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

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

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

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

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

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

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

\(2B=5\left(1-\frac{1}{103}\right)\)

\(2B=5.\frac{102}{103}\)

\(B=\frac{5.102}{103.2}\)

\(B=\frac{255}{103}\)

KL:...........................

xin lỗi b nhé. mk nhầm

b tham khảo bài của b trần tuấn anh nhé.

chúc bạn học tốt.

\(A=\frac{1}{1.4}+\frac{1}{2.7}+...+\frac{1}{67.70}\)

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{67.70}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)

\(3A=1-\frac{1}{70}=\frac{69}{70}\)

\(A=\frac{69}{70}:3=\frac{23}{70}\)

vì \(\frac{23}{70}< 1\)

nên \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{67.70}< 1\)

Vì nó bé hơn 1

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{67.70}\)

\(=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{67.70}\right)\)

\(=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{67.70}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)

\(=1-\frac{1}{70}\)

\(=\frac{69}{70}< 1\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

=>\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

=>\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

=>\(S=\frac{1}{2}.\left(1-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10}\right)\)

=>\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

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

=>\(S=\frac{11}{45}\)

lê chí cường dung 

A = 

A = \(1-\frac{1}{2018}\)

A = \(\frac{2017}{2018}\)

Có : 

2.B = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

2.B = \(1-\frac{1}{2017}\)

2.B = \(\frac{2016}{2017}\)

B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)

Có :

3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)

3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)

3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)

a=1/1-1/2+1/2-1/3+...+1/2017-1/2018

=1/1-1/2018

=kq

may bai duoi lam tuong tu nha

mình chưa điền kết quả ban tu dien nha 

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(A=\frac{1}{1}-\frac{1}{2018}=\frac{2017}{2018}\)

\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2015\cdot2017}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)

\(B=\frac{1008}{2017}\)

\(C=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{2017\cdot2020}\)

\(C=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{2017\cdot2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\cdot\frac{2019}{2020}\)

\(C=\frac{673}{2020}\)

a)\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\\ =\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\\ =\frac{1}{3}.\left(1-\frac{1}{103}\right)\\ =\frac{1}{3}.\frac{102}{103}\\ =\frac{34}{103}\)

@Băng Băng 2k6 lm ik

\(2x-3=x+\frac{1}{2}\\ \Leftrightarrow2x-x=3+\frac{1}{2}\\ x=\frac{7}{2}\)

Vậy ...

\(A=\frac{2}{3}\left[\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right]\)

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

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

\(A=\frac{2}{3}\left[1-\frac{1}{100}\right]=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

AI THẤY ĐÚNG ỦNG HỘ MIK NHÉ

 mn ơi \(2ab=200+ab\) nha không phải \(2\cdot ab\)

làm :                                                                                                                                                                                                                  

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{3}{8}\)

b, \(ab\cdot10-ab=2ab\)

\(ab\cdot10-ab\cdot1=2ab\)

\(ab\cdot\left(10-1\right)=2ab\)

\(ab\cdot9=2ab\)

\(ab\cdot9=200+ab\cdot1\)

\(ab\cdot9-ab\cdot1=200\)

\(ab\cdot\left(9-1\right)=200\)

\(ab\cdot8=200\)

\(ab=200:8\)

\(ab=25\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{1}{3}\left(\frac{99}{100}\right)=\frac{33}{100}\)

33/100 nha ban that do

tìm X cơ