tìm x :
2/1.3 + 2/3.5+2/5.7+...+2/x(x+2)=99/100
K
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NN
0
TT
6
D
21 tháng 5 2017
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)-x=-\frac{100}{99}\)
\(\Rightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)-x=-\frac{100}{99}\)
\(\Rightarrow\left(1-\frac{1}{99}\right)-x=-\frac{100}{99}\)
\(\Rightarrow\frac{98}{99}-x=-\frac{100}{99}\)
\(\Rightarrow x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)
\(\Rightarrow x=\frac{198}{99}=2\)
Vậy x = 2
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
3 tháng 5 2015
mình làm được bài tìm x
x.(2/1.3+2/3.5+2/5.7+...+2/97.99)-x=-100/99
x.(1-1/3+1/3-1/4+1/4-1/5+1/5+...+1/97-1/97-1/99)-x=-100/99
x.(1-1/99)-x=-100/99
x.98/99-x=-100/99
x.98/99=-100/99+x
x.x=-100/99-98/99
2x=-198/99
x=-198/99/2
x=-1
B
3
12 tháng 4 2018
Ta có :
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(\frac{98}{99}-x=\frac{-100}{99}\)
\(\Leftrightarrow\)\(x=\frac{98}{99}+\frac{100}{99}\)
\(\Leftrightarrow\)\(x=\frac{198}{99}\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
Chúc bạn học tốt ~
C
5 tháng 5 2017
\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\dfrac{98}{99}-x=-\dfrac{100}{99}\)
\(x=\dfrac{98}{99}-\left(-\dfrac{100}{99}\right)\)
\(x=\dfrac{198}{99}\)
Vậy \(x=\dfrac{198}{99}\)
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: \(x\ne0;x\ne2\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=19\)
\(\Leftrightarrow\)\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=19\)
\(\Leftrightarrow\)\(1-\frac{1}{x+2}=19\)
\(\Leftrightarrow\)\(\frac{1}{x+2}=-18\)
\(\Rightarrow\)\(x+2=-\frac{1}{18}\)
\(\Leftrightarrow\)\(x=-2\frac{1}{18}\)
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
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