Question

\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{5\cdot8}-...-\frac{1}{89\cdot92}-\frac{1}{92\cdot95}\)

Answer 1

mk sửa lại đề nha

\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)

= \(1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)

= \(1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)

= \(1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)

= \(1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)

= \(1-\frac{1}{3}.\frac{93}{190}\)

= \(1-\frac{31}{190}\)

= \(\frac{159}{190}\)

Answer 2

\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{89.92}-\frac{1}{92.95}\)

\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.9}\right)\)

\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\frac{93}{190}\)

\(=1-\frac{31}{190}\)

\(=\frac{159}{190}\)