K
Khách

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

26 tháng 7 2020

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)

=> x + 1 = 2013 => x = 2012

26 tháng 7 2020

Trả lời:

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Leftrightarrow x+1=2013\)

\(\Leftrightarrow x=2012\)

Vậy \(x=2012\)

21 tháng 1 2020

Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(=1-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\)x+1=4026

x=4026-1

x=4025

Vậy x=4025.

21 tháng 1 2020

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(1-\frac{1}{x+1}=\frac{2011}{4026}\)

=> \(\frac{1}{x+1}=\frac{2015}{4026}\Rightarrow x+1=\frac{4026}{2015}\Rightarrow x=\frac{2011}{2015}\)

25 tháng 1 2017

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2012\)

Vậy x = 2012

16 tháng 8 2018

a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)

\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)

b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)

c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)

\(\Rightarrow x=9\)

DD
10 tháng 1 2021

b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{15}\)

21 tháng 7 2015

\(\frac{1}{x}+\frac{1}{2.x}+\frac{1}{6x}+\frac{1}{12x}+\frac{1}{30x}\)

\(\frac{1}{x}\left(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}\right)\)

\(\frac{1}{x}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)

\(\frac{1}{x}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)

=\(\frac{1}{x}\left(1+1-\frac{1}{6}\right)\)

=\(\frac{1}{x}.\frac{11}{6}\)

=\(\frac{11}{6x}\)

1 tháng 6 2018

x + 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 = 1 

x + 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 = 1 

x + 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 = 1 

x + 1/1 - 1/7 = 1 

x + 6/7 = 1 

x = 1 - 6/7 

x = 1/7 

1 tháng 6 2018

x + 1/2 + 1/6 + 1/20 + 1/30 + 1/42 = 1

x + 65/84 = 1

x = 1 - 65/84

x = 19/84

7 tháng 6 2018

Bài 3: 

= 1- 1/2 + 1/2 -1/3 +...+ 1/98 -1/99

= 1- 1/99

= 98/99

Bài 4:

= 1/2*3 + 1/3*4 + 1/4*5 +...+  1/10*11

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/10 - 1/11

= 1/2 - 1/11= 9/22

29 tháng 7 2015

=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2010}{2011}\)

=> \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)

=>\(1-\frac{1}{x+1}=\frac{2010}{2011}\)

=> \(\frac{1}{x+1}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)

=> x + 1 = 2011

=> x = 2010

1 tháng 8 2019

đề lạ thế