\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\\ =>\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=2.\dfrac{49}{99}\\ =>\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{7.5}+...+\dfrac{\left(2x+1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{2.49}{99}\)
\(=>1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{49.2}{99}\\ =>1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ đk:x\ne-\dfrac{1}{2}\\ =>\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\\ =>\dfrac{2x}{2x+1}=\dfrac{98}{99}\\ =>2x.99=98.\left(2x+1\right)\\ =>198x-196x=98\\ =>2x=98\\ =>x=49\left(t/mđk\right)\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}+\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{1}{2x+1}=\dfrac{1}{99}\)
\(2x+1=99\)
\(2x=98\)
\(x=49\)
