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19 tháng 7 2018

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

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

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

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

20 tháng 7 2018

\(\frac{1}{1.3}+\frac{1}{3,5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{101-99}{99.101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

-.-

19 tháng 7 2018

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

=1-1/101

=100/101

k cho mình nha

19 tháng 7 2018

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

4 tháng 5 2016

 nhung ma ko cothoi gian giai

4 tháng 5 2016

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

25 tháng 4 2019

A=1 - 1/3+1/3 - 1/5+1/5 - 1/6+...+1/99 - 101+1/101 - 1/103

A=1 - 1/103

A=102/103

25 tháng 4 2019

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}+\frac{1}{101.103}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{2}.\frac{102}{103}\)

\(=\frac{51}{103}\)

5 tháng 1 2016

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(S=\frac{2}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(S=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(S=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

5 tháng 1 2016

nhân S cho 2 

Công thức \(\frac{2}{x.\left(x+2\right)}=\frac{1}{x}-\frac{1}{x+2}\)

23 tháng 1 2017

a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))

=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)

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

b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

=\(\frac{250}{101}\)

\(=\frac{5}{2}.\frac{100}{101}\)

3 tháng 5 2019

a,21.3+23.5+25.7+....+

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

=>\(\frac{1}{1}-\frac{1}{101}\)

=>\(\frac{100}{101}\)

b,

51.3+53.5+55.7+....+

=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}.\frac{100}{101}\)

=>\(\frac{250}{101}\)

6 tháng 1 2016

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2S=1-\frac{1}{101}\Rightarrow2S+\frac{1}{101}=1\)

2 tháng 6 2016

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

=1-1/101 

=100/101 

b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5 

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

=(1-1/101).2,5

=100/101.2,5 

=250/101 

dấu / là phần nhé. bạn có thể xem bài có dấu phần ở : Câu hỏi của Nguyễn Thị Hoài Anh 

2 tháng 6 2016

A)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

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

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

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

B) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{1}{99.101}\)

=5.(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))

=5.\(\frac{2}{2}.\)(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))

=5.\(\frac{1}{2}\).(\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\))

=5.\(\frac{1}{2}\).(1-\(\frac{1}{3}\)+\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

=5.\(\frac{1}{2}\).(1-\(\frac{1}{101}\))

=\(\frac{5}{2}.\frac{100}{101}=\frac{250}{100}\)

Chúc bạn học tốtleuleu

7 tháng 5 2016

\(a,=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)

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

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

\(b,=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)

7 tháng 5 2016

a,\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=1.\frac{99}{100}=\frac{99}{100}\)