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\(B=\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}\)
\(B=\frac{1}{5.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{670.671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{670}-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\frac{670}{671}=\frac{134}{2013}\)
Nguyễn Huy Thắngsoyeon_Tiểubàng giảiSilver bulletLê Nguyên HạoPhương AnVõ Đông Anh Tuấnsoyeon_Tiểubàng giảiLê Thị Linh ChiNguyễn Huy Tú
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
\(-A=\left(\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(-A=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+\frac{1}{8}-\frac{1}{7}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)
\(-A=\frac{1}{10}-1=\frac{-9}{10}\Rightarrow A=\frac{9}{10}\)
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
Vậy A=-79/90
a) \(-5,13:\left(5\frac{5}{28}-1\frac{8}{9}.1,25+1\frac{16}{63}\right)\)
\(=-5,13:\left(\frac{145}{28}-\frac{17}{9}.1,25+\frac{79}{63}\right)\)
\(=-5,13:\left(\frac{145}{28}-\frac{85}{36}+\frac{79}{63}\right)\)
\(=-5,13:\frac{57}{14}\)
\(=-\frac{63}{50}\)
b) \(\left(3\frac{1}{3}.1,9+19,5:4\frac{1}{3}.\left(\frac{62}{75}-\frac{4}{25}\right)\right)\)
\(=\left(\frac{10}{3}.1,9+19,5:\frac{13}{3}\right).\left(\frac{62}{75}-\frac{4}{25}\right)\)
\(=\left(\frac{19}{3}+\frac{9}{2}\right).\frac{2}{3}\)
\(=\frac{65}{6}.\frac{2}{3}\)
\(=\frac{65}{9}\)
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