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10 tháng 5 2017

\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)

10 tháng 5 2017

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)

\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)

\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(3M=5\left(1-\frac{1}{2017}\right)\)

\(3M=5.\frac{2016}{2017}\)

\(3M=\frac{10080}{2017}\)

\(\Rightarrow M=\frac{3360}{2017}\)

4 tháng 2 2020

A = \(\frac{1}{1.4}\)\(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+ \(\frac{1}{2014.2017}\)
3A = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2014.2017}\)
3A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{2014}-\frac{1}{2017}\)
3A= 1 - \(\frac{1}{2017}\)
A = \(\frac{1}{3}-\frac{1}{2017.3}\)
A = \(\frac{672}{2017}\)

4 tháng 2 2020

Ta có \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2014.2017}\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\frac{2016}{2017}=\frac{672}{2017}\)

Vậy \(A=\frac{672}{2017}\)

~ Học tốt

# Chiyuki Fujito

6 tháng 4 2017

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

6 tháng 4 2017

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

6 tháng 5 2018

Trả lời

\(B=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\)

\(\frac{3}{5}B=\frac{3}{5}\left(\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+...+\frac{5}{100\cdot103}\right)\)

\(\frac{3}{5}B=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{...3}{100\cdot103}\)

\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

\(\frac{3}{5}B=1-\frac{1}{103}\)

\(\frac{3}{5}B=\frac{102}{103}\)

\(B=\frac{102}{103}:\frac{3}{5}\)

\(B=\frac{170}{103}\)

6 tháng 5 2018

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)

\(3B=15\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(3B=15\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(3B=15\left(\frac{1}{1}-\frac{1}{100}\right)=15\left(\frac{100}{100}-\frac{1}{100}\right)=15.\frac{99}{100}\)

\(B=\frac{1}{3}.15-\frac{1}{3}.\frac{99}{100}=5-\frac{33}{100}=\frac{500}{100}-\frac{33}{100}=\frac{467}{100}\)

3 tháng 3 2017

\(B=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+...+\dfrac{4}{2014\cdot2017}\)

\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2014\cdot2017}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{2017}\right)=\dfrac{4}{3}\cdot\dfrac{2016}{2017}=\dfrac{8064}{6051}\)

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

15 tháng 5 2017

\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(2S=\frac{2.5}{3.13}+\frac{2.5}{13.23}+....+\frac{2.5}{83.93}\)

\(2S=\frac{10}{3.13}+\frac{10}{13.23}+.....+\frac{10}{83.93}\)

\(2S=\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\)

\(2S=\frac{1}{3}-\frac{1}{93}=\frac{30}{93}\)

\(S=\frac{30}{93}.\frac{1}{2}=\frac{15}{93}\)

15 tháng 5 2017

Sửa đề:

\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{31}{93}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\frac{10}{31}\)

\(S=\frac{5}{31}\)

22 tháng 4 2017

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(3B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)

\(3B=5\left(1-\frac{1}{103}\right)\)

\(3B=5.\frac{102}{103}\)

\(3B=\frac{510}{103}\)

\(\Rightarrow B=\frac{170}{103}\)

Ta có:

B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}+.....+\frac{5}{100.103}\)

B=\(\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{100.103}\right)\)

B=\(\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\right)\)

B=\(\frac{5}{3}\left(1-\frac{1}{103}\right)\)

B=\(\frac{5}{3}.\frac{102}{103}\)

B=\(\frac{170}{103}\)

Vậy B=\(\frac{170}{103}\)

nhớ k

3 tháng 8 2018

tớ ko biết

3 tháng 8 2018

S = 3 - \(\frac{3}{100}\)\(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)

25 tháng 8 2017

lp 6  lm bài lp 7 lm j

tí nữa lm cho đag mải

25 tháng 8 2017

đề dễ mà định thi vao đâu vậy

\(A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3\left(1-\frac{1}{100}\right)\)

\(A=\frac{297}{100}\)