K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

2 tháng 5 2017

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n\left(n+3\right)}\)

\(\Rightarrow S=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{\left(n+3\right)-n}{n\left(n+3\right)}\)

\(\Rightarrow S=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{n+3}{n\left(n+3\right)}-\dfrac{n}{n\left(n+3\right)}\)

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

\(\Rightarrow S=1-\dfrac{1}{n+3}< 1\Rightarrow S< 1\)

Vậy S < 1

2 tháng 7 2017

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)

\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)

\(\Rightarrow S< 1\rightarrowđpcm\)

2 tháng 7 2017

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)

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

\(S=1-\dfrac{1}{n+1}\)\(< 1\)

\(\Leftrightarrow S< 1\)

tik cho mik nhé

1 tháng 5 2021

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)

Vậy S < 1 (đpcm)

1 tháng 5 2021

cảm ơn cậu nhiều na

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cậu thấy mik xinh hum????

12 tháng 4 2017

\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{43\cdot46}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\)

\(S=1-\dfrac{1}{46}< 1\)

25 tháng 4 2017

S= \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)

S= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{42}-\dfrac{1}{46}\)

S= \(1-\dfrac{1}{46}\)

S= \(\dfrac{45}{46}\)

\(\dfrac{45}{46}< 1\)

\(\Rightarrow S< 1\)

Vậy S < 1

NV
4 tháng 1

\(\Leftrightarrow\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow\dfrac{1}{x+3}=1-\dfrac{375}{376}=\dfrac{1}{376}\)

\(\Rightarrow x+3=376\)

\(\Rightarrow x=373\)

9 tháng 4 2015

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

=>S= 1- 1/(n+3)

=>S + 1/(n+3) = 1

=>S<1

a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)

b: undefined

31 tháng 3 2017

a) \(4,5:\left[\left(\dfrac{9-10}{6}\right)-\dfrac{9}{5}+\dfrac{12}{5}\right]-\dfrac{1}{7}\)

\(=4,5:\left(\dfrac{-1}{6}-\dfrac{-3}{5}\right)-\dfrac{1}{7}\)

=\(4,5:\left(\dfrac{-5+18}{30}\right)-\dfrac{1}{7}\)

=\(4,5:\dfrac{13}{30}-\dfrac{1}{7}\)=\(\dfrac{135}{13}-\dfrac{1}{7}=\dfrac{932}{91}\)

b) \(\dfrac{13}{3}:\left(\dfrac{1}{4}+\dfrac{5}{4}\right)-\dfrac{20}{3}\)

=\(\dfrac{13}{3}.\dfrac{2}{3}-\dfrac{20}{3}\)=\(\dfrac{26}{9}-\dfrac{20}{3}=\dfrac{26}{9}-\dfrac{60}{9}=\dfrac{-34}{9}\)

c) \(5.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+.....+\dfrac{1}{91.94}\right)\)

\(=5.\left[\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\right]\)

\(=5.\left[\dfrac{1}{3}.\left(1-\dfrac{1}{94}\right)\right]\)

=\(5.\left(\dfrac{1}{3}.\dfrac{93}{94}\right)\)

\(=5.\dfrac{31}{94}=\dfrac{155}{94}\)

Chúc bạn học tốt hehe

2 tháng 4 2017

cảm ơn

21 tháng 8 2023

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)

\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=\dfrac{1}{103}\)

\(\Rightarrow x+3=103\)

\(x=103-3\)

\(x=100\)

Vậy x = 100