đặt \(A=\frac{3^4}{20\cdot23}+\frac{3^4}{23\cdot26}+...+\frac{3^4}{77\cdot80}\)

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

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

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

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

\(A=\frac{3^4}{80}=\frac{81}{80}>1\)

\(\frac{3^4}{20.23}+\frac{3^4}{23.26}+...+\frac{3^4}{77.80}\)

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

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

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

\(=\frac{3^3.3}{80}\)

\(=\frac{3^4}{80}\)

\(=\frac{81}{80}\)

\(=\frac{80}{80}+\frac{1}{80}\)

\(=1+\frac{1}{80}\)

=> Biểu thức trên lớn hơn 1

Đặt vế trái là B

\(3B=\frac{23-20}{20.23}+\frac{26-23}{23.26}+\frac{29-26}{26.29}+...+\frac{80-77}{77.80}\)

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

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

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

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

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

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

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

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

:3 B<1/79 mà 1/79<1/9=>B<1/9

Ta có \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)=\dfrac{1}{3}.\dfrac{3}{80}=\dfrac{1}{80}< \dfrac{1}{9}\)

Ta có:

\(\dfrac{1}{20.23}+\dfrac{1}{23.26}+\dfrac{1}{26.29}+...+\dfrac{1}{77.80}\)

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

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

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

=\(\dfrac{1}{80}\)

Vì \(\dfrac{1}{80}\)>\(\dfrac{1}{9}\)

Nên \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+\dfrac{1}{26.29}+...+\dfrac{1}{77.80}>\dfrac{1}{9}\)

Ta có:

\(\\ \dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}\\ =\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)\\ =\dfrac{1}{3}\left(\dfrac{3}{80}-\dfrac{1}{80}\right)\\ =\dfrac{1}{3}.\dfrac{3}{80}\\ =\dfrac{1}{80}\\ \)

Vậy \(\dfrac{1}{80}< \dfrac{1}{9}\)

Ta có : \(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)

\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)=3\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}< 1\)

Bài làm:

Mình nghĩ đề sai rồi, phải như vậy nè:

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

\(A=\frac{23-20}{20.23}+\frac{26-23}{23.26}+...+\frac{80-77}{77.80}\)

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

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

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

Vậy \(A=\frac{3}{80}\)

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

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

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

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

đề bé Sa đúng nha phần gạch phân số mình ko viết đc thông cảm

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

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

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

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

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

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

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