Chứng minh 3^2/ 20.23+3^2/ 23.26+...+ 3^2/ 77.80< 1
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29 tháng 4 2015
3^2= 9
Vậy thì sẽ là:
9/ 20.23+ 9/ 23.26+...9/77.80
cách nhau 3 bỏ 3 ra ngoài
= 3(3/20.23+...3/77.80)
=3(3/20-3/23+3/23-3/26+.....+3/77-3/80)
=3(3/20-3/80)
=3. 9/80
=27/80<1
27 tháng 4 2017
32=9
\(\frac{3^2}{20.23}\)+\(\frac{3^2}{23.26}\)+...+\(\frac{3^2}{77.80}\)
=\(\frac{9}{20.23}\)+\(\frac{9}{23.26}\)+...+\(\frac{9}{77.80}\)
=3(\(\frac{3}{20.23}\)+\(\frac{3}{23.26}\)+...+\(\frac{3}{77.80}\))
=3(\(\frac{1}{20}\)-\(\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\))
=3(\(\frac{1}{20}-\frac{1}{80}\))
=3(\(\frac{4}{80}-\frac{1}{80}\))
=3.\(\frac{3}{80}\)
=\(\frac{9}{80}\)<1
Vậy\(\frac{9}{80}< 1\)
DN
1 tháng 5 2017
\(\dfrac{3^2}{20.23}\)+\(\dfrac{3^2}{23.26}\)+...+\(\dfrac{3^2}{77.80}\)
=> \(\dfrac{9}{20.23}+...+\dfrac{9}{77.80}\)
= 9.\(\left(\dfrac{1}{20.23}+...+\dfrac{1}{77.80}\right)\)
\(=9.\left(\dfrac{1}{20.3}-\dfrac{1}{23.3}+\dfrac{1}{23.3}-\dfrac{1}{26.3}+...+\dfrac{1}{77.3}-\dfrac{1}{80.3}\right)\)= \(9.\left(\dfrac{1}{20.3}-\dfrac{1}{80.3}\right)\)
\(=9.\dfrac{1}{80}\)=\(\dfrac{9}{80}=0,1125< 1.\)
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)
ND
25 tháng 4 2017
Đặt A= ...(như trên)
=>\(\dfrac{1}{3}A=\dfrac{1}{3}.\left(\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+...+\dfrac{3^2}{77.80}\right)\)
=>\(\dfrac{1}{3}A=\dfrac{3}{20.23}+\dfrac{3}{23.26}+...+\dfrac{3}{77.80}\)
=>\(\dfrac{1}{3}A=\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\\ \)
=>\(\dfrac{1}{3}A=\dfrac{1}{20}-\dfrac{1}{80}\\ =>\dfrac{1}{3}A=\dfrac{4}{80}-\dfrac{1}{80}\\ =>\dfrac{1}{3}A=\dfrac{3}{80}=>A=\dfrac{3}{80}:\dfrac{1}{3}\\ =>A=\dfrac{3}{80}.3=\dfrac{9}{80}< 1\)
Vậy A<1 . Chúc bạn học tốt ! :)
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
LB
8 tháng 5 2017
Ta có
\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)
\(A=3^2\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(A=3^2\cdot\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3\cdot\frac{3}{80}=\frac{9}{80}< 1\left(9< 80\right)\)
NT
1
3 tháng 5 2016
\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}<\frac{1}{8}\)
\(=3\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=3.\frac{3}{80}=\frac{9}{80}\)
\(\Rightarrow\frac{9}{80}=\frac{1}{8}\)
