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nhân 2 vào 2 vế rồi bạn biến đổi ra( mình lười làm ắ)

tìm được x=50 ắ

\(\frac{x-2}{3}+\frac{x-2}{3.5}+\frac{x-2}{5.7}+...+\frac{x-2}{97.99}=\frac{-49}{99}\)

<=>\(\left(x-2\right)\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=-\frac{49}{99}\)

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

<=>\(\left(x-2\right)\cdot\frac{1}{2}\cdot\left(1-\frac{1}{99}\right)=-\frac{49}{99}\)

<=>\(\left(x-2\right)\cdot\frac{49}{99}=-\frac{49}{99}\)

<=>x-2=-1

<=>x=1

=>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

\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)

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

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