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NH
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18 tháng 6 2016
\(\frac{10}{56}+\frac{10}{140}+...+\frac{10}{1400}\)
\(=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{3}\left(\frac{3}{28}+\frac{3}{70}+...+\frac{3}{700}\right)\)
\(=\frac{5}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)
DP
23 tháng 7 2017
\(D=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(D=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)
\(D=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}\)
\(D=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(D=2\left(\frac{1}{4}-\frac{1}{10}\right)=2\cdot\frac{3}{20}=\frac{3}{10}\)
\(E=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(E=\frac{5}{28}+\frac{1}{14}+\frac{1}{26}+...+\frac{1}{140}\)
\(E=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(E=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(E=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(E=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
12 tháng 5 2016
=15.(1/90.94+1/94.98+....+1/146.150) =15.1/4(1/90-1/94+1/94-1/98+1/9 --......+1/150 =15/4.(1/90 - 1/150) =15/4.1/1225 =1/60 NHỚ nhấn đúng nha
6 tháng 5 2019
Tôi thi còn mỗi môn toán ngày mai nên tôi vào đây luyện tập giải toán
# USAS - 12 #
6 tháng 5 2019
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
đặt:\(S=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(S=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(3S=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\)
\(\frac{3S}{5}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...+\frac{1}{25}-\frac{1}{28}\)
\(\frac{3S}{5}=\frac{1}{4}-\frac{1}{28}\)
\(\frac{3S}{5}=\frac{3}{14}\)
\(S=\frac{3}{14}\times\frac{5}{3}\)
\(S=\frac{5}{14}\)
24 tháng 9 2021
\(\)\(\dfrac{10}{56}+\dfrac{10}{140}+...+\dfrac{10}{1400}\)
\(=\dfrac{5}{28}+\dfrac{5}{70}+...+\dfrac{5}{700}\)
\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{5}{3}\cdot\dfrac{6}{28}=2\cdot\dfrac{5}{28}=\dfrac{10}{28}=\dfrac{5}{14}\)
24 tháng 9 2021
\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\\ =\dfrac{5}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{25\cdot28}\right)\\ =\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\\ =\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}\cdot\dfrac{3}{14}=\dfrac{5}{14}\)
NT
1
20 tháng 4 2015
\(\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+\frac{5}{208}+\frac{5}{304}=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{16}-\frac{1}{19}\right)=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{19}\right)\)
\(=\frac{5}{3}.\frac{15}{74}=\frac{25}{74}\)
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{25.28}\right)=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)