\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\\ \Leftrightarrow2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\right)=2.\dfrac{49}{99}\\ \Leftrightarrow\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=1-\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\)
\(\Leftrightarrow2x=98\\ \Leftrightarrow x=49\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x-1}\right)=\dfrac{49}{99}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{2x-1}\right)=\dfrac{49}{99}\Rightarrow1-\dfrac{1}{2x-1}=\dfrac{98}{99}\Rightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\)
\(\Rightarrow2x+1=99\Leftrightarrow x=49\)
