Question

chứng minh rằng : \(\frac{1}{20\cdot23}\)+\(\frac{1}{23\cdot26}\)+\(\frac{1}{26\cdot29}\)+...+\(\frac{1}{77\cdot80}\)<\(\frac{1}{9}\)

giúp mình với mai mình thi rồi

Answer 1

#)Giải:

Đặt \(A=\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)

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

     \(A=\frac{1}{20}-\frac{1}{80}\)

     \(A=\frac{3}{80}< \frac{1}{9}\)

\(\Leftrightarrow A< \frac{1}{9}\)

          #~Will~be~Pens~#

Answer 2

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

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

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

\(=\frac{3}{240}=\frac{1}{80}\)

Vì \(\frac{1}{80}< \frac{1}{9}\)

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

Answer 3

Đặt \(A=\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+\frac{1}{26\cdot29}+...+\frac{1}{77\cdot80}\)

\(\frac{A}{3}=\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+\frac{3}{26\cdot29}+...+\frac{3}{77\cdot80}\)

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

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

\(\frac{A}{3}=\frac{3}{80}\)

\(A=\frac{3}{80}:3=\frac{3}{240}=\frac{1}{80}\)

\(\Rightarrow A=\frac{1}{80}< \frac{9}{720}< \frac{80}{720}=\frac{1}{9}\)

Vậy A < 1/9

Answer 4

Sửa lại phần chỗ này nhé :>

\(A=\frac{1}{80}=\frac{9}{720}< \frac{80}{720}=\frac{1}{9}\)

Answer 5

\(\text{Đặt }A=\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+\frac{1}{26\cdot29}+...+\frac{1}{77\cdot80}\)

        \(3A=\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+\frac{3}{26\cdot29}+...+\frac{3}{77\cdot80}\)

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

        \(3A=\frac{1}{20}-\frac{1}{80}=\frac{4}{80}-\frac{1}{80}\)

        \(3A=\frac{3}{80}\)

        \(A=\frac{3}{80}\text{ : }3=\frac{3}{80}\cdot\frac{1}{3}\)

       \(A=\frac{1}{80}\)