\(\frac{1}{1.3}\))(1+\(\frac{1}{2.4}\))(1+
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

19 tháng 8 2018

Ta có: \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)  

\(=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{9999}\right)\) 

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{10000}{9999}\) 

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{100.100}{99.101}\) 

\(=\frac{2.2.3.3.4.4.....100.100}{1.3.2.4.3.5.....99.101}\) 

\(=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\) 

\(=\frac{2.3.4.....100}{1.2.3.....99}.\frac{2.3.4.....100}{3.4.5.....101}\) 

\(=100.\frac{2}{101}\) 

\(=\frac{200}{101}\)

29 tháng 8 2016

A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/1999.2001
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/1999.2001
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/1999 - 1/2001 
2.A = 1 - 1/2001 

2.A = 2000/2001

Vậy A =1000/2001

B = 1/3.5 + 1/5.7 + 1/7.9 +........+ 1/99.101
2.A = 2/3.5 + 2/5.7 + 2/7.9 +........+ 2/99.101
2.A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/99 - 1/101 
2.A = 1/3 - 1/101 = 98/303 
Vậy A =49/303

29 tháng 8 2016

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{1999.2001}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{1999.2001}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{1999}-\frac{1}{2001}\)

\(2A=\frac{1}{1}-\frac{1}{2001}=\frac{2000}{2001}\)

\(A=\frac{2000}{2001}.\frac{1}{2}=\frac{1000}{2001}\)

20 tháng 4 2018

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

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)

\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(A=\frac{100}{101}:2\)

\(A=\frac{50}{101}< 1\left(đpcm\right)\)

Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{100^2}\)

\(\Rightarrow B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

\(\Rightarrow B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow B< 1-\frac{1}{100}\)

\(\Rightarrow B< \frac{99}{100}< 1\left(đpcm\right)\)

tích mình đi

làm ơn

rùi mình

tích lại

thanks

8 tháng 6 2017

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{1}{1}-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}\div2\)

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

8 tháng 6 2017

đề 

sai r bn ak

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,21.321.3+23.523.5+25.725.7+....+299.101

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

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

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

b,

51.351.3+53.553.5+55.755.7+....+599.101

=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)

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

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

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

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

2 tháng 4 2018

2/7A=2/1.3+2/3.5+...+2/99+101

2/7A=1-1/3+1/3-1/5+...+1/99-1/101

2/7A=1-1/101

2/7A=100/101

A=350/101

B=(1+1+1+1)-(1/2+1/6+1/12+1/20)

=4-(1/1.2+1/2.3+1/3.4+1/4.5)

=4-(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)

=4-(1-1/5)

=4-4/5

=16/5

8 tháng 5 2017

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

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

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

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

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

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

\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)

\(A=\frac{3^2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=3.\frac{9}{20}\)

\(A=\frac{27}{20}\)

k nhá bn!

8 tháng 5 2017

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

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

\(2A=1-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

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

\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)

\(\Rightarrow A=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=3.\frac{9}{20}\)

\(A=\frac{27}{20}\)