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22 tháng 7 2016

\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{19.20}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{5}-\frac{1}{20}\)

\(=\frac{4}{20}-\frac{1}{20}=\frac{3}{20}\)

16 tháng 7 2016

\(A=\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

16 tháng 7 2016

A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{49.50}\)

A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+  \(\frac{1}{3}\) -    \(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)

A=1-\(\frac{1}{50}\)

A=\(\frac{49}{50}\)

27 tháng 1 2018

C=\(\frac{7}{3.4}\)-\(\frac{9}{4.5}\)+\(\frac{11}{5.6}\)+\(\frac{13}{6.7}\)+\(\frac{15}{7.8}\)-\(\frac{17}{8.9}\)+\(\frac{19}{9.10}\) 

=\(\frac{1}{3}\)+\(\frac{1}{4}\)-\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{6}\)-\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)+\(\frac{1}{8}\)-\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)+\(\frac{1}{10}\)

=\(\frac{1}{3}\)+\(\frac{1}{10}\)=\(\frac{13}{30}\)

1 tháng 7 2019

fddddddddddddddddddddddddddddddddddddddddđ

15 tháng 9 2015

mk bít lm cách lớp 5, vừa học

Cần ko bn

18 tháng 8 2023

B = \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\) - \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\)

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))

B =  \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\)

B = \(\dfrac{1}{12}\) - \(\dfrac{3}{20}\) 

B = - \(\dfrac{1}{15}\)

12 tháng 8 2016

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

= 1 - 1/50

= 49/50

ỦNG HỘ NHA

12 tháng 8 2016

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

7 tháng 9 2017

\(A=\frac{4}{4.5}+\frac{4}{5.6}+\frac{4}{6.7}+...+\frac{4}{47.48}\)

\(A=4.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+......+\frac{1}{47.48}\right)\)

\(A=4.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{47}-\frac{1}{48}\right)\)

\(A=4.\left(\frac{1}{4}-\frac{1}{48}\right)\)

\(A=4.\frac{11}{48}\)

\(A=\frac{11}{12}\)

7 tháng 6 2016

a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25

= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25

= 1/5 - 1/25

= 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101

= 1 - 1/101

= 100/101

c) 3/1.4 + 3/4.7 + ... + 3/2002.2005

= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005

= 1 - 1/2005

= 2004/2005

d) 5/2.7 + 5/7.12 + ... + 5/1997.2002

= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002

= 1/2 - 1/2002

= 500/1001

7 tháng 6 2016

a,A =  \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)

A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)

B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

B=\(1-\frac{1}{101}=\frac{100}{101}\)

c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)

C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)

C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)

d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)

D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)

D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)