Question

1.tính hợp ly nếu có thể

E=\(\frac{2}{5.10}\)+\(\frac{2}{10.15}\)+............+\(\frac{2}{35.40}\)

Answer 1

E = 2/5.10 + 2/10.15 + ... + 2/35.40

E = 2/5.(1/5 - 1/10 + 1/10 - 1/15 + ... + 1/35 - 1/40)

E = 2/5.(1/5 - 1/40)

E = 2/5.7/40

E = 7/100

Answer 2

E = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)

   = \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\frac{7}{40}\)

   = \(\frac{7}{100}\)

Answer 3

E=\(\frac{2}{5\cdot10}+\frac{2}{10\cdot15}+.....+\frac{2}{35\cdot40}\)

E=\(2\cdot\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{35\cdot45}\right)\)

5E=\(2\cdot5\cdot\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{35\cdot45}\right)\)

5E=\(2\cdot\left(\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{35\cdot45}\right)\)

5E=\(2\cdot\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{45}\right)\)

5E=\(2\cdot\left(\frac{1}{5}-\frac{1}{40}\right)\)

5E=\(2\cdot\frac{7}{40}\)

5E=\(\frac{7}{20}\)

E=\(\frac{7}{20}:5\)

E=\(\frac{7}{100}\)

Answer 4

\(\frac{1}{5}\)E = \(\frac{2}{5}\)- \(\frac{2}{10}\)+\(\frac{2}{10}\)-\(\frac{2}{15}\)+ ... + \(\frac{2}{35}\)- \(\frac{2}{40}\)

             = \(\frac{2}{5}\)- \(\frac{2}{40}\)

              = \(\frac{7}{20}\)

=> E= \(\frac{7}{4}\)

Answer 5

\(E=\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)

\(\frac{5}{2}E=\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\)

\(\frac{5}{2}E=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\)

\(\frac{5}{2}E=\frac{1}{5}-\frac{1}{40}=\frac{8}{40}-\frac{1}{40}=\frac{7}{40}\)

\(=>E=\frac{7}{40}:\frac{5}{2}=\frac{7}{40}.\frac{2}{5}\)

\(E=\frac{7}{100}\)

Answer 6

\(E=\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)

\(E=\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)

\(E=\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)

\(E=\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)

\(E=\frac{2}{5}.\frac{7}{40}\)

\(E=\frac{7}{100}\)

Answer 7

\(E=\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)

\(=\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)

\(=\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)

\(=\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)

\(=\frac{2}{5}.\frac{7}{40}\)

\(=\frac{7}{100}\)