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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
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{80}\)
\(\frac{A}{3}=\frac{3}{80}\)
\(A=\frac{3}{80}.3=\frac{9}{80}< 1\)
Đặt A=32/20.23+32/23.26+....................+32/77.80
A=3(3/20.23+3/23.26+.........+3/77.80)
A=3(1/20-1/23+1/23-1/26+.+1/77-1/80)
A=3(1/20-1/80)
A=3.3/80
A=9/80 Mà A=9/80<1 =>A<1 (đpcm)
\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+\frac{3^2}{26.29}+...+\frac{3^2}{77.80}\)
=\(\frac{3.3}{20.23}+\frac{3.3}{23.26}+\frac{3.3}{26.29}+...+\frac{3.3}{77.80}\)
=\(\frac{3}{20}-\frac{3}{23}+\frac{3}{23}-\frac{3}{26}+\frac{3}{26}-\frac{3}{29}+....+\frac{3}{77}-\frac{3}{80}\)
=\(\frac{3}{20}-\frac{3}{80}\)
=\(\frac{9}{80}\)
Ta có:
\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+\frac{3^2}{26.29}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)=3.\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}\)
=\(3\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=3\left(\frac{4}{80}-\frac{1}{80}\right)\)
\(=3.\frac{3}{80}\)
\(=\frac{9}{80}\)
\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}<\frac{1}{8}\)
\(=3\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=3.\frac{3}{80}=\frac{9}{80}\)
\(\Rightarrow\frac{9}{80}=\frac{1}{8}\)
Ta có
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(A=3^2\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(A=3^2\cdot\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3\cdot\frac{3}{80}=\frac{9}{80}< 1\left(9< 80\right)\)
\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(\frac{A}{3}=\frac{23-20}{20.23}+\frac{26-23}{23.26}+...+\frac{80-77}{77.80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\Rightarrow A=\frac{9}{80}< 1\)
\(3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+....+\frac{3}{77\cdot80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+.....+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{3}{20}-\frac{3}{80}\)
\(< 1\)
Ta có B= 3(\(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\) )
B= 3\(\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)
B= 3.(\(\frac{1}{20}-\frac{1}{80}\))
B=3.\(\frac{3}{80}\)=\(\frac{9}{80}\)
\(\frac{3}{9}B=\frac{3}{9}.\left(\frac{9}{20.23}+\frac{9}{23.26}+\frac{9}{26.29}+...+\frac{9}{77.80}\right)\)
=> \(\frac{3}{9}B=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+....+\frac{3}{77.80}\)
=\(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+....+\frac{1}{77}-\frac{1}{80}\)
=\(\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)
=> \(B=\frac{3}{80}:\frac{3}{9}=\frac{3}{80}.\frac{9}{3}=\frac{9}{80}\)
Ta có : \(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)=3\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}< 1\)