x = \(\frac{2}{99}\)

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)-x=-\frac{100}{99}\)

\(\Rightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)-x=-\frac{100}{99}\)

\(\Rightarrow\left(1-\frac{1}{99}\right)-x=-\frac{100}{99}\)

\(\Rightarrow\frac{98}{99}-x=-\frac{100}{99}\)

\(\Rightarrow x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)

\(\Rightarrow x=\frac{198}{99}=2\)

Vậy x = 2

tách\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+......+\(\frac{2}{97.99}\)ra thành A

A=1/1-1/3+1/3-1/5+1/5-........+1/97-1/99

A=1-1/99=98/99

=>x=98/99-\(\frac{-100}{99}\)=2

k mình nha

=>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

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

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 ~ 

98/99 - x = -100/99

x = 98/99 - -100/99

x = 198/99

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)-X=\frac{-100}{99}\)

\(\left(1-\frac{1}{3}+\frac{1}{3}+-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-X=\frac{-100}{99}\)

\(\left(1-\frac{1}{99}\right)-X=\frac{-100}{99}\)

\(\frac{98}{99}-X=\frac{-100}{99}\)

             \(X=\frac{98}{99}-\frac{-100}{99}\)

             \(X=\frac{198}{99}\)

              \(X=2\)

x= -2

-Lớp lớn rồi nên không biết làm bài này, chỉ dùng máy tính tính thôi-

\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=-\dfrac{100}{99}\)

\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)

\(\left(1-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)

\(\dfrac{98}{99}-x=-\dfrac{100}{99}\)

\(x=\dfrac{98}{99}-\left(-\dfrac{100}{99}\right)\)

\(x=\dfrac{198}{99}\)

Vậy \(x=\dfrac{198}{99}\)

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 = \(-\frac{37}{19}\)

ĐK:  \(x\ne0;x\ne2\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=19\)

\(\Leftrightarrow\)\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=19\)

\(\Leftrightarrow\)\(1-\frac{1}{x+2}=19\)

\(\Leftrightarrow\)\(\frac{1}{x+2}=-18\)

\(\Rightarrow\)\(x+2=-\frac{1}{18}\)

\(\Leftrightarrow\)\(x=-2\frac{1}{18}\)

sai đề rồi

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

\(\frac{x}{1.3}+\frac{x}{3.5}+\frac{x}{5.7}+....+\frac{x}{97.99}=\frac{49}{99}\)

\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)=\frac{49}{99}\)

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

\(\Leftrightarrow\frac{x}{2}.\frac{98}{99}=\frac{49}{99}\)

\(\Leftrightarrow\frac{x}{2}=\frac{49}{99}\div\frac{98}{99}\)

\(\Leftrightarrow\frac{x}{2}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}\times2=1\)

\(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+...+\frac{x}{97\cdot99}=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{97}{99}\)

\(\Rightarrow\frac{x}{2}\left[1-\frac{1}{99}\right]=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}\cdot\frac{98}{99}=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}=\frac{1}{2}\)

=> x = 1/2 * 2 = 1