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1/5x6 + 1/6x7 + 1/7x8 + 1/8 x9 + 1/9x10 + 1 /10x11 + 1/11x12
1/5 - 1/6+ 1/6 - 1/7 + 1/7 - 1/8+ 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12
= 1/5 - 1/12
= 7/60
\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}+\dfrac{1}{156}+\dfrac{1}{182}+\dfrac{1}{210}\\ =\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{13}-\dfrac{1}{14}\\ =\dfrac{1}{5}-\dfrac{1}{14}\\ =\dfrac{14}{70}-\dfrac{5}{70}=\dfrac{9}{70}\)
Bài 1: Ta có:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)
\(=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+...+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
Vậy \(M=\dfrac{5}{16}\)
Bài 2: Ta có:
\(A=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+...+\dfrac{1}{210}\)
\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+...+\dfrac{1}{14.15}\)
\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{1}{10}\)
Vậy \(A=\dfrac{1}{10}\)
Giải:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}.\)
\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+\dfrac{5}{208}.\)
\(M=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+\dfrac{5}{13.16}.\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left[\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{10}-\dfrac{1}{10}\right)+\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\right].\)
\(M=\dfrac{5}{3}\left[0+0+0+\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\right]\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left(\dfrac{4}{16}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}.\dfrac{3}{16}.\)
\(M=\dfrac{15}{48}=\dfrac{5}{16}.\)
\(S=\dfrac{1}{2}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{11}\)
\(S=\dfrac{67}{110}\)
\(A=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(=\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{12}{60}-\dfrac{5}{60}=\dfrac{7}{60}\)
Vậy \(A=\dfrac{7}{60}\)
\(A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(A=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(A=\dfrac{1}{5}-\dfrac{1}{12}\)
\(A=\dfrac{7}{60}\)
D = 1 + \(\dfrac{-1}{20}\) + \(\dfrac{-1}{30}\) + \(\dfrac{-1}{42}\)+ \(\dfrac{-1}{56}\)+ \(\dfrac{-1}{72}\)+ \(\dfrac{-1}{90}\)
D = 1 - ( \(\dfrac{1}{4\times5}\) + \(\dfrac{1}{5\times6}\)+ \(\dfrac{1}{6\times7}\)+ \(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\))
D = 1 - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\))
D = 1 - \(\dfrac{3}{20}\)
D = \(\dfrac{17}{20}\)
D=1+(1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10)
D=1+(1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
D=1+(1/4-1/10)
D=1+3/5
D=8/5
\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(\Rightarrow B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\Rightarrow B=\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{10}{9.10}-\dfrac{9}{9.10}\)
\(\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\\ B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\\ B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ B=\dfrac{1}{2}-\dfrac{1}{10}\\ B=\dfrac{5}{10}-\dfrac{1}{10}\\ B=\dfrac{4}{10}\\ B=\dfrac{2}{5}\)
Ta có:
\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)
\(\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\right)-x=\dfrac{-19}{24}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\dfrac{7}{30}-x=\dfrac{-19}{24}\)
\(x=\dfrac{7}{30}-\dfrac{-19}{24}\)
\(x=\dfrac{41}{40}\)
\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\dfrac{7}{30}-x=\dfrac{-19}{24}\)
\(\Rightarrow x=\dfrac{7}{30}-\dfrac{-19}{24}\)
\(\Rightarrow x=\dfrac{41}{40}\)
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
\(B=\dfrac{-1}{20}+\dfrac{-1}{30}+...+\dfrac{-1}{132}\)
\(=-\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{11\cdot12}\right)\)
\(=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)=-\left(\dfrac{3}{12}-\dfrac{1}{12}\right)=-\dfrac{2}{12}=-\dfrac{1}{6}\)