ta nhân 3 cả hai vế, được : 

\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)

hay 

\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)

\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)

Ta có : 1/ 1.4 + 1/ 4.7 + .... + 1/ 2016.2019 .

      = 1 - 1/4 + 1/4 - 1/7 + ... + 1/2016 - 1/2019 .

      = 1 - 1/2019 .

      = 2018/2019 .

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2016.2019}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2016.2019}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2016}-\frac{1}{2019}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{2019}\right)\)

\(=\frac{1}{3}.\frac{2018}{2019}\)

\(=\frac{2018}{6057}\)

_Chúc bạn học tốt_

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\right).\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{97}\right)\)

\(=\frac{1}{3}.\frac{96}{97}\)

\(=\frac{32}{97}\)

học tốt 

3A = 3(1/1.4 + 1/4.7 + 1/7.10 + ...... + 1/94.97)

3A=1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - ........ - 1/97

3A = 1-1/97

3A = 96/97

A = 32/97

Oke nha bạn

A = \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{94.97}\)

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)

\(3A=1-\frac{1}{97}\)

\(3A=\frac{96}{97}\)

\(A=\frac{32}{97}\)

\(3B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}.\)

\(3B=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{103-100}{100.103}\)

\(3B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}=1-\frac{1}{103}=\frac{102}{103}\)

\(B=\frac{102}{3.103}=\frac{34}{103}\)

B=\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......+\frac{1}{100.103}\)

B=\(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{100.103}\right)\)

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

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

B=\(\frac{1}{3}.\frac{102}{103}\)

B=\(\frac{34}{103}\)

More images for 1−14 +14 −17 +...+197 −1100 =0,99·x2009 100100 −1100 =0,99x2009 99100 =0,99x2009 =>0,99x*100=2009*9999x=2009*99=>x=2009Vậy x=2009 Đúng 4 Sai 0 Diana Andrea đã chọn câu trả lời này.Đỗ Lê Tú Linh 26/12/2015 lúc 22:10 Báo cáo sai phạm

`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`

`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`

`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

Giúp mình đi

Đặt 2/3 ra ngoài  trong ngoặc còn :

1-1/4+1/4-1/7+...-1/97=96/97

Lấy 2/3 nhân với 96/97 sẽ ra đáp án nhé

100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 nha bạn hjhj o

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

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

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

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

  \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\) + \(\frac{1}{7.10}\)

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

= 1 -  \(\frac{1}{10}\)

=    \(\frac{9}{10}\)