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T
3
DP
26 tháng 7 2017
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)
MH
25 tháng 6 2020
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(S=\frac{1}{2}-\frac{1}{32}\)
\(S=\frac{17}{32}< 1\)
LN
31 tháng 1 2019
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
MV
22 tháng 4 2017
2)
S = \(\dfrac{6}{2.5}\) + \(\dfrac{6}{5.8}\) + ... + \(\dfrac{6}{26.29}\)+ \(\dfrac{6}{29.32}\)
= 2.\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{29.32}\right)\)
= \(2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{29}-\dfrac{1}{32}\right)\)
= 2.\(\left(\dfrac{1}{2}-\dfrac{1}{32}\right)\)
= 1 - \(\left(2.\dfrac{1}{32}\right)\)< 1
Vậy S < 1
BB
1
18 tháng 4 2019
\(S=2.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=1-\frac{1}{16}< 1\)
Vậy \(S< 1\)
NL
2
NM
3
15 tháng 4 2021
Đặt A = \(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{23.26}+\frac{1}{26.29}\)
3A = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}+\frac{3}{26.29}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}\)
= \(\frac{1}{2}-\frac{1}{29}\)\(=\frac{27}{58}\)
A = \(\frac{27}{58}:3=\frac{9}{58}\)
XO
15 tháng 4 2021
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{23.26}+\frac{1}{26.29}=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}+\frac{3}{26.29}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}\right)=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{29}\right)\)
\(=\frac{1}{3}.\frac{27}{58}=\frac{9}{58}\)
24 tháng 4 2018
Bài 1 :
S = \(\frac{6}{2.5}+\frac{6}{5.8}+...+\frac{6}{29.32}\)
= 2 . \(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{32}\right)\)= ....
30 tháng 3 2018
a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=1-\frac{1}{25}\)
\(=\frac{24}{25}\)
30 tháng 3 2018
đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)
\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)
\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)
\(2A=\frac{1}{9}-\frac{1}{45}\)
\(2A=\frac{4}{45}\)
\(A=\frac{4}{45}\div2\)
\(A=\frac{2}{45}\)
22 tháng 4 2017
a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?
\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)
b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)
=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)
=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)
S=6/2*5+6/5*8+...+6/29*32
=6/3*(3/2*5+3/5*8+...+3/29*32)
=2*(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2*(1/2-1/32)=2*15/32
=15/16<1
S=6/2*5+6/5*8+...+6/29*32,c
=6/3*(3/2*5+3/5*8+...+3/29*32)
=2*(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2*(1/2-1/32)
=2*15/32
=15/16<1