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\(A=\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{100\cdot103}\)

\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{98}{515}\)

28 tháng 3 2022

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\)
   = \(\dfrac{1}{2}\) - \(\dfrac{1}{20}\)
   = \(\dfrac{9}{20}\)

 

28 tháng 3 2022

cách làm như trên

=9/20

\(\Leftrightarrow x\cdot\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{32\cdot35}\right)=\dfrac{33}{70}\)

=>\(x\cdot\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)

=>x=3

2 tháng 8 2023

Mk bt kết quả là x =1 nhung ko bt cách trình bày 

14 tháng 10 2021

\(A=\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+...+\dfrac{3}{2009\times2012}\)

\(A=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{2009}-\dfrac{1}{2012}\)

\(=\dfrac{1}{5}-\dfrac{1}{2012}=\dfrac{2007}{10060}\)

1 tháng 11 2018

a/ \(\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)

\(=3\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\right)\)

\(=3\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\right)\)

\(=3\left(\dfrac{1}{11}-\dfrac{1}{15}\right)\)

\(=\dfrac{4}{55}\)

b/ \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}\)

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)

\(=\dfrac{2}{3}\)

c/ \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+.....+\dfrac{3}{97.100}\)

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

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

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

d/ \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.....+\dfrac{3}{100.103}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{1}{2}-\dfrac{1}{103}\)

\(=\dfrac{101}{206}\)

e/ Đặt :

\(A=\dfrac{1}{1.5}+\dfrac{1}{5.10}+....+\dfrac{1}{95.100}\)

\(\Leftrightarrow5A=\dfrac{5}{1.5}+\dfrac{5}{5.10}+....+\dfrac{5}{95.100}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+....+\dfrac{1}{95}-\dfrac{1}{100}\)

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

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

\(\Leftrightarrow A=\dfrac{99}{100}:5=\dfrac{99}{500}\)

Dấu . là dấu nhân nhé <3

1 tháng 11 2018

Cảm ơn ạyeu

A=1/5-1/8+1/8-1/11+...+1/602-1/605

=1/5-1/605

=24/121

AH
Akai Haruma
Giáo viên
9 tháng 7 2018

a)

\(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2006.2009}\)

\(=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+....+\frac{2009-2006}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)

AH
Akai Haruma
Giáo viên
9 tháng 7 2018

b)

\(B=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)

\(\Rightarrow 4B=\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\)

\(4B=\frac{10-6}{6.10}+\frac{14-10}{10.14}+...+\frac{406-402}{402.406}\)

\(4B=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\)

\(4B=\frac{1}{6}-\frac{1}{406}=\frac{100}{609}\Rightarrow B=\frac{25}{609}\)

15 tháng 7 2020

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}\)

\(=\frac{2004}{10045}\)

15 tháng 7 2020

Đề: Tính

\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)

Vậy \(A=\frac{2004}{10045}.\)

A=3/5x8+3/8x11+3/11x14+...+3/2009x2012+3/2012x2015

A=1/5-1/8+1/8-1/11+1/11-1/14+...+1/2009-1/2012+1/2012-1/2015

A=1/5-1/2015

A=403/2015-1/2015

A=402/2015

Dũng làm sai