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

Đặt A=\(\frac{1}{20.23}+\frac{1}{23.26}+....+\frac{1}{77.80}\)

=>A=\(\frac{1}{3}\).(\(\frac{3}{20.23}+\frac{3}{23.26}+....+\frac{3}{77.80}\))

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

=>A=\(\frac{1}{3}\).(\(\frac{1}{20}-\frac{1}{80}\))

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

=>A=\(\frac{1}{80}\)

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

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

Answer 2

ko bt

Answer 3

giúp mik nha đng cần gấp