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1 tháng 4 2017

B)

B = \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{29.31}\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...\dfrac{1}{29.31}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{29}-\dfrac{1}{31}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{31}\right)\)

= \(\dfrac{1}{2}.\dfrac{30}{31}\)

= \(\dfrac{30}{62}\) = \(\dfrac{15}{31}\)

2 tháng 4 2017

1A,47/20

B.13/11

C,427/330

2A,47/120

B,15/31

Câu 1: 

a: ĐKXĐ: x+5<>0

hay x<>-5

b: ĐKXĐ: x-2<>0

hay x<>2

25 tháng 4 2018

A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)

25 tháng 4 2018

B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)

B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)

a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)

b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)

3 tháng 4 2018

a)

\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)

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

\(=\dfrac{1}{5}-\dfrac{1}{25}\)

\(=\dfrac{4}{25}\)

b)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

3 tháng 4 2018

a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)

\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)

tương tự

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

6 tháng 5 2018

A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)

=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)

=2.(1-1/101)

=2.(101/101-1/101)

=2.100/101

200/101

6 tháng 5 2018

B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)

=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)

=2.(1/1+1/101)

=2.(101/101+1/101)

=2.102/101

=204/101

24 tháng 4 2017

cho minh xin yeu cau de bai

26 tháng 4 2017

trả hiểu yêu cầu đề bài là j cả

24 tháng 3 2017

B=1/2. (2/25.27+2/27.29+2/29.31+....+2/73.75) B=1/2. (1/25-1/27+1/27-1/29+1/29-1/31+....+1/73-1/75) B=1/2. (1/25-1/75) B=1/2. 2/75 B=1/75

29 tháng 3 2017

\(3A=\dfrac{3}{8.11}+\dfrac{3}{18.21}+..+\dfrac{3}{197.200}\)

27 tháng 3 2017

A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)

A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)

A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)

A=\(\dfrac{7}{24}\)

B=\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)

B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)

B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)

B=\(1+\left(-1\right)+\left(-1\right)=-1\)

C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)

C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)

D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)

D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)

10 tháng 5 2018

\(B=\dfrac{1}{1.3}\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2003.2005}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}\right)\\ =\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}.\dfrac{2004}{2005}\\ =\dfrac{1002}{2005}\)

10 tháng 5 2018

Hình như bn vt sai đề phải ko???

lolang