\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)

=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)

= \(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

= ​\(3\left(\frac{1}{2}-\frac{1}{17}\right)\)​​

=\(\frac{45}{34}\)

\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)​

=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

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)

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{95\cdot98}\)

\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{95\cdot98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\frac{48}{98}\)

\(A=\frac{16}{98}=\frac{8}{49}\)

\(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)

\(B=2\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{97\cdot100}\right)\)

\(B=2\left[\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\right]\)

\(B=2\left[\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\right]\)

\(B=2\left[\frac{1}{3}\left(1-\frac{1}{100}\right)\right]\)

\(B=2\left[\frac{1}{3}\cdot\frac{99}{100}\right]\)

\(B=2\cdot\frac{33}{100}\)

\(B=\frac{33}{50}\)

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

\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)

\(\Rightarrow A=\frac{24}{49}:3=\frac{8}{49}.\)

Vậy \(A=\frac{8}{49}.\)

\(\frac{3}{2}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow B=\frac{99}{100}:\frac{3}{2}=\frac{33}{50}.\)

Vậy \(B=\frac{33}{50}.\)

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#

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 \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}< \frac{1}{2}\left(đpcm\right)\)

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{14\cdot17}\)

=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

=\(\frac{1}{2}-\frac{1}{17}\)

=\(\frac{17}{34}-\frac{2}{34}\)

=\(\frac{15}{34}\)

Mà \(\frac{1}{2}=\frac{17}{34}>\frac{15}{34}\)

\(\Rightarrowđpcm\)

\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{8\cdot11}-...-\frac{1}{92\cdot95}\)

\(=1-\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}\right)\)

\(=1-\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{2}{92\cdot95}\right)\)

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}\cdot\frac{93}{190}\)

\(=1-\frac{31}{190}\)

\(=\frac{159}{190}\)

\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)

\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)

\(=1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)

\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\frac{93}{190}\)

\(=1-\frac{31}{190}\)

\(=\frac{159}{190}\)

\(1-\frac{1}{2.5}-\frac{1}{5.8}-..-\frac{1}{92.95}=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{92.95}\right)\)

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{92}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)

\(=1-\frac{1}{3}.\frac{93}{190}=1-\frac{31}{190}=\frac{159}{190}\)

   học tốt nha

\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}\right)=3.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=3.\left(\frac{11}{22}-\frac{2}{22}\right)=3.\frac{9}{22}=\frac{27}{22}\)

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

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

\(=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(=\frac{1}{3}.\frac{24}{49}=\frac{8}{49}\)

A=1/3.(1/2-1/5 + 1/5 - 1/8 +......+1/92 - 1/95 + 1/95 - 1/98)

A=1/3.(1/2 - 1/98)

A=1/3. 48/98

A=48/294

Theo mk thì như vậy

Chúc bạn hok tốt ^O^

\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)

\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)

\(a\)) 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: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

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

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=\(\frac{8}{49}\) 

SAI DE