\(\frac{3^2}{2.5}\)+ \(\frac{3^2}{5.8}\)+ \(\frac{3^2}{8.11}\)+ . . . . + \(\frac{3^2}{92.95}\)
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LT
6 tháng 3 2020
\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)
=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{17}\right)\)
=\(\frac{45}{34}\)
\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
13 tháng 7 2021
=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)
=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)
=3(1/2-1/17)
=45/34
cô Nhung ơi k đúng cho con đi cô pls
25 tháng 4 2019
Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)
\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)
Vậy:..........................................(đpcm)
21 tháng 5 2018
\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{95\cdot98}\)
\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{95\cdot98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\cdot\frac{48}{98}\)
\(A=\frac{16}{98}=\frac{8}{49}\)
\(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(B=2\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{97\cdot100}\right)\)
\(B=2\left[\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\right]\)
\(B=2\left[\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\right]\)
\(B=2\left[\frac{1}{3}\left(1-\frac{1}{100}\right)\right]\)
\(B=2\left[\frac{1}{3}\cdot\frac{99}{100}\right]\)
\(B=2\cdot\frac{33}{100}\)
\(B=\frac{33}{50}\)
21 tháng 5 2018
A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
3A = 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98
3A = 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98
3A = 1/2 - 1/98
3A = 24/49
A = 24/49 : 3
A = 72/49
B = 2/1.4 + 2/4.7 + 2/7.10 + ... + 2/97.100
3/2B = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100
3/2B = 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100
3/2B = 1 - 1/100
3/2B = 99/100
B = 99/100 : 3/2
B = 33/50
25 tháng 4 2019
xét vế trái
ta có:đề\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}< < \frac{1}{2}\)
vậy vế trái bé hơn \(\frac{1}{2}\)
P/S: \(< < \)là luôn luôn bé hơn nha
k mình nha bạn
Thiengl2015#
KN
25 tháng 4 2019
Ta có :
\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}\)
Mà \(\frac{1}{2}-\frac{1}{17}< \frac{1}{2}\)
Nên \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}< \frac{1}{2}\left(đpcm\right)\)
27 tháng 8 2019
\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{8\cdot11}-...-\frac{1}{92\cdot95}\)
\(=1-\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{2}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\cdot\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
XO
27 tháng 8 2019
\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)
\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
PQ
1
MT
11 tháng 8 2015
\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}\right)=3.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=3.\left(\frac{11}{22}-\frac{2}{22}\right)=3.\frac{9}{22}=\frac{27}{22}\)
26 tháng 4 2018
A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)
A = \(\frac{1}{2}-\frac{1}{98}\)
A = \(\frac{24}{49}\)
Vậy A = \(\frac{24}{49}\)
~~~
#Sunrise
KT
26 tháng 4 2018
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(=\frac{1}{3}.\frac{24}{49}=\frac{8}{49}\)
9 tháng 1 2020
\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
9 tháng 1 2020
\(a\)) Mình giải theo cách khác:
Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)
Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
16 tháng 4 2016
Đề sai rồi! Đề đúng nè: A = 1/2.5 + 1/5.8 +.......+ 1/92.95 + 1/95.98
Bài làm : A =.............Ghi lại đề
3A = 3/2.5 + 3/5.8 +........+ 3/95.98
3A = 1/2 - 1/5 + 1/5 - 1/8 +............+ 1/95 - 1/98
3A = 1/2 - 1/98
3A = 48/98
A = 16/98 = 8/49
16 tháng 4 2016
3A=\(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\)
3A= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{98}\)
3A=\(\frac{1}{2}-\frac{1}{98}\)
3A=\(\frac{49}{98}-\frac{1}{98}\)
3A=\(\frac{24}{49}\)
A=\(\frac{8}{49}\)
=3(3/2.5+3/5.8+...+3/92.95)
=3(1/2-1/5+1/5-1/8+...+1/92-1/95)
=3(1/2-1/95)
=3.93/190=279/190