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Khách

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Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)

\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)

\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)

\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)

Suy ra: x+2=82

hay x=80

26 tháng 4 2015

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}.\frac{x+1}{x+2}=\frac{20}{41}\)
\(\frac{x+1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\frac{x+1}{x+2}=\frac{40}{41}\)
\(x+1=40 \)
\(x=40-1\)
\(x=39\)
Đúng thì ****

30 tháng 11 2018

Lương Hồ Khánh Duy trả lời đúng nhưng đúng cảu bài khác

Ở đây, câu hỏi ghi x+1 bn ghi x+2

13 tháng 8 2019

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow20\left(x+2\right)=41\)

\(\Leftrightarrow x-2=\frac{41}{20}\)

\(\Leftrightarrow x=\frac{41}{20}+2\)

\(\Leftrightarrow x=\frac{81}{20}\)

13 tháng 8 2019

\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\) 

\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)

6 tháng 4 2018

Ta có :

\(\dfrac{1}{2}\)(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)

\(\dfrac{1}{2}\)(\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)

\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\)=\(\dfrac{40}{41}\)

\(\dfrac{1}{x+2}\)=\(\dfrac{1}{3}\)-\(\dfrac{40}{41}\)

6 tháng 4 2018

1/3-1/1+1/7-1/5+1/9-1/7...

8 tháng 4 2016

bạn nào giải giúp mình với

nếu đúng thì mình sẽ ***

8 tháng 4 2016

=1/2*(1-1/3+1/3-1/5+....+1/x+1/x+2)

=1/2*(1-1/x+2)

=>1/2*x+1/x+2=20/21

Đến đó đưa về giống tìm x nha

6 tháng 4 2017

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

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

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

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

\(\Rightarrow x+2=2011\)

\(x=2009\)

6 tháng 4 2017

Đặt biểu thứ là A

2A=2/1.3+2/2.5+...+2/x.x+2

2A=1-1/3+1/3-1/5+.......+1/x-1/x+2

2A=1-1/x+2

22 tháng 3 2017

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{n\left(n+2\right)}=\frac{20}{41}\)

\(\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{n\left(n+2\right)}\right)\cdot2=\frac{20}{41}\cdot2\)

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{n\left(n+2\right)}=\frac{40}{41}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{40}{41}\)

\(1-\frac{1}{n+2}=\frac{40}{41}\)

\(\frac{1}{n+2}=1-\frac{40}{41}\)

\(\frac{1}{n+2}=\frac{1}{41}\)

\(\Rightarrow n+2=41\)

\(n=41-2\)

\(n=39\)

11 tháng 1 2018

2/1.3+2/3.5+...+2/x(x+2)= 40/41

1-1/3+1/3-1/5+...+1/x-1/(x+2)=40/41

1-1/(x+2)=40/41

1/(x+2)=1-40/41=1/41

x+2=41

x=41-2=39

11 tháng 1 2018

x = 1235

17 tháng 4 2016

\(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+........+\frac{x}{39\cdot41}=\frac{1}{41}\)

\(\Rightarrow x\cdot\left[\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.........+\frac{1}{39\cdot41}\right)\right]=\frac{1}{41}\)

\(\Rightarrow x\cdot\left[\frac{1}{2}\cdot\left(1-\frac{1}{41}\right)\right]=\frac{1}{41}\)

\(\Rightarrow x\cdot\left(\frac{1}{2}\cdot\frac{40}{41}\right)=\frac{1}{41}\)

\(\Rightarrow x\cdot\frac{20}{41}=\frac{1}{41}\)

\(\Rightarrow x=\frac{1}{41}:\frac{20}{21}\)

\(\Rightarrow x=\frac{1}{41}\cdot\frac{21}{20}\)

\(\Rightarrow x=\frac{21}{820}\)

ai k mh mh k lại

k cho mh nha