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\(\frac{F}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)
\(\frac{F}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(\frac{F}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.5}+\frac{5-4}{4.5}+...+\frac{20-19}{19.20}\)
\(\frac{F}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(\frac{F}{2}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow F=\frac{18}{20}=\frac{9}{10}\)
\(\frac{\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)=\(\frac{1}{\frac{7}{2}}\)=\(\frac{2}{7}\)
\(=1+\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+\frac{1}{5.3}+\frac{1}{3.7}+\frac{1}{7.4}+\frac{1}{4.9}+\frac{1}{9.5}\)
\(=1+1-\frac{1}{5}\)
\(=\frac{10}{5}-\frac{1}{5}\)
\(=\frac{9}{5}\)
Ai thấy đúng thì
\(A=3+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{4}{3}}}}=3+\frac{1}{1+\frac{1}{1+\frac{3}{4}}}\)
\(=3+\frac{1}{1+\frac{1}{\frac{7}{4}}}=3+\frac{1}{1+\frac{4}{7}}=3+\frac{1}{\frac{11}{4}}=3+\frac{4}{11}=\frac{37}{11}\)
\(B=-5+\frac{1}{1-\frac{1}{2+\frac{1}{\frac{3}{4}}}}=-5+\frac{1}{1-\frac{1}{2+\frac{4}{3}}}\)
\(=-5+\frac{1}{1-\frac{1}{\frac{10}{3}}}=-5+\frac{1}{1-\frac{3}{10}}=-5+\frac{1}{\frac{7}{10}}=-5+\frac{10}{7}=\frac{-25}{7}\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(=0\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)
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
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 \(D=\frac{7}{15}\)