x/1.3+x/3.5+x/5.7+....+x/2022.2023
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NT
1
HM
2
DP
16 tháng 7 2017
\(\frac{x}{1.3}+\frac{x}{3.5}+\frac{x}{5.7}+....+\frac{x}{97.99}=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}.\frac{98}{99}=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}=\frac{49}{99}\div\frac{98}{99}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}\times2=1\)
16 tháng 7 2017
\(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+...+\frac{x}{97\cdot99}=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{97}{99}\)
\(\Rightarrow\frac{x}{2}\left[1-\frac{1}{99}\right]=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}\cdot\frac{98}{99}=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}=\frac{1}{2}\)
=> x = 1/2 * 2 = 1
PT
3
18 tháng 7 2018
Ta có : \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x\left(x+2\right)}=19\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{x}-\frac{1}{x+2}=19\)
\(\Leftrightarrow1-\frac{1}{x+2}=19\)
\(\Leftrightarrow\frac{x+2}{x+2}-\frac{1}{x+2}=19\)
\(\Leftrightarrow\frac{x+1}{x+2}=19\)
<=> 19(x + 2) = x + 1
<=> 19x + 38 = x + 1
=> 19x - x = 1 - 38
=> 19x = -37
=> x = \(-\frac{37}{19}\)
KT
18 tháng 7 2018
ĐK: \(x\ne0;x\ne2\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=19\)
\(\Leftrightarrow\)\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=19\)
\(\Leftrightarrow\)\(1-\frac{1}{x+2}=19\)
\(\Leftrightarrow\)\(\frac{1}{x+2}=-18\)
\(\Rightarrow\)\(x+2=-\frac{1}{18}\)
\(\Leftrightarrow\)\(x=-2\frac{1}{18}\)
T
0
HT
3
H
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!
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
M
1
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
NK
3
LV
12 tháng 8 2016
Ta có:
1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x.(x+2) = 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2)
= 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2
= 1/2.(1 - 1/x+2)
=> 1/2.(1 - 1/x+2) = 20/41
1 - 1/x+ 2 = 20/41 : 1/2
1 - 1/x+2 = 40/41
1/x+2 = 1/41
=>x + 2 = 41
=>x = 41 - 2
=>x = 39
Vậy x = 39
Ủng hộ nha
12 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=2.\frac{20}{41}\)
=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
=> \(1-\frac{1}{x+2}=\frac{40}{41}\)
=> \(\frac{1}{x+2}=1-\frac{40}{41}\)
=> \(\frac{1}{x+2}=\frac{1}{41}\)
=> \(x+2=41\)
=> \(x=41-2=39\)
NK
3
12 tháng 8 2016
Gọi tổng trên là A
1/2A= 2/1.3+1/3.5+...+1/x.(x+2)
1/2A= 1-1/x.(x+2)
A=\(\frac{1-\frac{1}{x.\left(x+2\right)}}{2}\)
L
1
29 tháng 7 2023
=>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
SC
3
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 ****
A = \(\dfrac{x}{1.3}\) + \(\dfrac{x}{3.5}\) + \(\dfrac{x}{5.7}\) +......+\(\dfrac{x}{2021.2023}\)
A = \(\dfrac{x}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + .......+ \(\dfrac{2}{2021.2023}\))
A = \(\dfrac{x}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) +......+ \(\dfrac{1}{2021}\) - \(\dfrac{1}{2023}\))
A = \(\dfrac{x}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{2023}\))
A = \(\dfrac{x}{2}\). \(\dfrac{2022}{2023}\)
A = \(\dfrac{1011x}{2023}\)