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5 tháng 6 2020

\(\frac{F}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

\(\frac{F}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(\frac{F}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.5}+\frac{5-4}{4.5}+...+\frac{20-19}{19.20}\)

\(\frac{F}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(\frac{F}{2}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow F=\frac{18}{20}=\frac{9}{10}\)

2 tháng 5 2015

1/2D=1/2(1/6+1/10+......+1/45)

1/2D=1/12+1/20+1/30+.....+1/90

1/2D=1/3.4+1/4.5+1/5.6+......+1/9.10

1/2D=1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10

1/2D=1/3-1/10

1/2D=7/30

D=7/30:1/2

D=7/15

2 tháng 5 2015

Ta có:\(D=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)

\(=\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)

\(=2.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=2.\left(\frac{1}{3}-\frac{1}{10}\right)=2.\frac{7}{30}=\frac{7}{15}\)

Vậy \(D=\frac{7}{15}\)

1 tháng 6 2020

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

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

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

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

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

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

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

29 tháng 7 2019

B = \(\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}=\frac{-240+270+60-144}{360}=\frac{-54}{360}=-0,15\)

26 tháng 7 2019

\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)

\(=0\)

26 tháng 7 2019

\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)