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\(\Rightarrow A=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)
\(\Rightarrow A=\frac{-1}{1}\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=\frac{-1}{1}\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{-1}{1}\left(\frac{5}{20}-\frac{2}{20}\right)=\frac{-1}{1}.\frac{3}{20}=\frac{-3}{20}\)
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}\right)=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{9.10}\right)\)
\(=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...-\frac{1}{10}\right)=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{1}{5}-\frac{1}{4}=\frac{-1}{20}\)
\(A=\frac{1}{10}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\)
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{1}{10}-\left[\left(\frac{1}{4}-\frac{1}{10}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-...-\left(\frac{1}{9}-\frac{1}{9}\right)\right]\)
\(A=\frac{1}{10}-\frac{1}{4}+\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{4}\)
\(A=-\frac{1}{20}\)
\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=-\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=-\frac{3}{20}\)
Bài làm:
Ta có: \(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=-\frac{3}{20}\)
\(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(A=-\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(A=\frac{-3}{20}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{3}{20}\)
Ta có : \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(\Rightarrow A=-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(\Rightarrow A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.......+\frac{1}{9.10}\right)\)
\(\Rightarrow A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{9}-\frac{1}{10}\right)\)\(\Rightarrow A=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(\Rightarrow A=-\frac{3}{20}\)
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
