Đáp án:
$h)
\dfrac{403}{808}\\
g)
\dfrac{49}{250}$
Giải thích các bước giải:
$h)
\dfrac{2}{4.8}+\dfrac{2}{8.12}+\dfrac{2}{12.16}+...+\dfrac{2}{400.404}\\
=\dfrac{1}{2}.\left ( \dfrac{4}{4.8}+\dfrac{4}{8.12}+\dfrac{4}{12.16}+...+\dfrac{4}{400.404} \right )\\
=\dfrac{1}{2}.\left (1- \dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+...+\dfrac{1}{400}-\dfrac{1}{404} \right )\\
=\dfrac{1}{2}.\left ( 1-\dfrac{1}{404} \right )\\
=\dfrac{1}{2}.\left ( \dfrac{404}{404}-\dfrac{1}{404} \right )\\
=\dfrac{1}{2}.\dfrac{403}{404}\\
=\dfrac{403}{808}\\
g)
\dfrac{1}{5.10}+\dfrac{1}{10.15}+\dfrac{1}{15.20}+...+\dfrac{1}{45.50}\\
=\dfrac{1}{5}.\left ( \dfrac{5}{5.10}+\dfrac{5}{10.15}+\dfrac{5}{15.20}+...+\dfrac{5}{45.50} \right )\\
=\dfrac{1}{5}.\left ( 1-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+...+\dfrac{1}{45}-\dfrac{1}{50} \right )\\
=\dfrac{1}{5}.\left ( 1-\dfrac{1}{50} \right )\\
=\dfrac{1}{5}.\left ( \dfrac{50}{50}-\dfrac{1}{50} \right )\\
=\dfrac{1}{5}.\dfrac{49}{50}\\
=\dfrac{49}{250}$
