Đáp án:
$a)A=\dfrac{4036}{2019}$
Giải thích các bước giải:
$a) A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2017.2019}\\
=2\left ( \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{2017.2019} \right )\\
=2\left ( \dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2017}-\dfrac{1}{2019} \right )\\
=2(1-\dfrac{1}{2019})\\
=2.(\dfrac{2019-1}{2019})=\dfrac{4036}{2019}\\
b)
S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\\
\Leftrightarrow S=\left ( \dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40} \right )+\left ( \dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50} \right )+\left ( \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60} \right )(1)$
Mà
$ \dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}<10.\dfrac{1}{30}=\dfrac{1}{3}\\
\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}<10.\dfrac{1}{40}=\dfrac{1}{4}\\
\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}<10.\dfrac{1}{50}=\dfrac{1}{5}\\
\Rightarrow S<\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}<\dfrac{48}{60}=\dfrac{4}{5}\\
\Rightarrow S<\dfrac{4}{5}$
