đặt A=9/20x23+9/23x26+...+9/77x80

<=>\(A=3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right)\)

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

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

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

\(\Rightarrow A=\frac{9}{80}\)

A=1/20*23+1/23*26+...+1/77*80

=1/3(1/20-1/23+1/23-1/26+...+1/77-1/80)

=1/3*3/80=1/80<1/79

\(\frac{1}{11×14}+\frac{1}{14×17}+\frac{1}{17×20}+\frac{1}{20×23}+\frac{1}{23×26}\)

\(=\frac{1}{3}×\left(\frac{3}{11×14}+\frac{3}{14×17}+\frac{3}{17×20}+\frac{3}{20×23}+\frac{3}{23×26}\right)\)

\(=\frac{1}{3}×\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{23}-\frac{1}{26}\right)\)

\(=\frac{1}{3}×\left(\frac{1}{11}-\frac{1}{26}\right)\)

\(=\frac{1}{3}×\frac{15}{286}\)

\(=\frac{5}{286}\)

\(\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}+\frac{1}{20\times23}+\frac{1}{23\times26}\)

\(=\frac{1}{3}\times\left(\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}+\frac{1}{20\times23}+\frac{1}{23\times26}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{11}-\frac{1}{26}\right)\)

  =    5/286

Đặt biểu thức trên là A. Ta có :

3A = \(\frac{3}{11x14}+\frac{3}{14x17}+\frac{3}{17x20}+\frac{3}{20x23}+\frac{3}{23x26}\)

3A = \(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+\frac{1}{23}-\frac{1}{26}\)

3A = \(\frac{1}{11}-\frac{1}{26}=\frac{15}{286}\)

A = \(\frac{15}{286}:3=\frac{5}{286}\)

                                  Vậy A = \(\frac{15}{286}\)

Đặt \(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\) ta có : 

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

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

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

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

\(A=\frac{9}{80}< 1\) ( tử bé hơn mẫu ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}\)

\(=\frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right]\)

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

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

\(=\frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)

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

Mà \(\frac{1}{80}< \frac{1}{9}\)nên \(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+...+\frac{1}{77\cdot80}< \frac{1}{9}\)

Vậy : ...

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

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

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

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

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

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

32+(32:4)=32+8=40

Nguyễn Minh Vũ mình biết rồi nhé cảm ơn bạn 

Chọn B