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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=\(\frac{8}{49}\) 

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

26 tháng 4 2018

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

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

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

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

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

8 tháng 5 2018

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

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

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

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

\(A=\frac{1}{3}\left(\frac{49}{98}-\frac{1}{98}\right)\)

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

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

Vậy A = \(\frac{8}{49}\)

8 tháng 5 2018

Phân tích: 1/2.5 = 1/2 - 1/5
1/5.8 = 1/5 - 1/8
1/8.11 = 1/8 - 1/11
...
1/92.95 = 1/92 - 1/95
1/95.98 = 1/95 - 1/98
Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98
3 =  1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
= 1 - 1/98
= 97/98 : 3 = 97/98 x 1/3 = (tự tính)

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

4 tháng 5 2017

A=1/3x(1/2x5+1/5x8+......+1/95x98)

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

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

A=1/3x24/49

A=8/49

4 tháng 5 2017

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

A = \(\frac{1.3}{2.5.3}+\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+...+\frac{1.3}{92.95.3}+\frac{1.3}{95.98.3}\)

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

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

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

A =\(\frac{1}{3}.\frac{97}{98}\)

A =\(\frac{97}{294}\)

4 tháng 4 2016

3B = 3/2.5 + 3/5.8 +......+3/35.38

3B = 1/2 - 1/5 + 1/5 - 1/8 +......+ 1/35 - 1/38

3B = 1/2 - 1/38

3B = 18/38

B = 6/38 = 2/19

Vậy B = 2/19 nhé!

4 tháng 4 2016

Bằng 3/19 mới đúng nhé!

14 tháng 2 2016

ủng hộ mình nha

14 tháng 2 2016

  \(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)

\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)=\frac{1}{2}\left(\frac{34}{68}-\frac{1}{68}\right)=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)

 

8 tháng 6 2019

#)Giải :

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)

\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{99.101}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{101}\)

\(\Rightarrow3A=\frac{99}{202}\)

\(\Leftrightarrow A=\frac{33}{202}\)

8 tháng 6 2019

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)

28 tháng 3 2017

\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)

\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

mk đầu tiên đó

28 tháng 3 2017

=\(\frac{3}{20}=0,15\)

12 tháng 5 2017

A=...

<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)

12 tháng 5 2017

sai ùi