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=>2/1*3+2/3*5+...+2/(2x-1)(2x+1)=98/99

=>1-1/3+1/3-1/5+...+1/(2x-1)-1/(2x+1)=98/99

=>1-1/(2x+1)=98/99

=>1/(2x+1)=1/99

=>2x+1=99

=>x=49

20 tháng 8 2023

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)

=> 2x=98

=> x=49

29 tháng 4 2018

\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left\{\left(2x+1\right).\left(2x+3\right)\right\}}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\cdot\left(\frac{2x+3}{2x+3}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\frac{2x+2}{2x+3}=\frac{49}{99}\)

\(\frac{2x+2}{2x+3}=\frac{49}{99}:\frac{1}{2}\)

\(\frac{2x+2}{2x+3}=\frac{98}{99}\)

=) \(2x+2=98\)và \(2x+3=99\)

TH1 : \(2x+2=98\)

\(2x=98-2\)

\(2x=96\)

\(x=96:2\)

\(x=48\)( THỎa mãn )

TH2 : 
\(2x+3=99\)

\(2x=99-3\)

\(2x=96\)

\(x=96:2\)

\(x=48\)( THỎa mãn )

Vậy x = 48

29 tháng 4 2018

Đặt A=

15 tháng 5 2017

Có:

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(1-0-0-0...-0-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{5}{11}:\dfrac{1}{2}=\dfrac{10}{11}\)

\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{10}{11}\)

\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{11}\)

\(\Rightarrow x+2=11\)

\(\Rightarrow x=11-2=9\)

Vậy x = 9.

Chúc bạn học tốt!ok

15 tháng 5 2017

1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2)

= 1/2.(1/1 - 1/3) + 1/2.(1/3 - 1/5) + 1/2.(1/5 - 1/7) + ... + 1/2.(1/x -1/x+2)

= 1/2.(1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 )

= 1/2.(1/1 - 0 - 1/x+2 )

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

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

= 1/2 . x+1/x+2

Mà 1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2) = 5/11

=> 1/2 . x+1/x+2 = 5/11

=> x+1/x+2 = 5/11 : 1/2

=> x+1/x+2 = 10/11

=> x+1/x+2-1 = 10/11-1

=> x+1/x+1 = 10/10

=> x + 1 = 10

=> x = 10 - 1

=> x = 9

Vậy x = 9

18 tháng 3 2016

Gọi \(A=\frac{1005}{2011}\)

A=1/3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)

A=1/1.3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)

A . 2=2/1.3 + 2/3.5 + 2/5.7 +......................+2/x.(x+2)

A . 2=1/1-1/3+1/3-1/5+1/5-1/7+..............+1/x-1/x+2

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

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

Suy gia:1005/2011 . 2=1/1-1/x+2

             2010/2011    =1/1-1/x+2

             1/x+2           =1/1-2010/2011

              1/x+2          =1/2011

Suy gia:x+2=2011

            x    =2011-2

            x    =2009

26 tháng 7 2016

\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)

\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)

\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)

\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)

\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)

8 tháng 4 2016

bạn gom các số vào tách số 1/2 ra ngoài làm thừa số,tử 1 chuyển thành 2 lập hiệu xuất hiện tích đối nhau trừ đi phân phối ra là xong

8 tháng 4 2016

1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x.(x+2) = 30/61

( 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x.(x+2) x 2 = 30/61 x 2

2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2) = 60/61

1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x + 2 = 60/61

1 - 1/x+2 = 60/61

1/x+2 = 1 - 60/61

1/x+2 = 1/61

x + 2 = 61

x       = 61 - 2

x       = 59 

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