tính \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{100.102}\)
ai nhanh đúng tick
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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2019
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{100}{2}=50\)
\(\frac{-17}{2.4}-\frac{17}{4.6}-\frac{17}{6.8}-...-\frac{17}{100.102}\)
\(=-\frac{17}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)
\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(=-\frac{17}{2}.\frac{25}{51}=-\frac{25}{6}\)
NN
2
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
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{30}\)
\(2.A=\frac{1}{2}-\frac{1}{30}\)
\(2.A=\frac{7}{15}\)
\(A=\frac{7}{15}:2=\frac{7}{30}\)
NT
1
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!
28 tháng 4 2018
\(\Rightarrow2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2x-2\right).2x}\right)=\frac{1}{8}.2\).2
\(\Rightarrow\frac{2}{2.4}+\frac{2}{4.6}+...\frac{2}{\left(2x-2\right).2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\Rightarrow\frac{1}{2x}=\frac{1}{2.2}\)
\(\Rightarrow x=2\)
TP
14 tháng 8 2017
Ta có:
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{1}{4,6}+\frac{1}{6.8}+...+\frac{1}{98.100}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)
T
14 tháng 8 2017
Đặt \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(4-2=2;6-4=2;...\)
\(2A=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(2A=\frac{1}{2}-\frac{1}{100}\)
\(2A=\frac{49}{100}\)
HH
4
3 tháng 9 2017
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
\(2S=\frac{1}{2}-\frac{1}{10}\)
\(2S=\frac{2}{5}\)
\(S=\frac{2}{5}:2\)
\(S=\frac{1}{5}\)
HN
3 tháng 9 2017
S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)
AN
3
19 tháng 3 2017
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{38.40}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{38}-\frac{1}{40}\)
=\(\frac{1}{2}-\frac{1}{40}\)
=\(\frac{19}{40}\)
19 tháng 3 2017
= 2 *[1/2 * 1/4 +1/4 *1/6 +1/6*1/8+...+1/38*1/40
=2*[1/2 - 1/40]
=2 * (-19/40)
= -380
8 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
MA
7 tháng 9 2016
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
HT
7 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
Đặt BT trên là A
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)
\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(2A=\frac{1}{2}-\frac{1}{102}=\frac{50}{102}\Rightarrow A=\frac{25}{102}\)
Đặt A là biểu thức trên ta có :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(=\frac{1}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{50}{102}=\frac{25}{102}\)