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17 tháng 4 2017

\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)

\(\Rightarrow x+1=2017\)

\(\Rightarrow x=2017-1=2016\)

Vậy x = 2016

17 tháng 4 2017

\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{2016}{2017}\)

1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)

\(\dfrac{3}{4}\)+\(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)

\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{2013}{8068}\)

Bn tự lm tiếp nhé!!! Sorry mk đang vội

27 tháng 3 2017

Bài 2:

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2016}{2017}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)

\(\Leftrightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)

\(\Leftrightarrow x+1=2017\Leftrightarrow x=2016\)

Vậy \(x=2016\)

25 tháng 12 2018

2.x=2016

17 tháng 4 2017

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)

\(\dfrac{1}{x+1}=\dfrac{1}{600}\)

=> x + 1 = 600

x = 600 - 1

x = 599

Vậy x = 599

14 tháng 8 2017

a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)

\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)

\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)

\(x=\dfrac{-9198}{4400}\)

14 tháng 8 2017

a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)

\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)

\(x+\dfrac{206}{100}=5\)

\(x=5-\dfrac{206}{100}\)

\(x=\dfrac{147}{50}\)

Vậy \(x=\dfrac{147}{50}\)

18 tháng 3 2018

|2x - 1|.\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1996.1997}\right)\)= 1996

|2x - 1|.\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1996}-\dfrac{1}{1997}\right)\)= 1996

|2x - 1|.\(\left(1-\dfrac{1}{1997}\right)\)= 1996

|2x - 1|. \(\dfrac{1996}{1997}\)= 1996

|2x - 1| = 1996 : \(\dfrac{1996}{1997}\)

|2x - 1| = 1996 . \(\dfrac{1997}{1996}\)

|2x - 1| = 1997

2x - 1 = ± 1997

TH1:

2x -1 = 1997

2x = 1997 +1

2x= 1998

x= 1998:2

x=999

TH2:

2x-1= -1997

2x= -1997+1

2x= -1996

x= -1996:2

x= -998

Vậy x {999; -998}

18 tháng 3 2018

Phân phối phép nhân với phép cộng: v

20 tháng 3 2022

\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)

20 tháng 3 2022

\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)

9 tháng 5 2017

\(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2016}-\dfrac{1}{2017}\)

\(B=1-\dfrac{1}{2017}\)

\(B=\dfrac{2017}{2017}-\dfrac{1}{2017}\)

\(B=\dfrac{2016}{2017}\)

9 tháng 5 2017

câu này truong minh lm hoai a

hihahiha

21 tháng 3 2017

a, đặt đề bài là A

Ta có : A=( 1-1/2+1/2-1/3+...+1/9-1/10).(x-1)+1/10.x=x-9/10

= (1-1/10).(x-1)+1/10.x

= 9/10 .( x-1 )+1/10.x

=1.x-9/10

nên x= 0 hoặc 1

21 tháng 3 2017

với -1 nữa nha