Tính tổng:
a) \(\frac{3^2}{20.23}+\frac{3^2}{23.26}+\frac{3^2}{26.29}+...+\frac{3^2}{77.80}\)
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PT
2
E
11 tháng 4 2017
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.\(\frac{3}{80}\)=\(\frac{9}{80}\)
D
11 tháng 4 2017
\(\frac{3}{9}B=\frac{3}{9}.\left(\frac{9}{20.23}+\frac{9}{23.26}+\frac{9}{26.29}+...+\frac{9}{77.80}\right)\)
=> \(\frac{3}{9}B=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+....+\frac{3}{77.80}\)
=\(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+....+\frac{1}{77}-\frac{1}{80}\)
=\(\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)
=> \(B=\frac{3}{80}:\frac{3}{9}=\frac{3}{80}.\frac{9}{3}=\frac{9}{80}\)
NV
2
AH
29 tháng 3 2018
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{80}\)
\(\frac{A}{3}=\frac{3}{80}\)
\(A=\frac{3}{80}.3=\frac{9}{80}< 1\)
29 tháng 3 2018
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)
TT
Chứng minh rằng
\(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)
3
6 tháng 5 2019
Đặt A=\(\frac{1}{20.23}+\frac{1}{23.26}+....+\frac{1}{77.80}\)
=>A=\(\frac{1}{3}\).(\(\frac{3}{20.23}+\frac{3}{23.26}+....+\frac{3}{77.80}\))
=>A=\(\frac{1}{3}\).(\(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+.....+\frac{1}{77}-\frac{1}{80}\))
=>A=\(\frac{1}{3}\).(\(\frac{1}{20}-\frac{1}{80}\))
=>A=\(\frac{1}{3}.\frac{3}{80}\)
=>A=\(\frac{1}{80}\)
Do \(\frac{1}{80}\)<\(\frac{1}{9}\)
Nên \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+....+\frac{1}{77.80}< \frac{1}{9}\)
TT
2
6 tháng 5 2019
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
6 tháng 5 2019
Ta có: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
= \(\frac{1}{3.}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+....+\frac{1}{77}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)
DP
3
12 tháng 5 2015
=\(3\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=3\left(\frac{4}{80}-\frac{1}{80}\right)\)
\(=3.\frac{3}{80}\)
\(=\frac{9}{80}\)
NT
1
3 tháng 5 2016
\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}<\frac{1}{8}\)
\(=3\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=3.\frac{3}{80}=\frac{9}{80}\)
\(\Rightarrow\frac{9}{80}=\frac{1}{8}\)
NT
1
LB
8 tháng 5 2017
Ta có
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(A=3^2\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(A=3^2\cdot\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3\cdot\frac{3}{80}=\frac{9}{80}< 1\left(9< 80\right)\)
HH
2
23 tháng 4 2018
đặt \(A=\frac{3^4}{20\cdot23}+\frac{3^4}{23\cdot26}+...+\frac{3^4}{77\cdot80}\)
\(A=3^3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+...+\frac{3}{77\cdot80}\right)\)
\(A=3^3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3^3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3^3\cdot\frac{3}{80}\)
\(A=\frac{3^4}{80}=\frac{81}{80}>1\)
23 tháng 4 2018
\(\frac{3^4}{20.23}+\frac{3^4}{23.26}+...+\frac{3^4}{77.80}\)
\(=3^3\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=3^3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3^3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{3^3.3}{80}\)
\(=\frac{3^4}{80}\)
\(=\frac{81}{80}\)
\(=\frac{80}{80}+\frac{1}{80}\)
\(=1+\frac{1}{80}\)
=> Biểu thức trên lớn hơn 1
\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+\frac{3^2}{26.29}+...+\frac{3^2}{77.80}\)
=\(\frac{3.3}{20.23}+\frac{3.3}{23.26}+\frac{3.3}{26.29}+...+\frac{3.3}{77.80}\)
=\(\frac{3}{20}-\frac{3}{23}+\frac{3}{23}-\frac{3}{26}+\frac{3}{26}-\frac{3}{29}+....+\frac{3}{77}-\frac{3}{80}\)
=\(\frac{3}{20}-\frac{3}{80}\)
=\(\frac{9}{80}\)
Ta có:
\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+\frac{3^2}{26.29}+...+\frac{3^2}{77.80}=3\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)=3.\left(\frac{1}{20}-\frac{1}{80}\right)=3.\frac{3}{80}=\frac{9}{80}\)