Cho A= \(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{46.48}\)+\(\dfrac{4}{48.50}\)
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L3
4
5 tháng 7 2021
Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
SI
5 tháng 7 2021
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)
IA
2
17 tháng 6 2017
Ta có :
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+..................+\dfrac{4}{2008.2010}\)
\(\Rightarrow F=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.............+\dfrac{2}{2008.2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)
17 tháng 6 2017
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+......+\dfrac{4}{2008.2010}\)
\(F=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{2008.2010}\right)\)
\(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)\(F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)\(F=2.\dfrac{502}{1005}\)
\(F=\dfrac{1004}{1005}\)
A
4
28 tháng 4 2017
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
= \(\dfrac{2}{2}.\left(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+....+\dfrac{5}{48.50}\right)\)
\(\)\(=\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{48.50}\right)\)
\(=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\dfrac{12}{25}\)
=\(\dfrac{6}{5}\)=\(1\dfrac{1}{5}\)
Nếu bạn không biết cách giải bài này có thể bảo mình viết cách giải giúp!!!
Chúc bạn làm tốt!!!
TT
28 tháng 4 2017
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
=\(\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{48.50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{48}\right)\)
=\(\dfrac{5}{2}.\dfrac{23}{48}\) = \(\dfrac{115}{96}\)
28 tháng 1 2022
p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)
\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)
\(=\dfrac{11}{13}\)
25 tháng 4 2017
a)\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2008\cdot2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
b)\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)
25 tháng 4 2017
a) gọi biểu thức đó là A
Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức trên, ta có
\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2009}\right)\)
\(A=2.\left(\dfrac{2007}{4018}\right)=\dfrac{2007}{2009}\)
b) dễ quá bạn tự làm. (không phải mink không biết làm đâu nha)
NL
4
P
1
13 tháng 2 2022
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
1 tháng 10 2017
\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)
1 tháng 10 2017
a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)
F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)
F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))
F=\(\dfrac{1004}{1005}\)
\(A=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(=2\times\dfrac{12}{25}=\dfrac{24}{25}\)
\(=>A=4.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{46}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=4.\left(\dfrac{25}{50}-\dfrac{1}{50}\right)=\dfrac{4.24}{50}=\dfrac{48}{25}\)