h) 2/4.8+2/8.12+2/12.16+….+2/400.404 g)1/5.10+1/10.15+1/15.20+…..+1/45.100 Làm hộ mình hai câu trên với ạ 28/07/2021 Bởi Kennedy h) 2/4.8+2/8.12+2/12.16+….+2/400.404 g)1/5.10+1/10.15+1/15.20+…..+1/45.100 Làm hộ mình hai câu trên với ạ
Đáp án: $h)\dfrac{403}{808}\\g)\dfrac{49}{250}$ Giải thích các bước giải: $h)\dfrac{2}{4.8}+\dfrac{2}{8.12}+\dfrac{2}{12.16}+…+\dfrac{2}{400.404}\\=\dfrac{1}{2}.\left ( \dfrac{4}{4.8}+\dfrac{4}{8.12}+\dfrac{4}{12.16}+…+\dfrac{4}{400.404} \right )\\=\dfrac{1}{2}.\left (1- \dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+…+\dfrac{1}{400}-\dfrac{1}{404} \right )\\=\dfrac{1}{2}.\left ( 1-\dfrac{1}{404} \right )\\=\dfrac{1}{2}.\left ( \dfrac{404}{404}-\dfrac{1}{404} \right )\\=\dfrac{1}{2}.\dfrac{403}{404}\\=\dfrac{403}{808}\\g)\dfrac{1}{5.10}+\dfrac{1}{10.15}+\dfrac{1}{15.20}+…+\dfrac{1}{45.50}\\=\dfrac{1}{5}.\left ( \dfrac{5}{5.10}+\dfrac{5}{10.15}+\dfrac{5}{15.20}+…+\dfrac{5}{45.50} \right )\\=\dfrac{1}{5}.\left ( 1-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+…+\dfrac{1}{45}-\dfrac{1}{50} \right )\\=\dfrac{1}{5}.\left ( 1-\dfrac{1}{50} \right )\\=\dfrac{1}{5}.\left ( \dfrac{50}{50}-\dfrac{1}{50} \right )\\=\dfrac{1}{5}.\dfrac{49}{50}\\=\dfrac{49}{250}$ Bình luận
Đáp án:
$h)
\dfrac{403}{808}\\
g)
\dfrac{49}{250}$
Giải thích các bước giải:
$h)
\dfrac{2}{4.8}+\dfrac{2}{8.12}+\dfrac{2}{12.16}+…+\dfrac{2}{400.404}\\
=\dfrac{1}{2}.\left ( \dfrac{4}{4.8}+\dfrac{4}{8.12}+\dfrac{4}{12.16}+…+\dfrac{4}{400.404} \right )\\
=\dfrac{1}{2}.\left (1- \dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+…+\dfrac{1}{400}-\dfrac{1}{404} \right )\\
=\dfrac{1}{2}.\left ( 1-\dfrac{1}{404} \right )\\
=\dfrac{1}{2}.\left ( \dfrac{404}{404}-\dfrac{1}{404} \right )\\
=\dfrac{1}{2}.\dfrac{403}{404}\\
=\dfrac{403}{808}\\
g)
\dfrac{1}{5.10}+\dfrac{1}{10.15}+\dfrac{1}{15.20}+…+\dfrac{1}{45.50}\\
=\dfrac{1}{5}.\left ( \dfrac{5}{5.10}+\dfrac{5}{10.15}+\dfrac{5}{15.20}+…+\dfrac{5}{45.50} \right )\\
=\dfrac{1}{5}.\left ( 1-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+…+\dfrac{1}{45}-\dfrac{1}{50} \right )\\
=\dfrac{1}{5}.\left ( 1-\dfrac{1}{50} \right )\\
=\dfrac{1}{5}.\left ( \dfrac{50}{50}-\dfrac{1}{50} \right )\\
=\dfrac{1}{5}.\dfrac{49}{50}\\
=\dfrac{49}{250}$