`2/(1.4) + 2/(4.7) + 2/(7.10) + 2/(10.13)`

`= 1/3.2(3/(1.4) + 3(4.7) + 3/(7.10) + 3/(10.13))`

`= 2/3 . (1 - 1/4 + 1/4 -1/7 + 1/7 - 1/10 + 1/10 - 1/13)`

`= 2/3 .(1 - 1/13)`

`= 2/3 . 12/13`

`= 8/13`

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

\(A=2\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=2\left(1-\frac{1}{100}\right)\)

\(A=2.\frac{99}{100}=..............\)

Tự làm nốt nha

what?

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)

\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)

\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)

\(\Leftrightarrow376x+752=375x+1125\)

\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)

mk đc thầy cho làm bài này rồi nên cảm thấy nó dễ mà

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)

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

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

Còn lại thì dễ rồi bạn nhé

lộn đề

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

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

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

Ta có 

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)

\(S=1-\frac{1}{n+3}< 1\)(vì n thuộc N*)

_Kudo_

Cảm ơn bn

Bài làm

S= 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/n. (n+3)

  =1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... +1/(n-3) - 1/n +1/n - 1/(n+3)

  =1/1 + ( - 1/4 + 1/4 - 1/7 + 1/7 - ... -1/n + 1/n ) -1/ (n+3)

  = 1 + 0 - 1/(n+3)

  = 1 - 1/(n+3)

Mà 1 - 1/(n+3) < 1

Vậy S < 1

Mk trình bày ko đc chi tiết lắm ,sorry bạn nha

không thích trả lời đấy

.................................

mik chỉ hỏi thui mà, mik có bắt bạn giúp âu

họ hỏi ko trả lời thì hoi mặc chửi ngt :)

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

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

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

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

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

ai trả lời nhanh mình tick nha 

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

\(B=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2008}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{3}.\dfrac{2010}{2011}=\dfrac{2010}{6033}\)

Lại có : \(1=\dfrac{6033}{6033}\Rightarrow B< 1\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2008.2011}\)

\(=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2008}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{3}.\dfrac{2010}{2011}\)

\(=\dfrac{2010}{6033}=\dfrac{670}{2011}\)

Vì phân số \(\dfrac{670}{2011}\) có tử số nhỏ hơn mẫu số ⇒ \(\dfrac{670}{2011}< 1\) hay \(B< 1\)