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KL
0
MA
9 tháng 9 2016
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\)
\(A=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}.\frac{7}{15}\)
\(A=\frac{7}{30}\)
9 tháng 9 2016
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{30}\)
\(2.A=\frac{1}{2}-\frac{1}{30}\)
\(2.A=\frac{7}{15}\)
\(A=\frac{7}{15}:2=\frac{7}{30}\)
W
4 tháng 8 2018
\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)
= \(\frac{5}{2}-\frac{5}{100}\)
= \(\frac{49}{50}\)
4 tháng 8 2018
\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
\(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)
\(\Rightarrow Q=\frac{49}{40}\)
6 tháng 3 2016
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...........+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
cho mình nha!
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
20 tháng 6 2020
Đặt BT trên là A
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)
\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(2A=\frac{1}{2}-\frac{1}{102}=\frac{50}{102}\Rightarrow A=\frac{25}{102}\)
20 tháng 6 2020
Đặt A là biểu thức trên ta có :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(=\frac{1}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{50}{102}=\frac{25}{102}\)
22 tháng 6 2019
\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+...+\frac{5}{298\cdot300}\)
\(=\frac{5}{2}\cdot\left(\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\right)\)
\(=\frac{5}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\cdot\left(\frac{1}{4}-\frac{1}{300}\right)\)
\(=\frac{37}{60}\)
CR
3
28 tháng 3 2019
\(A=\frac{-1}{2.4}+\frac{-1}{4.6}+\frac{-1}{6.8}+...+\frac{-1}{98.100}\Leftrightarrow.\)\(-2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\Leftrightarrow.\)
\(-2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\Leftrightarrow.\)
\(-2A=\frac{1}{2}-\frac{1}{100}\Leftrightarrow-2A=\frac{49}{100}\Leftrightarrow A=\frac{-49}{200}.\)
ĐÁP SỐ : \(A=\frac{-49}{200}.\)
6 tháng 6 2015
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)