Question

[LÊ TRẦN BÁCH]

Answer 1

[Akai Haruma]

Lời giải:

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

$3A=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}$

$=\frac{23-20}{20.23}+\frac{26-23}{23.26}+\frac{29-26}{26.29}+...+\frac{80-77}{77.80}$

$=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}$

$=\frac{1}{20}-\frac{1}{80}$

$A=\frac{1}{3}(\frac{1}{20}-\frac{1}{80})=\frac{1}{60}-\frac{1}{240}< \frac{1}{60}< \frac{1}{9}$

Answer 2

[Nguyễn Việt Lâm]

Ta có:

\(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}\)

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

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

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

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