a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)

\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)

\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)

Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)

\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)

\(2A=\frac{12}{3}-\frac{12}{99}\)

\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)

1.5/1-5/4+5/4-5/7+5/7-5/9 +....+5/27-5/30

=5/1-5/30

=145/30=29/6.

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 = \(\frac{672}{2017}\)

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

672/2017

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

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

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

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

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

=5/3.(3/1.4+3/4.7+3/7.10+...+3/100.103)

=5/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)

=5/3.(1-1/103)=5/3.102/103=170/103

                                       đáp số : 170/103

\(\text{Ta có :}\)

\(\frac{3}{1.4}=1-\frac{1}{4}\)

\(\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\)

\(\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\)

\(......\)

\(\frac{19}{81.100}=\frac{1}{81}-\frac{1}{100}\)

\(\text{Cộng vế với vế ta có:}\)

\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Ko biết có đc k ko ta!?

Nguyễn Xuân Anh cảm ơn cậu nha ^^

Cảm ơn mà toàn k sai hoài là sao!! ahuhu

Trả lời

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 \(B=\frac{33}{50}.\)

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

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

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

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

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

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

Sửa đề:

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

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

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

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

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

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

bn lớp 5 ak

Trả lời

3/1.4+3/4.7+3/7.10

=3/1.3/10

=3/10

Chúc bạn học tốt #

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}\)

\(=\frac{9}{10}\)