\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\) \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)\(\Leftrightarrow\dfrac{1}{2}\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=1-\dfrac{98}{99}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\)
\(\Leftrightarrow2x+1=99\)
\(\Rightarrow x=\dfrac{99-1}{2}\)
\(\Rightarrow x=49\)
Vậy \(x=49\)
