Question

Chứng minh rằng

\(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)

Answer 1

\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=\frac{1}{3}.\frac{3}{80}\)

\(=\frac{1}{80}< \frac{1}{9}\)

Answer 2

Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)

= \(\frac{1}{3.}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)

= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+....+\frac{1}{77}-\frac{1}{80}\right)\)

= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)

= \(\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)