CMR: 1/20.23+1/23.26+1/26.29+....+1/77.80 < 1/9
K
Khách
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Những câu hỏi liên quan
30 tháng 5 2021
Ta có \(\dfrac{1}{20.23}+\dfrac{1}{23.26}+...+\dfrac{1}{77.80}=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\right)=\dfrac{1}{3}\left(\dfrac{1}{20}-\dfrac{1}{80}\right)=\dfrac{1}{3}.\dfrac{3}{80}=\dfrac{1}{80}< \dfrac{1}{9}\)
DQ
1
NG
14 tháng 6 2020
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80} \)
\(=\frac{1}{3}.(\frac{1}{20}-\frac{1}{23})+\frac{1}{3}.(\frac{1}{23}-\frac{1}{26})+...+\frac{1}{3}.(\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80})\)
=\(\frac{1}{3}.(\frac{1}{20}-\frac{1}{80})\)
=\(\frac{1}{3}.\frac{3}{80}\)
=\(\frac{1}{80}\)<\(\frac{1}{9}\)
Vậy tổng trên nhỏ hơn \(\frac{1}{9}\)
PM
3
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
22 tháng 6 2020
Đặt vế trái là B
\(3B=\frac{23-20}{20.23}+\frac{26-23}{23.26}+\frac{29-26}{26.29}+...+\frac{80-77}{77.80}\)
\(3B=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}\)
\(3B=\frac{3}{80}\Rightarrow B=\frac{1}{80}< \frac{1}{9}\)
22 tháng 6 2020
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 \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}< \frac{1}{9}\)
K
1
PT
Phạm Trần Hoàng Anh
CTVHS
22 tháng 6 2020
1/20.23+1/23.26+1/26.29+...+1/77.80 < 3/20.23+3/23.26+3/26.29+...+3/77.80=1/20-1/23+1/23-1/26+...+1/77-1/80
=1/20-1/80=3/80 < 1/9 = 3/27
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: \(\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
= \(\frac{1}{3.}\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+....+\frac{1}{77}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\frac{3}{80}=\frac{1}{80}< \frac{1}{9}\)
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
\(\frac{3}{9}B=\frac{3}{9}.\left(\frac{9}{20.23}+\frac{9}{23.26}+\frac{9}{26.29}+...+\frac{9}{77.80}\right)\)
=> \(\frac{3}{9}B=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+....+\frac{3}{77.80}\)
=\(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+....+\frac{1}{77}-\frac{1}{80}\)
=\(\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)
=> \(B=\frac{3}{80}:\frac{3}{9}=\frac{3}{80}.\frac{9}{3}=\frac{9}{80}\)
7 tháng 8 2019
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{27.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
\(\frac{1}{20.23} +\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
= \(\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
= \(\frac{1}{3}.\frac{3}{80}\)
= \(\frac{1}{80}\) < \(\frac{1}{9}\)
⇒ \(\frac{1}{20.23} +\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\) < \(\frac{1}{9}\) (ĐPCM)