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26 tháng 1 2016

hình như bài này là dạng chuỗi

29 tháng 1 2016

Bước 1: \(\left(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{19\times21}\right)=\frac{1}{7}\)

Bước 2: \(x=\frac{9}{7}\div\frac{1}{7}=9\)

24 tháng 6 2017

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{19.21}\right).x=\frac{9}{7}\)

\(\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{21-19}{19.21}\right).x=\frac{9}{7}\)

\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)

 \(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)

 \(\left[\frac{1}{2}.\frac{2}{7}\right].x=\frac{9}{7}\)

\(\frac{1}{7}.x=\frac{9}{7}\)

\(\Rightarrow x=\frac{9}{7}\div\frac{1}{7}=9\)

24 tháng 6 2017

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)

\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)x=\frac{9}{7}\)

\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{21}\right)x=\frac{9}{7}\)

\(\Leftrightarrow\frac{2}{7}x=\frac{9}{7}\)

\(\Leftrightarrow x=\frac{9}{2}\)

25 tháng 8 2015

1. A = 2

2. x = 9

25 tháng 8 2015

2,  tính vế trc 

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{19.21}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{19}-\frac{1}{21}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)=\frac{1}{2}.\frac{2}{7}=\frac{1}{7}\)

=> 1/7 . x = 9/7

          x    = 9/7 . 7 

          x    = 9 

6 tháng 7 2016

a)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{2}{9}\)

\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+2}=\frac{2}{9}:2\)

\(\frac{1}{x+2}=\frac{1}{6}-\frac{1}{9}\)

\(\frac{1}{x+2}=\frac{1}{18}\)

=>x+2=18

=>x=16

b tương tự nhân nó với 1/2

6 tháng 7 2016

Cám ơn bạn

26 tháng 3 2018

\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

=\(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

=\(2\left(1-\frac{1}{21}\right)\)

=\(\frac{2.20}{21}=\frac{40}{21}\)

3 tháng 5 2018

a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)

   \(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)

\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)

\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)

\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)

b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)

\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)

\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)

\(\Rightarrow x=10\cdot\)