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\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=-\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=-\frac{3}{20}\)
Bài làm:
Ta có: \(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=-\frac{3}{20}\)
A = \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)= \(\frac{1}{3}-\frac{1}{10}=\frac{7}{30}\)
\(\frac{1}{110}+\frac{1}{90}+.....+\frac{1}{20}=\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{10.11}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)
Vậy tổng bằng 7/44
\(\Rightarrow A=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)
\(\Rightarrow A=\frac{-1}{1}\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=\frac{-1}{1}\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{-1}{1}\left(\frac{5}{20}-\frac{2}{20}\right)=\frac{-1}{1}.\frac{3}{20}=\frac{-3}{20}\)
B = 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90
B = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)+ \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
B = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\)\(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
B = \(\frac{1}{2}-\frac{1}{10}\)
B = \(\frac{2}{5}\)
B=1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
B=1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10
B=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
B=1/2-1/10
B=2/5
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{12.13}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{4}-\frac{1}{13}=\frac{9}{52}\)
\(=\frac{1}{90}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\right)=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\left(\frac{9}{9}-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=\frac{1}{90}-\frac{80}{90}=-\frac{79}{90}\)
\(-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{3}{20}\)
Ta có : \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(\Rightarrow A=-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(\Rightarrow A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.......+\frac{1}{9.10}\right)\)
\(\Rightarrow A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{9}-\frac{1}{10}\right)\)\(\Rightarrow A=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(\Rightarrow A=-\frac{3}{20}\)