2) \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\) (ĐK: \(x\in N\))
\(\Rightarrow2\cdot\left[\dfrac{1}{1\cdot3}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\right]=2\cdot\dfrac{49}{99}\)
\(\Rightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\)
\(\Rightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\Rightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\Rightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\Rightarrow\dfrac{2x}{2x+1}=\dfrac{98}{99}\)
\(\Rightarrow2x\cdot99=98\cdot\left(2x+1\right)\)
\(\Rightarrow198x=196x+98\)
\(\Rightarrow198-196x=98\)
\(\Rightarrow2x=98\)
\(\Rightarrow x=\dfrac{98}{2}\)
\(\Rightarrow x=49\left(tm\right)\)
\(B=2023:\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-1\dfrac{1}{6}+0,875-0,7}{\dfrac{1}{3}-0,25+\dfrac{1}{5}}\right)\)
\(B=2023:\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}\right)\)
\(B=2023:\left[\dfrac{2\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\cdot\dfrac{-7\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{2\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\right]\)
\(B=2023:\left(\dfrac{2}{7}\cdot\dfrac{-7}{2}\right)\)
\(B=2023:-1\)
\(B=-2023\)
