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 

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

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

`#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`

Đặt biểu thức trên là A. Ta có:

3A = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/2016/2019

3A = 1-1/4 +1/4-1/7+1/7-1/10/+ ... + 1/2016-1/2019

3A = 1-1/2019=2018/2019

A =1009/2019

Ta có:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

điều kiện n thuộc N hay khác 0 gì không bạn?

a: =1/2-1/3+1/3-1/4+...+1/99-1/100

=1/2-1/100=49/100

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

=5/3*102/103

=510/309=170/103

c: =1/2(1/3-1/5+1/5-1/7+...+1/49-1/51)

=1/2*16/51=8/51

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

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

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

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

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

\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2021.2014}\)

\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2014}\right)\)

\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{2014}\right)\)

\(\Rightarrow B=\dfrac{1}{3}.\dfrac{2013}{2014}=\dfrac{671}{2014}\)

\(B=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{2021\cdot2024}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2021\cdot2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\dfrac{2023}{2024}\\ =\dfrac{2023}{6072}\)

ai trả lời nhanh mình tick nha 

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{97\cdot100}=\frac{0,33\cdot x}{2009}\cdot3\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,99\cdot x}{2009}\)

\(\frac{100}{100}-\frac{1}{100}=\frac{0,99x}{2009}\)

\(\frac{99}{100}=\frac{0,99x}{2009}\)

=>0,99x*100=2009*99

99x=2009*99

=>x=2009

Vậy x=2009

\(0,33\cdot\frac{x}{2009}\) hay \(\frac{0,33\cdot x}{2009}\)

Băng vệ sinh diana