K
Khách

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

1 tháng 4 2016

4/42+4/56+4/72+4/90+4/110

=4*(1/42+1/56+1/72+1/90+1/110)

=4*(1/(6*7)+1/(7*8)+1/(8*9)+1/(9*10)+1/(10*11))

=4*(1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11)

=4*(1/6-1/11)

=4*5/66

=10/33

1 tháng 4 2016

\(\frac{4}{6\cdot7}+\frac{4}{7\cdot8}+\frac{4}{8\cdot9}+\frac{4}{9\cdot10}+\frac{4}{10\cdot11}\)

\(4\cdot\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\right)\)

\(4\cdot\frac{5}{66}\)

\(\frac{10}{33}\)

Ta có:

\(C=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{40.41}+\frac{2}{41.42}\)

\(\Rightarrow C=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}+\frac{1}{41.42}\right)\)

\(\Rightarrow C=2\left(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{41-40}{40.41}+\frac{42-41}{41.42}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}+\frac{1}{41}-\frac{1}{42}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{42}\right)=\frac{13}{21}\)

\(D=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(\Rightarrow D=\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{111-107}{107.111}\)

\(\Rightarrow D=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)\(E=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow E=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

\(\Rightarrow E=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}+\frac{11-10}{10.11}\)

\(\Rightarrow E=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)

31 tháng 5 2017

c) x=-2 nha

d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)

=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)

=\(\frac{1}{5}\)-\(\frac{1}{12}\)\(\frac{7}{60}\)

31 tháng 5 2017

bạn ơi kết quả là 7/60

7 tháng 8 2016

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

7 tháng 8 2016

A=\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

=\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{14}-\frac{1}{15}\)

=\(\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

20 tháng 7 2016

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+...+\frac{1}{210}=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{14.15}\)

\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{14}-\frac{1}{15}\)

\(=\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

20 tháng 7 2016

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

23 tháng 1 2020

\(a,A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

\(=\frac{-1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}+\frac{1}{10}\)

\(=-\frac{1}{4}+\frac{1}{10}\)

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

\(b,B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{11.7}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-....-\frac{1}{28}\)

\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

23 tháng 1 2020

a) \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(\Rightarrow-1.A=\frac{1}{20}+\frac{1}{30}+........+\frac{1}{90}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+........+\frac{1}{9.10}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{9}-\frac{1}{10}=\frac{1}{4}-\frac{1}{10}=\frac{3}{20}\)

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

b) \(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(\Rightarrow\frac{1}{7}B=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

\(\Rightarrow B=\frac{13}{28}:\frac{1}{7}=\frac{13}{28}.7=\frac{13}{4}\)

6 tháng 9 2015

\(\frac{1}{110}+\frac{1}{90}+.....+\frac{1}{20}=\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{10.11}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)

Vậy tổng bằng 7/44

5 tháng 5 2017

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)

\(A=\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{5}...+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}\)

\(A=\frac{7}{60}\)

5 tháng 5 2017

A = \(\frac{1}{5.6}+\frac{1}{6.7}+...+\)\(\frac{1}{10.11}+\frac{1}{11.12}\)

A = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\)\(\frac{1}{11}-\frac{1}{12}\)

A = \(\frac{1}{5}-\frac{1}{12}\)

A = \(\frac{7}{60}\)

20 tháng 4 2020

dài thế ai mà tính đc

20 tháng 4 2020

\(\frac{1}{3}\)

10 tháng 5 2017

Bài 1:

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

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

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

Bài 2:

Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)

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

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

Vậy A < 2

Bài 3:

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

Bài 4:

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)

\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)

10 tháng 5 2017

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

A=1-1/50

A=49/50