

a ) 0

b ) 17/49

c )1

d) -79/30