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\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{14}\)
Bạn làm tương tự như bài này nhé! /hoi-dap/question/28380.html
Ta có :\(A=10\left(\frac{1}{1.2}+\frac{5}{2.3}+...+\frac{89}{9.10}\right)\)
\(\Rightarrow10\left(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\right)\)
\(\Rightarrow10\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\right]\)
\(\Rightarrow10\left[9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\right]\)
\(\Rightarrow10\left[9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\right]\)
\(\Rightarrow10\left[9-\left(1-\frac{1}{10}\right)\right]\)
\(\Rightarrow\)\(10\left[9-\frac{9}{10}\right]\)
\(\Rightarrow10.\frac{81}{10}\)
\(\Rightarrow A=81\)
Làm rờ đó sai thì thôi nha.
H=\(\frac{5}{28}\) + \(\frac{5}{70}\) + \(\frac{5}{130}\) +...+ \(\frac{5}{700}\)
H= \(\frac{5}{4.7}\) + \(\frac{5}{7.10}\) + \(\frac{5}{10.13}\) +...+\(\frac{5}{25.28}\)
H= \(\frac{5}{3}\) (\(\frac{1}{4}\) - \(\frac{1}{7}\) + \(\frac{1}{7}\) - \(\frac{1}{10}\) + \(\frac{1}{10}\) - \(\frac{1}{13}\) +...+ \(\frac{1}{25}\) - \(\frac{1}{28}\))
H= \(\frac{5}{3}\) (\(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) - \(\frac{1}{10}\)+...+ \(\frac{1}{25}\)- \(\frac{1}{25}\) - \(\frac{1}{28}\))
H= \(\frac{5}{3}\) ( \(\frac{1}{4}\) - \(\frac{1}{28}\)) = \(\frac{5}{3}\) . \(\frac{3}{14}\)= \(\frac{5}{14}\)