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\(S=\dfrac{1}{5.6}+\dfrac{1}{10.9}+\dfrac{1}{15.12}+...+\dfrac{1}{3350.2013}\\ \Rightarrow S=\dfrac{1}{5.3}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{670.671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{670}-\dfrac{1}{671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}\left(1-\dfrac{1}{671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}.\dfrac{670}{671}\\ \Rightarrow S=\dfrac{1}{15}.\dfrac{670}{671}=\dfrac{134}{2013}\)
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
\(-A=\left(\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(-A=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+\frac{1}{8}-\frac{1}{7}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)
\(-A=\frac{1}{10}-1=\frac{-9}{10}\Rightarrow A=\frac{9}{10}\)
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
Vậy A=-79/90
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)
Vậy \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(\Rightarrow A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)\(\left(dpcm\right)\)
Bài này lớp 6 mik kiểm tra chất lượng đầu năm nè
\(B=\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}\)
\(B=\frac{1}{5.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{670.671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{670}-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\frac{670}{671}=\frac{134}{2013}\)
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